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Theorem 13.12.1 * \( \mathcal{P}\left( A\right) \) is the set of all internal subsets of \( {}^{ * }A \), i.e.,\n\n\[ \n{}^{ * }\mathcal{P}\left( A\right) = \mathcal{P}\left( {{}^{ * }A}\right) \cap {}^{ * }\mathbb{U} = \{ B \subseteq {}^{ * }A : B\text{ is internal }\} .\n\] | Proof. We have \( {}^{ * }\mathcal{P}\left( A\right) \subseteq \mathcal{P}\left( {{}^{ * }A}\right) \) in general. Moreover, if \( B \in {}^{ * }\mathcal{P}\left( A\right) \), then \( B \) belongs to a standard entity, so \( B \) is internal.\n\nFor the converse, let \( B \) be an internal subset of \( {}^{ * }A \) . Then \( B \in {}^{ * }C \) for some \( C \in \mathbb{U} \) . Now, the sentence\n\n\[ \n\left( {\forall x \in C}\right) \left\lbrack {\left( {\forall y \in x}\right) \left( {y \in A}\right) \rightarrow x \in \mathcal{P}\left( A\right) }\right\rbrack\n\]\n\nis true, since it asserts of any \( x \in C \) that if \( x \subseteq A \), then \( x \) belongs to \( \mathcal{P}\left( A\right) \) . By transfer it follows that for any \( x \in {}^{ * }C \), if \( x \subseteq {}^{ * }A \), then \( x \) belongs to * \( \mathcal{P}\left( A\right) \) . But \( B \in {}^{ * }C \) and \( B \subseteq {}^{ * }A \), i.e.,\n\n\[ \n\left( {\forall y \in B}\right) \left( {y \in {}^{ * }A}\right) ,\n\]\n\nso we get \( B \in {}^{ * }\mathcal{P}\left( A\right) \) as desired. | Yes |
Theorem 13.14.1 The image \( {}^{\mathrm{{im}}}A \) of any infinite set \( A \in \mathbb{U} \) is external. | Proof. The method of proof was hinted at in Exercise 12.2(8), which is itself the special case in which \( A \) is an infinite subset of \( \mathbb{R} \) (cf. also Section 11.7).\n\nIn general, if \( A \) is infinite, then there is an injection \( f : \mathbb{N} \rightarrow A \) . Put \( X = \{ f\left( n\right) : n \in \mathbb{N}\} \subseteq A \) . Then * \( X \) is internal (indeed standard), and so if \( {}^{\mathrm{{im}}}A \) were internal, then so too would be \( {}^{\mathrm{{im}}}A \cap {}^{ * }X \) (Section 13.11). Observe that\n\n\[ \n{}^{\mathrm{{im}}}A \cap {}^{ * }X = \left\{ {{}^{ * }a : a \in A\text{ and }{}^{ * }a \in {}^{ * }X}\right\} = \left\{ {{}^{ * }a : a \in X}\right\} = {}^{\mathrm{{im}}}X.\n\]\n\nSince the transform \( {}^{ * }f : {}^{ * }\mathbb{N} \rightarrow {}^{ * }A \) is internal, this would then imply that the inverse image of \( {}^{\mathrm{{im}}}X \) under \( {}^{ * }f \) is internal (Section 13.11). But by transfer \( {}^{ * }f \) is injective, and from this it can be shown that \( {}^{ * }{f}^{-1}\left( {{}^{\mathrm{{im}}}X}\right) \) is equal to the external set \( \mathbb{N} \) .\n\nTherefore \( {}^{\mathrm{{im}}}A \) cannot be internal. | Yes |
Theorem 14.2.1 If \( \mathbb{U}\overset{ * }{ \rightarrow }{\mathbb{U}}^{\prime } \) is a universe embedding, then the following are equivalent.\n\n(1) \( {\mathbb{U}}^{\prime } \) is an enlargement of \( \mathbb{U} \) relative to \( \overset{ * }{ \rightarrow } \).\n\n(2) For any concurrent relation \( R \in \mathbb{U} \) there exists an entity \( b \in {\mathbb{U}}^{\prime } \) such that \( {}^{ * }x\left( {{}^{ * }R}\right) b \) for all \( x \in \operatorname{dom}R \).\n\n(3) For each set \( A \in \mathbb{U} \) there exists a hyperfinite subset \( B \) of \( {}^{ * }A \) that contains all the standard entities of \( {}^{ * }A \):\n\n\[ \n{}^{\mathrm{{im}}}A = \left\{ {{}^{ * }a : a \in A}\right\} \subseteq B \in {}^{ * }{\mathcal{P}}_{F}\left( A\right) .\n\] | Proof. First assume (1). If \( R \) is a binary relation in \( \mathbb{U} \), for each \( x \in \operatorname{dom}R \) let \( R\left\lbrack x\right\rbrack = \{ y \in \operatorname{ran}R : {xRy}\} \) . Then if \( R \) is concurrent, the collection\n\n\[ \n\{ R\left\lbrack x\right\rbrack : x \in \operatorname{dom}R\}\n\]\n\nhas the finite intersection property. Also, this collection is a subset of the power set \( \mathcal{P}\left( {\operatorname{ran}R}\right) \), so belongs to \( \mathbb{U} \) . Hence by (1) there is a \( b \in {\mathbb{U}}^{\prime } \) that belongs to every \( {}^{ * }\left( {R\left\lbrack x\right\rbrack }\right) \) . By transfer of \( \left( {\forall y \in R\left\lbrack x\right\rbrack }\right) \left( {xRy}\right) \) we then have \( {}^{ * }x\left( {{}^{ * }R}\right) b \) for all \( x \in \operatorname{dom}R \), establishing (2).\n\nThe proof that (2) implies (3) was indicated in the discussion of the last example above.\n\nTo show that (3) implies (1), let \( A \in \mathbb{U} \) be a collection of sets with the finite intersection property. Take a transitive \( T \in \mathbb{U} \) with \( A \subseteq T \) . Then the sentence\n\n\[ \n\left( {\forall Z \in {\mathcal{P}}_{F}\left( A\right) }\right) \left( {\exists y \in T}\right) \left( {\forall z \in Z}\right) \left( {y \in z}\right)\n\]\n\nis true. But assuming (3), there exists a hyperfinite \( B \subseteq {}^{ * }A \) containing \( {}^{\mathrm{{im}}}A \) . Then by transfer of this sentence, since \( B \in {}^{ * }{\mathcal{P}}_{F}\left( A\right) \), there exists some \( b \in * T \) with \( b \in z \) for all \( z \in B \), and hence \( b \in * C \) for all \( C \in A \) . Therefore (1) holds. | Yes |
Theorem 15.1.1 Let \( \varphi \left( x\right) \) be an internal \( {\mathcal{L}}_{{\mathbb{U}}^{\prime }} \) -formula with only the variable \( x \) free. Then\n\n(1) (Overflow) If there exists \( k \in \mathbb{N} \) such that \( \varphi \left( n\right) \) is true for all \( n \in \mathbb{N} \) with \( k \leq n \), then there exists \( K \in * \mathbb{N} - \mathbb{N} \) such that \( \varphi \left( n\right) \) is true for all \( n \in {}^{ * }\mathbb{N} \) with \( k \leq n \leq K \) . | (1) We adapt the proof of Theorem 11.4.1. Formula \( \left( {k < x}\right) \land \neg \varphi \left( x\right) \) is internal, and \( {}^{ * }\mathbb{N} \) is internal, so by the version in Section 13.15 of the internal set definition principle,\n\n\[ Y = \left\{ {n \in {}^{ * }\mathbb{N} : k < n\text{ and not }\varphi \left( n\right) }\right\} \]\n\nis an internal set. Now, if \( \varphi \left( n\right) \) is true for all \( * \mathbb{N} - \mathbb{N} \), then any \( K \in {}^{ * }\mathbb{N} \) – \( \mathbb{N} \) gives the desired result (remember we assume \( {}^{ * }\mathbb{N} \) – \( \mathbb{N} \neq \varnothing \) ). Otherwise \( Y \) is nonempty, and so by the internal least number principle has a least element \( H \) . Then \( H \) is unlimited and \( K = H - 1 \) gives the result. | Yes |
Lemma 15.2.1 If \( s : {}^{ * }\mathbb{N} \rightarrow {}^{ * }\mathbb{R} \) is an internal hypersequence such that \( {s}_{n} \) is infinitesimal for all standard \( n \in \mathbb{N} \), then there is an unlimited \( K \in {}^{ * }\mathbb{N} \) such that \( {s}_{n} \) is infinitesimal for all hypernatural \( n \leq K \) . | Proof. We cannot just apply overflow to \( \left\{ {n \in {}^{ * }\mathbb{N} : {s}_{n} \simeq 0}\right\} \), since we do not know that this set is internal. Instead we use \( \left\{ {n \in {}^{ * }\mathbb{N} : \left| {s}_{n}\right| < \frac{1}{n}}\right\} \) . Because \( s \) is internal, the formula\n\n\[ \left| {x \cdot s\left( x\right) }\right| < 1 \]\n\nis internal, and for each \( n \in \mathbb{N} \), since \( {s}_{n} \simeq 0 \) we have \( n \cdot {s}_{n} \simeq 0 \), and so \( \left| {n \cdot {s}_{n}}\right| < 1 \) . Hence by overflow there is a \( K \in {\mathbb{N}}_{\infty } \) such that if \( n \leq K \), then \( \left| {n \cdot {s}_{n}}\right| < 1 \) and so \( \left| {s}_{n}\right| < \frac{1}{n} \) . But when such an \( n \) is unlimited, \( \frac{1}{n} \simeq 0 \) and so \( {s}_{n} \simeq 0 \) . | Yes |
Theorem 15.2.3 If \( f \) is an internal \( {}^{ * }\mathbb{R} \) -valued function and \( f\left( x\right) \) is infinitesimal for all limited hyperreals \( x \), then there is an unlimited \( b \) such that \( f\left( x\right) \) is infinitesimal for all \( x \in \left\lbrack {-b, b}\right\rbrack \subseteq * \mathbb{R} \) . | Proof. As for the proof of Lemma 15.2.1, but using the internal formula \( \left| {x \cdot f\left( x\right) }\right| < 1 \) and the fact that if an internal set includes \( \mathbb{L} \) then it includes \( \left\lbrack {-b, b}\right\rbrack \) for some unlimited \( b \) (Section 11.6). | No |
Theorem 15.4.3 In a sequentially comprehensive enlargement, if \( X \) is any countable set of unlimited hypernaturals, then there is an unlimited hypernatural \( K \) less than every member of \( X \) . | Proof. Write \( X = \left\langle {{s}_{n} : n \in \mathbb{N}}\right\rangle \) and extend this to an internal hyperse-quence \( \left\langle {{s}_{n} : n \in * \mathbb{N}}\right\rangle \) by sequential comprehensiveness. Put\n\n\[ Y = \left\{ {k \in {}^{ * }\mathbb{N} : \left( {\forall n \in {}^{ * }\mathbb{N}}\right) \left( {n \leq k \rightarrow k < {s}_{n}}\right) }\right\} .\n\]\n\nThen \( Y \) is internal, as its defining formula is internal. Moreover, \( Y \) contains each standard \( k \in \mathbb{N} \), since in that case \( {s}_{1},\ldots ,{s}_{k} \) are all in \( X \), hence unlimited and so greater than \( k \) . By overflow, then, there is some unlimited \( K \) in \( Y \) . Then \( K < {s}_{n} \) whenever \( n \leq K \), which includes all \( n \in \mathbb{N} \) . Hence \( K < x \) whenever \( x \in X \) . | Yes |
Lemma 16.5.1 If \( B \) is Loeb measurable with respect to \( \mu \), then\n\n\[ \n{\mu }_{L}\left( B\right) = \inf \left\{ {{\mu }_{L}\left( A\right) : B \subseteq A \in \mathcal{A}}\right\} \n\] | Proof. By monotonicity, \( {\mu }_{L}\left( B\right) \) is a lower bound of the \( {\mu }_{L}\left( A\right) \) ’s for \( B \subseteq \) \( A \in \mathcal{A} \) . If \( {\mu }_{L}\left( B\right) = \infty \), then the result follows. If, however, \( {\mu }_{L}\left( B\right) < \infty \) , to show that it is the greatest lower bound it suffices to show that for any \( \varepsilon \in {\mathbb{R}}^{ + } \) there is some set \( {A}_{\varepsilon } \in \mathcal{A} \) with \( B \subseteq {A}_{\varepsilon } \) and \( {\mu }_{L}\left( {A}_{\varepsilon }\right) \leq {\mu }_{L}\left( B\right) + \varepsilon \) .\n\nNow, for such an \( \varepsilon \), by properties of outer measure there is an increasing sequence \( {A}_{1} \subseteq {A}_{2} \subseteq \cdots \) of \( \mathcal{A} \) -elements whose union includes \( B \) and has\n\n\[ \n{\mu }_{L}^{ + }\left( {\mathop{\bigcup }\limits_{{n \in \mathbb{N}}}{A}_{n}}\right) < {\mu }_{L}\left( B\right) + \varepsilon .\n\]\n\nThe sequence \( \left\langle {{A}_{n} : n \in \mathbb{N}}\right\rangle \) extends by sequential comprehensiveness to an internal sequence \( \left\langle {{A}_{n} : n \in * \mathbb{N}}\right\rangle \) of elements of \( \mathcal{A} \) . Then for each \( k \in \mathbb{N} \) we have\n\n\[ \n\left( {\forall n \in \mathbb{{N}^{d}}}\right) \left( {n \leq k\text{ implies }{A}_{n} \subseteq {A}_{k}\text{ and }\mu \left( {A}_{n}\right) < {\mu }_{L}\left( B\right) + \varepsilon }\right)\n\]\n\n(i)\n\n\( \left( {\mathrm{{since}}\mu \left( {A}_{n}\right) \simeq {\mu }_{L}\left( {A}_{n}\right) \leq {\mu }_{L}^{ + }\left( {\mathop{\bigcup }\limits_{{n \in \mathbb{N}}}{A}_{n}}\right) }\right) . \) But (i) is an internal assertion, since \( \mu \) and the extended sequence are internal, while \( k,\varepsilon \), and \( {\mu }_{L}\left( B\right) \) are fixed internal entities (real numbers). Therefore by overflow (i) must be true with some unlimited \( K \in {}^{ * }\mathbb{N} \) in place of \( k \) . For such a \( K \) we have \( {A}_{K} \in \mathcal{A} \) and \( {A}_{n} \subseteq {A}_{K} \) for all \( n \in \mathbb{N} \), so that\n\n\[ \nB \subseteq \mathop{\bigcup }\limits_{{n \in \mathbb{N}}}{A}_{n} \subseteq {A}_{K}\n\]\n\nwhile \( \mu \left( {A}_{K}\right) < {\mu }_{L}\left( B\right) + \varepsilon \) . Hence as \( \mu \left( {A}_{K}\right) \simeq {\mu }_{L}\left( {A}_{K}\right) \),\n\n\[ \n{\mu }_{L}\left( {A}_{K}\right) \leq {\mu }_{L}\left( B\right) + \varepsilon\n\]\n\nestablishing that \( {A}_{K} \) is the set \( {A}_{\varepsilon } \) we are looking for. | Yes |
Lemma 16.5.2 If \( B \) is Loeb measurable and \( {\mu }_{L} \) -finite, then\n\n\[{\mu }_{L}\left( B\right) = \sup \left\{ {{\mu }_{L}\left( A\right) : A \subseteq B}\right. \text{and}\left. {A \in \mathcal{A}}\right\} \text{.} | Proof. Given any \( \varepsilon \in {\mathbb{R}}^{ + } \), we will show that there is some set \( {A}_{\varepsilon } \in \mathcal{A} \) such that \( {A}_{\varepsilon } \subseteq B \) and \( {\mu }_{L}\left( B\right) - \varepsilon < \mu \left( {A}_{\varepsilon }\right) \). \n\nSince \( {\mu }_{L}\left( B\right) < \infty , \) we know from Lemma 16.5.1 that there is some \( D \in \mathcal{A} \) with \( B \subseteq D \) and \( {\mu }_{L}\left( D\right) < \infty \). The desired result is obtained by using complementation relative to \( D \). Firstly, \( D - B \) is Loeb measurable and \( {\mu }_{L} \) -finite, so by Lemma 16.5.1 there is a set \( C \) with \( D - B \subseteq C \in \mathcal{A} \) and\n\n\[{\mu }_{L}\left( C\right) < {\mu }_{L}\left( {D - B}\right) + \varepsilon .\n\nWe may assume \( C \subseteq D \) (since we could replace \( C \) by \( C \cap D \) here). Let \( {A}_{\varepsilon } = D - C \in \mathcal{A} \). Then \( {A}_{\varepsilon } \subseteq B \), and \( C \) is the disjoint union of \( D - B \) and \( B - {A}_{\varepsilon } \), so\n\n\[{\mu }_{L}\left( {D - B}\right) + {\mu }_{L}\left( {B - {A}_{\varepsilon }}\right) = {\mu }_{L}\left( C\right) < {\mu }_{L}\left( {D - B}\right) + \varepsilon ,\n\nimplying that \( {\mu }_{L}\left( {B - {A}_{\varepsilon }}\right) < \varepsilon \). Therefore\n\n\[{\mu }_{L}\left( B\right) = {\mu }_{L}\left( {A}_{\varepsilon }\right) + {\mu }_{L}\left( {B - {A}_{\varepsilon }}\right) < {\mu }_{L}\left( {A}_{\varepsilon }\right) + \varepsilon ,\n\nso \( {\mu }_{L}\left( B\right) - \varepsilon < \mu \left( {A}_{\varepsilon }\right) \) as desired. | Yes |
Lemma 16.6.1 If \( B \) is Loeb measurable with \( {\mu }_{L}\left( B\right) < \infty \), then \( B \) is \( \mu \) - approximable. | Proof. Given \( \varepsilon \in {\mathbb{R}}^{ + } \), by Lemmas 16.5.1 and 16.5.2 there are \( {C}_{\varepsilon },{D}_{\varepsilon } \in \mathcal{A} \) with \( {C}_{\varepsilon } \subseteq B \subseteq {D}_{\varepsilon } \) and \( {\mu }_{L}\left( {D}_{\varepsilon }\right) < {\mu }_{L}\left( B\right) + \frac{\varepsilon }{2} \) while \( {\mu }_{L}\left( B\right) < {\mu }_{L}\left( {C}_{\varepsilon }\right) + \frac{\varepsilon }{2} \) . Then\n\n\[ \n{\mu }_{L}\left( {{D}_{\varepsilon } - {C}_{\varepsilon }}\right) = {\mu }_{L}\left( {D}_{\varepsilon }\right) - {\mu }_{L}\left( B\right) + {\mu }_{L}\left( B\right) - {\mu }_{L}\left( {C}_{\varepsilon }\right) < \varepsilon .\n\] | Yes |
Lemma 16.6.3 If \( B \) is \( \mu \) -approximable, then \( B \) is Loeb measurable. | Proof. We have to show that any \( E \subseteq S \) is split \( {\mu }_{L}^{ + } \) -additively by \( B \), for which it suffices that\n\n\[{\mu }_{L}^{ + }\left( E\right) \geq {\mu }_{L}^{ + }\left( {E \cap B}\right) + {\mu }_{L}^{ + }\left( {E - B}\right)\]\n\nNow, by Lemma 16.6.2 there is an \( A \in \mathcal{A} \) that is almost equal to \( B \). The desired inequality holds with \( A \) in place of \( B \), since \( A \) is Loeb measurable, and so as \( {\mu }_{L}^{ + }\left( {B\Delta A}\right) = 0 \), we can show it holds for \( B \) as well. Formally, let\n\n\[C = \left( {E \cap B}\right) - A,\]\n\n\[D = \left( {E \cap A}\right) - B\n\n\[G = E - \left( {A \cup B}\right)\n\n\[H = E \cap A \cap B\n\n(draw a Venn diagram!). Then \( C \) and \( D \) are included in \( {B\Delta A} \), so \( {\mu }_{L}^{ + }\left( C\right) = \) \( {\mu }_{L}^{ + }\left( D\right) = 0 \). Thus\n\n\[{\mu }_{L}^{ + }\left( {E \cap B}\right) = {\mu }_{L}^{ + }\left( {C \cup H}\right) \leq {\mu }_{L}^{ + }\left( C\right) + {\mu }_{L}^{ + }\left( H\right) = {\mu }_{L}^{ + }\left( H\right) ,\]\n\nand similarly\n\n\[{\mu }_{L}^{ + }\left( {E - B}\right) = {\mu }_{L}^{ + }\left( {D \cup G}\right) \leq {\mu }_{L}^{ + }\left( D\right) + {\mu }_{L}^{ + }\left( G\right) = {\mu }_{L}^{ + }\left( G\right) .\n\nHence\n\n\[{\mu }_{L}^{ + }\left( {E \cap B}\right) + {\mu }_{L}^{ + }\left( {E - B}\right) \leq {\mu }_{L}^{ + }\left( H\right) + {\mu }_{L}^{ + }\left( G\right)\]\n\n\(\leq {\mu }_{L}^{ + }\left( {E \cap A}\right) + {\mu }_{L}^{ + }\left( {E - A}\right) \) by monotonicity\n\n\[= {\mu }_{L}^{ + }\left( E\right)\]\n\nwith the last step given by the \( {\mu }_{L}^{ + } \) -measurability of \( A \) . | Yes |
Lemma 16.6.4 If \( {\mu }_{L}^{ + }\left( B\right) < \infty \), then \( B \) is Loeb measurable with respect to \( \mu \) if and only if it is \( \mu \) -approximable. | Proof. By Lemmas 16.6.1 and 16.6.3. | No |
An arbitrary subset \( B \) of \( S \) is Loeb measurable with respect to \( \mu \) if and only if \( B \cap A \) is \( \mu \) -approximable for all \( \mu \) -finite \( A \in \mathcal{A} \) . | Proof. Let \( B \) be Loeb measurable. If \( A \in \mathcal{A} \) is \( \mu \) -finite, then \( B \cap A \) is Loeb measurable, since the Loeb measurable sets are \( \cap \) -closed, and \( {\mu }_{L}^{ + } \) -finite since \( B \cap A \subseteq A \), so by Corollary 16.6.4 \( B \cap A \) is \( \mu \) -approximable.\n\nConversely, let \( B \cap A \) be \( \mu \) -approximable for all \( \mu \) -finite \( A \in \mathcal{A} \) . To show that \( B \) is Loeb measurable, we have to show that for any \( {\mu }_{L}^{ + } \) -finite \( E \subseteq S \) ,\n\n\[{\mu }_{L}^{ + }\left( E\right) \geq {\mu }_{L}^{ + }\left( {E \cap B}\right) + {\mu }_{L}^{ + }\left( {E - B}\right)\]\n\n(iii)\n\nBut in such a case \( {\mu }_{L}^{ + }\left( {E \cap B}\right) < \infty \), so there must be a sequence \( \langle {A}_{n} : n \in \mathbb{N}\rangle \) of \( \mathcal{A} \) -elements that covers \( E \cap B \) with each \( {A}_{n} \) being \( \mu \) -finite. Each \( B \cap {A}_{n} \) is then \( \mu \) -approximable, by hypothesis, and so is Loeb measurable by Lemma 16.6.3. Hence if\n\n\[A = \mathop{\bigcup }\limits_{{n \in \mathbb{N}}}\left( {B \cap {A}_{n}}\right)\]\n\nit follows that \( A \) is Loeb measurable, and so\n\n\[{\mu }_{L}^{ + }\left( E\right) \geq {\mu }_{L}^{ + }\left( {E \cap A}\right) + {\mu }_{L}^{ + }\left( {E - A}\right)\]\n\n(iv)\n\nBut\n\n\[E \cap A = E \cap \left( {\mathop{\bigcup }\limits_{{n \in \mathbb{N}}}\left( {B \cap {A}_{n}}\right) }\right) = E \cap B\]\n\nas the \( {A}_{n} \) ’s cover \( E \cap B \) . Then \( E - A = E - B \), and hence (iii) follows from (iv). | Yes |
Theorem 16.8.1 For any \( a, b \in \mathbb{R} \) with \( a < b \) , \[ {\mu }_{L}\left( {\{ s \in S : a < s < b\} }\right) = b - a. \] | Proof. Let \( A = \{ s \in S : a < s < b\} \) . Then \( A = S \cap {}^{ * }\left( {a, b}\right) \), so \( A \) is internal and belongs to \( \mathcal{A} \), hence is Loeb measurable. Moreover, \( A \) is hyperfinite, so has smallest and greatest elements, say \( s \) and \( t \) . Since \( a \) and \( b \) can be approximated infinitely closely by members of \( S \), we must then have \( a \simeq s \) and \( b \simeq t \) . Also, we can put \( s = \frac{K + 1}{N} \) and \( t = \frac{L}{N} \) for some \( K, L \in * \mathbb{Z} \) . Thus \[ \left. {A = \left\{ {\frac{K + 1}{N},\frac{K + 2}{N},\ldots ,\frac{L}{N}}\right. }\right\} = \left\{ {\frac{M}{N} : K < M \leq L}\right\} , \] which is hyperfinite of cardinality \( L - K \), since the internal function \( f\left( x\right) = \) \( \frac{K + x}{N} \) is a bijection from \( \{ 1,\ldots, L - K\} \) onto \( A \) . It follows that \[ \frac{\left| A\right| }{N} = \frac{L - K}{N} = \frac{L}{N} - \frac{K}{N} \simeq b - a, \] and so \( {\mu }_{L}\left( A\right) = b - a \) as desired. | Yes |
A subset \( B \) of \( \mathbb{R} \) is Lebesgue measurable if and only if \( {\operatorname{sh}}^{-1}\left( B\right) \) is Loeb measurable. When this holds, the Lebesgue measure of \( B \) is equal to the Loeb measure of the set of grid points infinitely close to points of \( B \) : | Proof. Let \( \mathcal{M} = \left\{ {B \subseteq \mathbb{R} : {\operatorname{sh}}^{-1}\left( B\right) \in \mathcal{A}\left( {\mu }_{L}\right) }\right\} \) . For \( B \in \mathcal{M} \), put\n\n\[ \nu \left( B\right) = {\mu }_{L}\left( {{\operatorname{sh}}^{-1}\left( B\right) }\right) \]\n\nOur task is to show that \( \mathcal{M} \) is the class \( {\mathcal{C}}_{\mathbb{R}}\left( \lambda \right) \) of Lebesgue measurable sets, and that \( \nu \) is the Lebesgue measure \( \lambda \) .\n\nBy properties of inverse images of functions,\n\n\[ {\operatorname{sh}}^{-1}\left( \varnothing \right) = \varnothing \]\n\n\[ {\operatorname{sh}}^{-1}\left( {A - B}\right) = {\operatorname{sh}}^{-1}\left( A\right) - {\operatorname{sh}}^{-1}\left( B\right) \]\n\n\[ {\operatorname{sh}}^{-1}\left( {\mathop{\bigcup }\limits_{{n \in \mathbb{N}}}{A}_{n}}\right) = \mathop{\bigcup }\limits_{{n \in \mathbb{N}}}{\operatorname{sh}}^{-1}\left( {A}_{n}\right) \]\n\nSince \( \mathcal{A}\left( {\mu }_{L}\right) \) contains \( \varnothing \) and is closed under set differences and countable unions, these facts imply that \( \mathcal{M} \) has the same closure properties. Since \( {\operatorname{sh}}^{-1}\left( \mathbb{R}\right) \in \mathcal{A}\left( {\mu }_{L}\right) \), as was shown above, we also have \( \mathbb{R} \in \mathcal{M} \) . Altogether then \( \mathcal{M} \) is a \( \sigma \) -algebra, on which \( \nu \) proves to be a measure. | Yes |
Lemma 16.8.3 \( \mathcal{M} \) includes the Borel algebra \( {\mathcal{B}}_{\mathbb{R}} \), and \( \nu \) agrees with Lebesgue measure on all Borel sets. | Proof. Each open interval \( \left( {a, b}\right) \subseteq \mathbb{R} \) belongs to \( \mathcal{M} \), since \( {\operatorname{sh}}^{-1}\left( \left( {a, b}\right) \right) \) is the union of the sequence \( \left\langle {{A}_{n} : n \in \mathbb{N}}\right\rangle \), where\n\n\[ \n{A}_{n} = S \cap {}^{ * }\left( {a + \frac{1}{n}, b - \frac{1}{n}}\right) \in \mathcal{A}.\n\]\n\nBut \( {\mathcal{B}}_{\mathbb{R}} \) is the smallest \( \sigma \) -algebra containing all open intervals \( \left( {a, b}\right) \), so this implies that \( {\mathcal{B}}_{\mathbb{R}} \subseteq \mathcal{M} \) . Also, by Theorem 16.8.1,\n\n\[ \n{\mu }_{L}\left( {A}_{n}\right) = \left( {b - \frac{1}{n}}\right) - \left( {a + \frac{1}{n}}\right) = b - a - \frac{2}{n},\n\]\n\nand hence as the \( {A}_{n} \) ’s form an increasing sequence,\n\n\[ \n\nu \left( \left( {a, b}\right) \right) = {\mu }_{L}\left( {{\operatorname{sh}}^{-1}\left( \left( {a, b}\right) \right) = \mathop{\lim }\limits_{{n \rightarrow \infty }}{\mu }_{L}\left( {A}_{n}\right) = b - a.}\right.\n\]\n\nThus \( \nu \) is a measure on \( {\mathcal{B}}_{\mathbb{R}} \) that agrees with \( \lambda \) on all open intervals. But any such measure must agree with \( \lambda \) on all Borel sets (16.4(1)). | Yes |
Theorem 17.1.1 If \( \\left( {P, \\leq }\\right) \) is an infinite partially ordered set, then \( P \) contains a sequence \( \\left\\langle {{p}_{n} : n \\in \\mathbb{N}}\\right\\rangle \) that is\n\n(1) strictly increasing: \( {p}_{1} < {p}_{2} < \\cdots \\; \) or\n\n(2) strictly decreasing: \( {p}_{1} > {p}_{2} > \\cdots \\; \) or\n\n(3) an antichain, i.e., \( {p}_{n} \) and \( {p}_{m} \) are incomparable under the ordering \( \\leq \) for all \( n \\neq m. \) | Proof. Take a colouring\n\n\[ \n{\\left\\lbrack P\\right\\rbrack }^{2} = {C}_{b} \\cup {C}_{w} \n\]\n\nof the 2-element subsets of \( P \) in which\n\n\[ \n{C}_{b} = \\{ \\{ p, q\\} : p \\leq q\\text{ or }q \\leq p\\} \n\]\n\nand\n\n\[ \n{C}_{w} = {\\left\\lbrack P\\right\\rbrack }^{2} - {C}_{b} \n\]\n\nThus \( {C}_{b} \) (the black sets) consists of pairs of elements that are comparable, and \( {C}_{w} \) (white) consists of the incomparable pairs.\n\nBy Ramsey’s theorem there is an infinite set \( Q \\subseteq P \) that is monochromatic for this colouring. If \( {\\left\\lbrack Q\\right\\rbrack }^{2} \\subseteq {C}_{w} \), then any sequence of distinct points in \( Q \) will be an antichain fulfilling (3). If, however, \( {\\left\\lbrack Q\\right\\rbrack }^{2} \\subseteq {C}_{b} \), then \( Q \) is an infinite chain, from which we can extract a sequence satisfying (1) or (2). Indeed, if \( Q \) is not well-ordered by \( \\leq \), then it must contain an infinite decreasing sequence (2), and if it is well-ordered, then each of its nonempty subsets has a least element, and we can use this fact to construct an increasing sequence (1) inductively. | Yes |
Theorem 18.3.1 The nonstandard hull \( \left( {\widehat{\mathbb{X}}, d}\right) \) is complete. | Proof. Let \( \left\langle {\operatorname{hal}\left( {x}_{n}\right) : n \in \mathbb{N}}\right\rangle \) be a Cauchy sequence in \( \widehat{\mathbb{X}} \) . The sequence \( \left\langle {{x}_{n} : n \in \mathbb{N}}\right\rangle \) of points in \( {}^{ * }{\mathbb{X}}^{\text{lim }} \) extends to an internal hypersequence \( \left\langle {{x}_{n} : n \in {}^{ * }\mathbb{N}}\right\rangle \) in \( {}^{ * }\mathbb{X} \), by sequential comprehensiveness. We will show that \( \left\langle {\operatorname{hal}\left( {x}_{n}\right) : n \in \mathbb{N}}\right\rangle \) converges to \( \operatorname{hal}\left( {x}_{K}\right) \) for some \( K \in {}^{ * }{\mathbb{N}}_{\infty } \) .\n\nNow, for each \( n \in \mathbb{N} \), by the Cauchy property there exists \( {k}_{n} \in \mathbb{N} \) such that for all standard \( m \geq {k}_{n} \) ,\n\n\[ d\left( {\operatorname{hal}\left( {x}_{m}\right) ,\operatorname{hal}\left( {x}_{{k}_{n}}\right) }\right) < \frac{1}{2n} \]\n\nand hence\n\n\[ {}^{ * }d\left( {{x}_{m},{x}_{{k}_{n}}}\right) < \frac{1}{2n} \]\n\n(i)\n\nBut the set \( \left\{ {m \in {}^{ * }\mathbb{N} : {}^{ * }d\left( {{x}_{m},{x}_{{k}_{n}}}\right) < 1/\left( {2n}\right) }\right\} \) is internal, so by overflow we conclude that there is some unlimited \( {K}_{n} \in {}^{ * }\mathbb{N} \) such that (i) holds for all \( m \in {}^{ * }\mathbb{N} \) with \( {k}_{n} \leq m \leq {K}_{n} \) .\n\nInvoking sequential comprehensiveness again, there is some unlimited \( K \in {}^{ * }\mathbb{N} \) that is smaller than every \( {K}_{n} \) (cf. Theorem 15.4.3). Then \( {x}_{K} \) is limited (e.g., \( {}^{ * }d\left( {{x}_{K},{x}_{{k}_{1}}}\right) < \frac{1}{2} \) and \( {x}_{{k}_{1}} \) is limited), so \( \operatorname{hal}\left( {x}_{K}\right) \in \widehat{\mathbb{X}} \) . To show that \( \left\langle {\operatorname{hal}\left( {x}_{n}\right) : n \in \mathbb{N}}\right\rangle \) converges to \( \operatorname{hal}\left( {x}_{K}\right) \) it is enough to show that for each \( n \in \mathbb{N} \) we get\n\n\[ {}^{ * }d\left( {{x}_{m},{x}_{K}}\right) < \frac{1}{n} \]\n\nwhenever \( m \in \mathbb{N} \) and \( {k}_{n} \leq m \) . But for such \( m \) we have \( {k}_{n} \leq m, K < {K}_{n} \) , so by two applications of (i),\n\n\[ {}^{ * }d\left( {{x}_{m},{x}_{K}}\right) \leq {}^{ * }d\left( {{x}_{m},{x}_{{k}_{n}}}\right) + {}^{ * }d\left( {{x}_{{k}_{n}},{x}_{K}}\right) < \frac{1}{2n} + \frac{1}{2n} = \frac{1}{n}. \] | Yes |
Prove, conversely to Corollary 18.3.3, that if a metric space \( \left( {\mathbb{X}, d}\right) \) is complete, then in \( {}^{ * }\mathbb{X} \) every point approachable from \( \mathbb{X} \) is near to \( \mathbb{X} \) . | Consider for example a hyperrational \( x \in {}^{ * }\mathbb{Q} \) that is infinitely close to \( \sqrt{2} \) (recall that every real number is the shadow of some hyperrational). Then \( x \) is not near to \( \mathbb{Q} \), because it is not infinitely close to any standard rational number, but \( x \) is approachable from \( \mathbb{Q} \), since there is a sequence in \( \mathbb{Q} \) converging to \( \sqrt{2} \) . This is a manifestation of the fact that \( \mathbb{Q} \) is not complete under the Euclidean metric. | No |
Corollary 18.3.3 If in \( {}^{ * }\mathbb{X} \) every point approachable from \( \mathbb{X} \) is actually near to \( \mathbb{X} \), then \( \left( {\mathbb{X}, d}\right) \) is a complete metric space. | Null | No |
Theorem 18.4.1 For any nonzero hyperinteger \( x \in {}^{ * }\mathbb{Z} \), the following are equivalent.\n\n(1) \( x \in {}^{ * }{\mathbb{Z}}^{{\text{inf }}_{p}} \).\n\n(2) \( {o}_{p}\left( x\right) \) is unlimited.\n\n(3) \( x \) is divisible by \( {p}^{n} \) in \( {}^{ * }\mathbb{Z} \) for all \( n \in \mathbb{N} \).\n\n(4) \( x \) is divisible by \( {p}^{N} \) for some unlimited \( N \in * \mathbb{N} \) . | Proof. \( {\left| x\right| }_{p} \) is the reciprocal of \( {p}^{{o}_{p}\left( x\right) } \), so is infinitesimal iff \( {p}^{{o}_{p}\left( x\right) } \) is unlimited, which holds iff \( {o}_{p}\left( x\right) \) is unlimited, as \( p \) is standard. Thus (1) and (2) are equivalent.\n\nSince the divisibility relation | is defined in \( \mathbb{Z} \) by\n\n\[ x \mid y\;\text{ iff }\;\left( {\exists z \in \mathbb{Z}}\right) y = {xz} \]\n\nit follows by transfer that \( x \mid y \) for hyperintegers in \( {}^{ * }\mathbb{Z} \) iff \( y = {xz} \) for some \( z \in * \mathbb{Z} \) . Now, the statement\n\n\[ {p}^{n} \mid x\;\text{ iff }\;n \leq {o}_{p}\left( x\right) \]\n\n(ii)\n\nholds for all \( x \in {}^{ * }\mathbb{Z} \) and \( n \in {}^{ * }{\mathbb{Z}}^{ \geq } \), again by transfer. Hence if (2) holds, then for every \( n \in \mathbb{N} \) we have \( n \leq {o}_{p}\left( x\right) \), as \( {o}_{p}\left( x\right) \) is unlimited, and so \( {p}^{n} \mid x \) by (ii). Thus (2) implies (3).\n\nNext, observe that for each \( x \in * \mathbb{Z} \) the set\n\n\[ \left\{ {n \in {}^{ * }\mathbb{N} : {p}^{n} \mid x}\right\} \]\n\nis internal, by the internal set definition principle, so if (3) holds, this set contains all members of \( \mathbb{N} \), and hence by overflow it contains some unlimited \( N \), establishing (4).\n\nFinally, if \( {p}^{N} \mid x \) with \( N \) unlimited, then (ii) gives \( N \leq {o}_{p}\left( x\right) \), so \( {o}_{p}\left( x\right) \) is also unlimited. Thus (4) implies (2).\n\nThis result shows that\n\n\[ {}^{ * }{\mathbb{Z}}^{{\text{inf }}_{p}} = \left\{ {{p}^{N}q : N\text{ is unlimited and }q \in {}^{ * }\mathbb{Z}}\right\} . \] | Yes |
Theorem 18.5.1 Let \( y, z \in * \mathbb{Z} \) . If \( {\left| z\right| }_{p} \) is not infinitesimal, then \( y/z \) is p-adically limited. | Proof. \( \;{o}_{p}\left( z\right) \) is a nonnegative hyperinteger, so if \( {\left| z\right| }_{p} = {p}^{-{o}_{p}\left( z\right) } \neq 0 \), then \( {o}_{p}\left( z\right) \) must be limited, i.e., \( {o}_{p}\left( z\right) \in \mathbb{N} \cup \{ 0\} \) . But then since \( {o}_{p}\left( y\right) \geq 0 \) ,\n\n\[ \n{o}_{p}\left( {y/z}\right) = {o}_{p}\left( y\right) - {o}_{p}\left( z\right) \n\]\n\ncannot be negative unlimited. Hence as above, \( y/z \in {}^{ * }{\mathbb{Q}}^{\mathop{\lim }\limits_{p}} \) . | Yes |
Theorem 18.5.2 Let \( y, z \) be hyperintegers that have no common factors of the form \( {p}^{N} \) with \( N \in {\mathcal{N}}_{\infty } \) . If \( {\left| y/z\right| }_{p} \) is limited, then \( {\left| z\right| }_{p} \) is not infinitesimal. | Proof. Suppose that \( {\left| y/z\right| }_{p} \) is limited, but \( {\left| z\right| }_{p} \simeq 0 \) . Then \( {o}_{p}\left( z\right) \) is positive unlimited (Theorem 18.4.1), while \( {o}_{p}\left( {y/z}\right) = {o}_{p}\left( y\right) - {o}_{p}\left( z\right) \) is not negative unlimited. But this can be so only if \( {o}_{p}\left( y\right) \) is also positive unlimited. Then if \( N \) is the smaller of \( {o}_{p}\left( y\right) \) and \( {o}_{p}\left( z\right) \), we have \( N \in {}^{ * }{\mathbb{N}}_{\infty } \) and \( {p}^{N} \) a factor of both \( y \) and \( z \) . However, this contradicts the hypothesis. | Yes |
Theorem 18.6.1 \( a \in {}^{ * }R\left\lbrack x\right\rbrack \) is approachable from \( R\left\lbrack x\right\rbrack \) if and only if the coefficient \( {a}_{n} \) belongs to \( R \) for all standard \( n \) . | Proof. Fix a standard \( n \in {\mathbb{Z}}^{ \geq } \) . Then if two polynomials \( a, b \in R\left\lbrack x\right\rbrack \) are closer than \( {2}^{-n} \) to each other (i.e., \( \left| {a - b}\right| < {2}^{-n} \) ), the order of \( a - b \) must be at least \( n + 1 \), so \( {\left( a - b\right) }_{n} = 0 \) and hence \( {a}_{n} = {b}_{n} \) . Thus the statement\n\n\[ \left| {a - b}\right| < {2}^{-n} \rightarrow {a}_{n} = {b}_{n} \]\n\nholds for all \( a, b \in R\left\lbrack x\right\rbrack \), and so by transfer holds for all \( a, b \in {}^{ * }R\left\lbrack x\right\rbrack \) .\n\nNow suppose that \( a \) is in \( {}^{ * }R{\left\lbrack x\right\rbrack }^{\mathrm{{ap}}} \), the set of all members of \( {}^{ * }R\left\lbrack x\right\rbrack \) approachable from \( R\left\lbrack x\right\rbrack \) . Then for each standard \( n \) there must be some polynomial \( b \in R\left\lbrack x\right\rbrack \) with \( \left| {a - b}\right| < {2}^{-n} \) . From the previous paragraph it then follows that \( {a}_{n} = {b}_{n} \in R \) . Thus the coefficient \( {a}_{n} \) is in \( R \) for each standard \( n \) .\n\nConversely, suppose \( a \in {}^{ * }R\left\lbrack x\right\rbrack \) has \( {a}_{n} \in R \) for all standard \( n \) . For each such \( n \), the polynomial\n\n\[ a \upharpoonright n = {a}_{0} + {a}_{1}x + {a}_{2}{x}^{2} + \cdots + {a}_{n}{x}^{n} \]\n\nbelongs to \( R\left\lbrack x\right\rbrack \) . But \( a \upharpoonright n \) is within \( {2}^{-n} \) of \( a \), because the statement\n\n\[ \left| {a - a \upharpoonright n}\right| < {2}^{-n} \]\n\nholds for all \( a \in R\left\lbrack x\right\rbrack \) (see above) so holds for all \( a \in {}^{ * }R\left\lbrack x\right\rbrack \) by transfer. This shows that \( a \) is approachable from \( R\left\lbrack x\right\rbrack \) . | Yes |
Theorem 18.6.2 For any nonzero \( a \in {}^{ * }R\left\lbrack x\right\rbrack \), the following are equivalent.\n\n(1) \( \left| a\right| \simeq 0 \) .\n\n(2) \( o\left( a\right) \) is unlimited.\n\n(3) There is an unlimited \( N \in {}^{ * }\mathbb{N} \) such that \( {a}_{n} = 0 \) for all \( n < N \) .\n\n(4) \( {a}_{n} = 0 \) for all standard \( n \) . | Proof. In general, \( \left| a\right| = {2}^{-o\left( a\right) } \) and \( o\left( a\right) \) is a nonnegative hyperinteger, so \( \left| a\right| \) will be appreciable iff \( o\left( a\right) \) is limited, or equivalently, \( \left| a\right| \) will be infinitesimal iff \( o\left( a\right) \) is unlimited. Thus (1) and (2) are equivalent.\n\nNow, by transfer we have that for any nonzero \( a \in {}^{ * }R\left\lbrack x\right\rbrack \) ,\n\n\[ \left( {\forall m \in {}^{ * }{\mathbb{Z}}^{ \geq }}\right) \left\lbrack {m < o\left( a\right) \leftrightarrow \left( {\forall n \in {}^{ * }{\mathbb{Z}}^{ \geq }}\right) \left( {n \leq m \rightarrow {a}_{n} = 0}\right) }\right\rbrack .\n\]\n\nFrom this,(2) implies (3) by putting \( N = o\left( a\right) \) . It is immediate that (3) implies (4). Finally, if (4) holds, then the above transferred sentence ensures that each standard \( m \) is smaller than \( o\left( a\right) \), so (2) follows. | Yes |
Theorem 18.7.1 Two positive hyperintegers are p-adically infinitely close precisely when their base p expansions have identical coefficients of standard degree: | Now, we saw in Section 18.4 that if\n\n\[ a = {z}_{0} + {z}_{1}p + {z}_{2}{p}^{2} + \cdots + {z}_{n}{p}^{n} + \cdots \]\n\nis a \( p \) -adic integer, then there exists a positive hyperinteger \( x \) with\n\n\[ a = {\theta }_{p}\left( x\right) = \left\langle {x{\;\operatorname{mod}\;p}, x{\;\operatorname{mod}\;{p}^{2}},\ldots, x{\;\operatorname{mod}\;{p}^{n}},\ldots }\right\rangle .\n\]\n\nHence \( x{\;\operatorname{mod}\;{p}^{n + 1}} = {z}_{0} + {z}_{1}p + {z}_{2}{p}^{2} + \cdots + {z}_{n}{p}^{n} \) for all \( n \in {\mathbb{Z}}^{ \geq } \).\n\nBut \( x \) has a base- \( p \) expansion\n\n\[ x = {x}_{0} + {x}_{1}p + \cdots + {x}_{N}{p}^{N} \]\n\nfor some \( N \in {}^{ * }{\mathbb{Z}}^{ \geq } \), and for each standard \( n \geq 0 \) we get by result (v) that\n\n\[ x \equiv \mathop{\sum }\limits_{{i = 0}}^{n}{x}_{i}{p}^{i}\left( {\;\operatorname{mod}\;{p}^{n + 1}}\right) \]\n\nso\n\n\[ \mathop{\sum }\limits_{{i = 0}}^{n}{x}_{i}{p}^{i} = x{\;\operatorname{mod}\;{p}^{n + 1}} = \mathop{\sum }\limits_{{i = 0}}^{n}{z}_{i}{p}^{i}, \]\n\nand therefore \( {x}_{i} = {z}_{i} \) for all \( i \leq n \). Thus the \( p \) -adic integer \( a \) and the hyperinteger \( x \) have the same coefficients of standard degree in these base- \( p \) expansions. | Yes |
Theorem 19.8.2 If \( V \) is a real vector space, then in any enlargement of a universe over \( V \) there is a hyperreal subspace \( {V}^{ + } \) of \( {}^{ * }V \) with\n\n\[ V \subseteq {V}^{ + } \in {}^{ * }\operatorname{Fin}\left( V\right) \] | Proof. Let \( R \) be the membership relation from \( V \) to \( \operatorname{Fin}\left( V\right) \), i.e.,\n\n\[ {xRW}\text{ iff }x \in W \in \operatorname{Fin}\left( V\right) . \]\n\nThen \( R \) is concurrent, for if \( {x}_{1},\ldots ,{x}_{n} \) are vectors in \( V \), and \( W \) is the subspace they span (i.e., the set of all linear combinations \( \mathop{\sum }\limits_{1}^{n}{\lambda }_{i}{x}_{i} \) with real scalars), then \( W \) has dimension at most \( n \), and so \( {x}_{i}{RW} \) for all \( 1 \leq i \leq n \) . This also shows that the domain of \( R \) is \( V \) .\n\nSince \( {}^{ * }R \subseteq {}^{ * }V \times {}^{ * }\operatorname{Fin}\left( V\right) \), it follows that in the enlargement there is some \( {V}^{ + } \in {}^{ * }\operatorname{Fin}\left( V\right) \) with \( x\left( {{}^{ * }R}\right) {V}^{ + } \) for all \( x \in V \), and hence \( V \subseteq {V}^{ + } \) . | Yes |
Lemma 19.10.1 Given the hypothesis of the Hahn-Banach theorem, if \( x \in V - W \), then there exists a linear functional \( h \) on a subspace of \( V \) including \( W \cup \{ x\} \) such that \( h \) extends \( f \) and is dominated by \( p \) . | Proof. We give only a sketch of the proof of this standard piece of linear algebra. The subspace of \( V \) generated by \( W \cup \{ x\} \) is the set\n\n\[ \n\{ y + {\lambda x} : y \in W\text{ and }\lambda \in \mathbb{R}\} .\n\]\n\nA functional \( h \) is defined on this subspace by putting \( h\left( {y + {\lambda x}}\right) = f\left( y\right) + {\lambda c} \) , where \( c \) is a real constant. Any such \( c \) will make \( h \) a linear functional extending \( f \), but \( c \) has to be chosen suitably to ensure that \( h \) is dominated by \( p \) . It turns out that for this purpose \( c \) can any number bounded above by the infimum of the numbers \( p\left( {y + x}\right) - f\left( y\right) \) and below by the supremum of the numbers \( - p\left( {-y - x}\right) - f\left( y\right) \) as \( y \) ranges over \( W \) . | No |
Theorem 2.6.1 Any collection of subsets of \( I \) that has the finite intersection property can be extended to an ultrafilter on \( I \) . | Proof. If \( \mathcal{H} \) has the fip, then the filter \( {\mathcal{F}}^{\mathcal{H}} \) generated by \( \mathcal{F} \) is proper \( \left( {{2.5}\left( 7\right) }\right) \) . Let \( P \) be the collection of all proper filters on \( I \) that include \( {\mathcal{F}}^{\mathcal{H}} \) , partially ordered by set inclusion \( \subseteq \) . Then every linearly ordered subset of \( P \) has an upper bound in \( P \), since by \( {2.4}\left( 4\right) \) the union of this chain is in \( P \) . Hence by Zorn’s lemma \( P \) has a maximal element, which is thereby a maximal proper filter on \( I \) and thus an ultrafilter by \( {2.5}\left( 6\right) \) . | Yes |
Corollary 2.6.2 Any infinite set has a nonprincipal ultrafilter on it. | Proof. If \( I \) is infinite, the cofinite filter \( {\mathcal{F}}^{co} \) is proper and has the finite intersection property, and so is included in an ultrafilter \( \mathcal{F} \) . But for any \( i \in I \) we have \( I - \{ i\} \in {\mathcal{F}}^{co} \subseteq \mathcal{F} \), so \( \{ i\} \notin \mathcal{F} \), whereas \( \{ i\} \in {\mathcal{F}}^{i} \) . Hence \( \mathcal{F} \neq {\mathcal{F}}^{i} \) . Thus \( \mathcal{F} \) is nonprincipal. | Yes |
Theorem 3.6.1 The structure \( \langle * \mathbb{R}, + , \cdot , < \rangle \) is an ordered field with zero [0] and unity \( \left\lbrack \mathbf{1}\right\rbrack \) . | Proof. (Sketch) As a quotient ring of \( {\mathbb{R}}^{\mathbb{N}} \) ,* \( \mathbb{R} \) is readily shown to be a commutative ring with zero \( \left\lbrack \mathbf{0}\right\rbrack \) and unity \( \left\lbrack \mathbf{1}\right\rbrack \), and additive inverses given by\n\n\[ - \left\lbrack \left\langle {{r}_{n} : n \in \mathbb{N}}\right\rangle \right\rbrack = \left\lbrack \left\langle {-{r}_{n} : n \in \mathbb{N}}\right\rangle \right\rbrack ,\]\n\nor more briefly, \( - \left\lbrack {r}_{n}\right\rbrack = \left\lbrack {-{r}_{n}}\right\rbrack \) . To show that it has multiplicative inverses, suppose \( \left\lbrack r\right\rbrack \neq \left\lbrack \mathbf{0}\right\rbrack \) . Then \( r ≢ \mathbf{0} \), i.e., \( \left\{ {n \in \mathbb{N} : {r}_{n} = 0}\right\} \notin \mathcal{F} \), so as \( \mathcal{F} \) is an ultrafilter, \( J = \left\{ {n \in \mathbb{N} : {r}_{n} \neq 0}\right\} \in \mathcal{F} \) . Define a sequence \( s \) by putting\n\n\[ {s}_{n} = \left\{ \begin{array}{ll} \frac{1}{{r}_{n}} & \text{if }n \in J \\ 0 & \text{otherwise.} \end{array}\right.\]\n\nThen \( \llbracket r \odot s = \mathbf{1}\rrbracket \) is equal to \( J \), so \( \llbracket r \odot s = \mathbf{1}\rrbracket \in \mathcal{F} \), giving \( r \odot s \equiv \mathbf{1} \) and hence\n\n\[ \left\lbrack r\right\rbrack \cdot \left\lbrack s\right\rbrack = \left\lbrack {r \odot s}\right\rbrack = \left\lbrack 1\right\rbrack \]\n\nin * \( \mathbb{R} \) . But this means that \( \left\lbrack s\right\rbrack \) is the multiplicative inverse \( {\left\lbrack r\right\rbrack }^{-1} \) of \( \left\lbrack r\right\rbrack \) .\n\nTo see that the ordering \( < \) on \( {}^{ * }\mathbb{R} \) is linear, observe that \( \mathbb{N} \) is the disjoint union of the three sets\n\n\[ \llbracket r < s\rrbracket ,\;\llbracket r = s\rrbracket ,\;\llbracket s < r\rrbracket ,\]\n\nso exactly one of the three belongs to \( \mathcal{F} \) (by \( {2.5}\left( 4\right) \) ), and so exactly one of\n\n\[ \left\lbrack r\right\rbrack < \left\lbrack s\right\rbrack ,\;\left\lbrack r\right\rbrack = \left\lbrack s\right\rbrack ,\;\left\lbrack s\right\rbrack < \left\lbrack r\right\rbrack \]\n\nis true. It remains to show that the set \( \left\{ {\left\lbrack r\right\rbrack : \left\lbrack 0\right\rbrack < \left\lbrack r\right\rbrack }\right\} \) of \ | No |
Theorem 3.7.1 The map \( r \mapsto {}^{ * }r \) is an order-preserving field isomorphism from \( \mathbb{R} \) into * \( \mathbb{R} \) . | Null | No |
Theorem 3.9.1 Any infinite subset of \( \mathbb{R} \) has nonstandard members. | Proof. Note first that this result must depend on \( \mathcal{F} \) being nonprincipal, because if \( \mathcal{F} \) were principal, there would be no nonstandard elements of \( {}^{ * }\mathbb{R} \) at all.\n\nNow, if \( A \subseteq \mathbb{R} \) is infinite, then there is a sequence \( r \) of elements of \( A \) whose terms are all distinct. Then \( \llbracket r \in A\rrbracket = \mathbb{N} \in \mathcal{F} \), so \( \left\lbrack r\right\rbrack \in {}^{ * }A \) . But for each \( s \in A,\left\{ {n : {r}_{n} = s}\right\} \) is either \( \varnothing \) or a singleton, neither of which can belong to \( \mathcal{F}\left( {{2.5}\left( 5\right) }\right) \), so \( \left\lbrack r\right\rbrack \neq {}^{ * }s \) . Hence \( \left\lbrack r\right\rbrack \in {}^{ * }A - A \) . | Yes |
Theorem 5.6.1 Every limited hyperreal \( b \) is infinitely close to exactly one real number, called the shadow of \( b \), denoted by \( \operatorname{sh}\left( b\right) \) . | Proof. Let \( A = \{ r \in \mathbb{R} : r < b\} \) . Since \( b \) is limited, there exist real \( r, s \) with \( r < b < s \), so \( A \) is nonempty and bounded above in \( \mathbb{R} \) by \( s \) . By the completeness of \( \mathbb{R} \), it follows that \( A \) has a least upper bound \( c \in \mathbb{R} \) .\n\nTo show \( b \simeq c \), take any positive real \( \varepsilon \in \mathbb{R} \) . Since \( c \) is an upper bound of \( A \), we cannot have \( c + \varepsilon \in A \) ; hence \( b \leq c + \varepsilon \) . Also, if \( b \leq c - \varepsilon \), then \( c - \varepsilon \) would be an upper bound of \( A \), contrary to the fact that \( c \) is the smallest such upper bound. Hence \( b \nleqslant c - \varepsilon \) . Altogether then, \( c - \varepsilon < b \leq c + \varepsilon \), so \( \left| {b - c}\right| \leq \varepsilon \) . Since this holds for all positive real \( \varepsilon, b \) is infinitely close to \( c \) .\n\nFinally, for uniqueness, if \( b \simeq {c}^{\prime } \in \mathbb{R} \), then as \( b \simeq c \), we get \( c \simeq {c}^{\prime } \), and so \( c = {c}^{\prime } \), since both are real. | Yes |
Theorem 5.6.2 If \( b \) and \( c \) are limited and \( n \in \mathbb{N} \), then\n\n(1) \( \operatorname{sh}\left( {b \pm c}\right) = \operatorname{sh}\left( b\right) \pm \operatorname{sh}\left( c\right) \) ,\n\n(2) \( \operatorname{sh}\left( {b \cdot c}\right) = \operatorname{sh}\left( b\right) \cdot \operatorname{sh}\left( c\right) \) ,\n\n(3) \( \operatorname{sh}\left( {b/c}\right) = \operatorname{sh}\left( b\right) /\operatorname{sh}\left( c\right) \) if \( \operatorname{sh}\left( c\right) \neq 0 \) (i.e., if \( c \) is appreciable),\n\n(4) \( \operatorname{sh}\left( {b}^{n}\right) = \operatorname{sh}{\left( b\right) }^{n} \) ,\n\n(5) \( \operatorname{sh}\left( \left| b\right| \right) = \left| {\operatorname{sh}\left( b\right) }\right| \) ,\n\n(6) \( \operatorname{sh}\left( \sqrt[n]{b}\right) = \sqrt[n]{\operatorname{sh}\left( b\right) } \) if \( b \geq 0 \) ,\n\n(7) if \( b \leq c \) then \( \operatorname{sh}\left( b\right) \leq \operatorname{sh}\left( c\right) \) . | Proof. Exercise. | No |
Theorem 5.6.3 The quotient ring \( \mathbb{L}/\mathbb{I} \) is isomorphic to the real number field \( \mathbb{R} \) by the correspondence \( \operatorname{hal}\left( b\right) \mapsto \operatorname{sh}\left( b\right) \) . Hence \( \mathbb{I} \) is a maximal ideal of the ring \( \mathbb{L} \) . | Null | No |
Theorem 6.1.1 A real-valued sequence \( \left\langle {{s}_{n} : n \in \mathbb{N}}\right\rangle \) converges to \( L \in \mathbb{R} \) if and only if \( {s}_{n} \simeq L \) for all unlimited \( n \) . | Proof. Suppose \( \left\langle {{s}_{n} : n \in \mathbb{N}}\right\rangle \) converges to \( L \), and fix an \( N \in {}^{ * }{\mathbb{N}}_{\infty } \) . In order to show that \( {s}_{N} \simeq L \) we have to show that \( \left| {{s}_{N} - L}\right| < \varepsilon \) for any positive real \( \varepsilon \) . But given such an \( \varepsilon \), the standard convergence condition implies that there is an \( {m}_{\varepsilon } \in \mathbb{N} \) such that the standard tail beyond \( {s}_{{m}_{\varepsilon }} \) is within \( \varepsilon \) of \( L \) :\n\n\[ \left( {\forall n \in \mathbb{N}}\right) \left( {n > {m}_{\varepsilon } \rightarrow \left| {{s}_{n} - L}\right| < \varepsilon }\right) .\n\]\n\nThen by (universal) transfer this holds for the extended tail as well:\n\n\[ \left( {\forall n \in {}^{ * }\mathbb{N}}\right) \left( {n > {m}_{\varepsilon } \rightarrow \left| {{s}_{n} - L}\right| < \varepsilon }\right) .\n\]\n\nBut in fact, \( N > {m}_{\varepsilon } \) because \( N \) is unlimited and \( {m}_{\varepsilon } \) is limited, and so this last sentence implies \( \left| {{s}_{N} - L}\right| < \varepsilon \) as desired.\n\nFor the converse, suppose \( {s}_{n} \simeq L \) for all unlimited \( n \) . We have to show that any given interval \( \left( {L - \varepsilon, L + \varepsilon }\right) \) in \( \mathbb{R} \) contains some standard tail of the sequence. The essence of the argument is to invoke the fact that the extended tail is infinitely close to \( L \), hence contained in \( {}^{ * }\left( {L - \varepsilon, L + \varepsilon }\right) \), and then apply transfer.\n\nTo spell this out, fix an unlimited \( N \in {}^{ * }{\mathbb{N}}_{\infty } \) . Then for any \( n \in {}^{ * }\mathbb{N} \) , if \( n > N \), it follows that \( n \) is also unlimited, so \( {s}_{n} \simeq L \) and therefore \( \left| {{s}_{n} - L}\right| < \varepsilon \) . This shows that\n\n\[ \left( {\forall n \in {}^{ * }\mathbb{N}}\right) \left( {n > N \rightarrow \left| {{s}_{n} - L}\right| < \varepsilon }\right) .\n\]\n\nHence the sentence\n\n\[ \left( {\exists z \in {}^{ * }\mathbb{N}}\right) \left( {\forall n \in {}^{ * }\mathbb{N}}\right) \left( {n > z \rightarrow \left| {{s}_{n} - L}\right| < \varepsilon }\right)\n\]\n\nis true. But this is the \( * \) -transform of\n\n\[ \left( {\exists z \in \mathbb{N}}\right) \left( {\forall n \in \mathbb{N}}\right) \left( {n > z \rightarrow \left| {{s}_{n} - L}\right| < \varepsilon }\right) ,\n\]\n\nso by (existential) transfer the latter holds true, giving the desired conclusion.\n\nThus convergence to \( L \) amounts to the requirement that the extended tail of the sequence is contained in the halo of \( L \) . In this characterisation the role of the standard tails is taken over by the extended tail, while the standard open neighbourhoods \( \left( {L - \varepsilon, L + \varepsilon }\right) \) are replaced by the \ | Yes |
Theorem 6.2.1 A real-valued sequence \( \left\langle {{s}_{n} : n \in \mathbb{N}}\right\rangle \) converges in \( \mathbb{R} \) if either\n\n(1) it is bounded above in \( \mathbb{R} \) and nondecreasing: \( {s}_{1} \leq {s}_{2} \leq \cdots \) ; or\n\n(2) it is bounded below in \( \mathbb{R} \) and nonincreasing: \( {s}_{1} \geq {s}_{2} \geq \cdots \) . | Proof. Consider case (1). Let \( {s}_{N} \) be an extended term. We will show that \( {s}_{N} \) has a shadow, and that this shadow is a least upper bound of the set \( \left\{ {{s}_{n} : n \in \mathbb{N}}\right\} \) in \( \mathbb{R} \) . Since a set can have only one least upper bound, this implies that all extended terms have the same shadow, and so by Theorem 6.1.1 the original sequence converges to this shadow in \( \mathbb{R} \) .\n\nNow, by hypothesis there is a real number \( b \) that is an upper bound for \( \left\{ {{s}_{n} : n \in \mathbb{N}}\right\} \) . Then the statement \( {s}_{1} \leq {s}_{n} \leq b \) holds for all \( n \in \mathbb{N} \), so it holds for all \( n \in {}^{ * }\mathbb{N} \) by universal transfer. In particular, \( {s}_{1} \leq {s}_{N} \leq b \) , showing that \( {s}_{N} \) is limited, so indeed has a shadow \( L \) .\n\nNext we show that \( L \) is an upper bound of the real sequence. Since this sequence is nondecreasing, by universal transfer we have\n\n\[ n \leq m \rightarrow {s}_{n} \leq {s}_{m} \]\n\nfor all \( n, m \in * \mathbb{N} \) . In particular, if \( n \in \mathbb{N} \), then \( n \leq N \), so \( {s}_{n} \leq {s}_{N} \simeq L \) , giving \( {s}_{n} \leq L \), as both numbers are real.\n\nFinally, we show that \( L \) is the least upper bound in \( \mathbb{R} \) . For if \( r \) is any real upper bound of \( \left\{ {{s}_{n} : n \in \mathbb{N}}\right\} \), then by transfer, \( {s}_{n} \leq r \) for all \( n \in {}^{ * }\mathbb{N} \) , so \( L \simeq {s}_{N} \leq r \), giving \( L \leq r \), as both are real. | Yes |
Theorem 6.3.1 If \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{s}_{n} = L \) and \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{t}_{n} = M \) in \( \mathbb{R} \), then\n\n(1) \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {{s}_{n} + {t}_{n}}\right) = L + M \) ,\n\n(2) \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {c{s}_{n}}\right) = {cL} \), for any \( c \in \mathbb{R} \) ,\n\n(3) \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {{s}_{n}{t}_{n}}\right) = {LM} \) ,\n\n(4) \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {{s}_{n}/{t}_{n}}\right) = L/M \), if \( M \neq 0 \) . | Proof. Use Exercise 5.7(1). | No |
Theorem 6.4.1 A real-valued sequence \( \left\langle {s}_{n}\right\rangle \) is bounded in \( \mathbb{R} \) if and only if its extended terms are all limited. | Proof. To say that \( \left\langle {{s}_{n} : n \in \mathbb{N}}\right\rangle \) is bounded in \( \mathbb{R} \) means that it is contained within some real interval \( \left\lbrack {-b, b}\right\rbrack \), or equivalently that its absolute values \( \left| {s}_{n}\right| \) have some real upper bound \( b \) :\n\n\[ \left( {\forall n \in \mathbb{N}}\right) \left| {s}_{n}\right| < b. \]\n\nThen by universal transfer the extended sequence is contained in \( {}^{ * }\left\lbrack {-b, b}\right\rbrack \) , i.e., \( \left| {s}_{n}\right| < b \) for all \( n \in * \mathbb{N} \) ; hence \( {s}_{n} \) is limited in general.\n\nFor the converse, if \( {s}_{n} \) is limited for all unlimited \( n \in {}^{ * }{\mathbb{N}}_{\infty } \), then it is limited for all \( n \in \mathbb{N} \) . Hence if \( r \in {\mathbb{R}}_{\infty }^{ + } \) is any positive unlimited hyperreal, we observe that the entire extended sequence lies in the interval \( \{ x \in \mathbb{{R}^{n}} : \) \( - r < x < r\} \) and apply transfer. More formally, we have \( \left| {s}_{n}\right| < r \) for all \( n \in * \mathbb{N} \), so the sentence\n\n\[ \left( {\exists y \in {}^{ * }\mathbb{R}}\right) \left( {\forall n \in {}^{ * }\mathbb{N}}\right) \left| {s}_{n}\right| < y \]\n\n\nis true. But then by existential transfer it follows that there is some real number that is an upper bound to \( \left| {s}_{n}\right| \) for all \( n \in \mathbb{N} \) . | Yes |
Theorem 6.4.2 A real-valued sequence\n\n(1) diverges to infinity if and only if all of its extended terms are positive unlimited; and\n\n(2) diverges to minus infinity if and only if all of its extended terms are negative unlimited. | Proof. Exercise. | No |
Theorem 6.5.1 A real-valued sequence \( \left\langle {s}_{n}\right\rangle \) is Cauchy in \( \mathbb{R} \) if and only if all its extended terms are infinitely close to each other, i.e., iff \( {s}_{m} \simeq {s}_{n} \) for all \( m, n \in {}^{ * }{\mathbb{N}}_{\infty } \) . | Proof. Exercise. | No |
Theorem 6.5.2 (Cauchy’s Convergence Criterion). A real-valued sequence converges in \( \mathbb{R} \) if and only if it is Cauchy. | Proof. If \( \left\langle {{s}_{n} : n \in \mathbb{N}}\right\rangle \) is Cauchy, then it is bounded (standard result-why is it true?). Thus taking an unlimited number \( m \in {}^{ * }{\mathbb{N}}_{\infty } \), we have that \( {s}_{m} \) is limited (Theorem 6.4.1) and so it has a shadow \( L \in \mathbb{R} \) . But all extended terms of the sequence are infinitely close to each other (Theorem 6.5.1), hence are infinitely close to \( {s}_{m} \), and therefore are infinitely close to \( L \) as \( {s}_{m} \simeq L \) . This shows that the extended tail of the sequence is contained in the halo of \( L \), implying by Theorem 6.1.1 that \( \left\langle {s}_{n}\right\rangle \) converges to \( L \in \mathbb{R} \) . | No |
Theorem 6.6.1 \( L \in \mathbb{R} \) is a cluster point of the real-valued sequence \( \left\langle {s}_{n}\right. \) : \( n \in \mathbb{N}\rangle \) if and only if the sequence has an extended term infinitely close to \( L \), i.e., iff \( {s}_{N} \simeq L \) for some unlimited \( N \) . | Proof. Assume that (i) holds. Let \( \varepsilon \) be a positive infinitesimal and \( m \in \) \( {}^{ * }{\mathbb{N}}_{\infty } \) . Then by transfer of (i), there is some \( n \in {}^{ * }\mathbb{N} \) with \( n > m \), and hence \( n \) is unlimited, and\n\n\[ \left| {{s}_{n} - L}\right| < \varepsilon \simeq 0. \]\n\nThus \( {s}_{n} \) is an extended term infinitely close to \( L \) . (Indeed, the argument shows that any interval of infinitesimal width around \( L \) contains terms arbitrarily far along the extended tail.)\n\nConversely, suppose there is an unlimited \( N \) with \( {s}_{N} \simeq L \) . To prove (i), take any positive \( \varepsilon \in \mathbb{R} \) and \( m \in \mathbb{N} \) . Then \( N > m \) and \( \left| {{s}_{N} - L}\right| < \varepsilon \) . This shows that\n\n\[ \left( {\exists n \in * \mathbb{N}}\right) \left( {n > m \land \left| {{s}_{n} - L}\right| < \varepsilon }\right) . \]\n\nThus by existential transfer, \( \left| {{s}_{n} - L}\right| < \varepsilon \) for some \( n \in \mathbb{N} \), with \( n > m \) . | Yes |
Theorem 6.8.2 A real number \( L \) is equal to \( \overline{\lim }s \) if and only if\n\n(1) \( {s}_{n} < L \) or \( {s}_{n} \simeq L \) for all unlimited \( n \) ; and\n\n(2) \( {s}_{n} \simeq L \) for at least one unlimited \( n \) . | Proof. The condition \ | No |
Theorem 6.8.3 A bounded real-valued sequence \( s \) converges to \( L \in \mathbb{R} \) if and only if\n\n\[ \mathop{\limsup }\limits_{{n \rightarrow \infty }}{s}_{n} = \mathop{\liminf }\limits_{{n \rightarrow \infty }}{s}_{n} = L. \] | Proof. Since \( \overline{\lim }s \) and \( \underline{\lim }s \) are the maximum and minimum elements of \( {C}_{s} \), requiring that they both be equal to \( L \) amounts to requiring that \( {C}_{s} = \{ L\} \) . But that just means that the shadow of every extended term is equal to \( L \), which is equivalent to having \( s \) converge to \( L \) by Theorem 6.1.1. | Yes |
Theorem 6.8.4 If \( s \) is a bounded real-valued sequence with limit superior \( \overline{\lim } \), then for any positive real \( \varepsilon \) :\n\n(1) some standard tail of \( s \) has all its terms smaller than \( \overline{\lim } + \varepsilon \), i.e., \( {s}_{n} < \overline{\lim } + \varepsilon \) for all but finitely many \( n \in \mathbb{N} \) ;\n\n(2) \( \overline{\lim } - \varepsilon < {s}_{n} \) for infinitely many \( n \in \mathbb{N} \) . | Proof.\n\n(1) If \( m \in * \mathbb{N} \) is unlimited, then \( \operatorname{sh}\left( {s}_{m}\right) \leq \overline{\lim } \), so\n\n\[ \n{s}_{m} \simeq \operatorname{sh}\left( {s}_{m}\right) < \overline{\lim } + \varepsilon \n\]\n\nshowing that \( {s}_{m} < \overline{\lim } + \varepsilon \) because \( \operatorname{sh}\left( {s}_{m}\right) \) and \( \overline{\lim } + \varepsilon \) are both real. Thus all extended terms are smaller than \( \overline{\lim } + \varepsilon \), and in particular, this holds for all terms after \( {s}_{N} \) for any fixed unlimited \( N \) :\n\n\[ \n\left( {\forall m \in * \mathbb{N}}\right) \left( {m \geq N \rightarrow {s}_{m} < \overline{\lim } + \varepsilon }\right) .\n\]\n\nExistential transfer then provides an \( n \in \mathbb{N} \) such that all of\n\n\[ \n{s}_{n},{s}_{n + 1},{s}_{n + 2},\ldots \n\]\n\nare smaller than \( \overline{\lim } + \varepsilon \) .\n\n(2) \( \overline{\lim } \) is the shadow of some extended term \( {s}_{N} \) with \( N \) unlimited. Then \( \overline{\lim } - \varepsilon < \overline{\lim } \simeq {s}_{N} \), so \( \overline{\lim } - \varepsilon < {s}_{N} \) . But now for any limited \( m \in \mathbb{N} \) we have \( m < N \) and \( \overline{\lim } - \varepsilon < {s}_{N} \) . Existential transfer then ensures that there is a limited \( n \) with \( m < n \) and \( \overline{\lim } - \varepsilon < {s}_{n} \) . This shows that \( \overline{\lim } - \varepsilon < {s}_{n} \) for arbitrarily large \( n \in \mathbb{N} \), giving the desired conclusion. | Yes |
Theorem 6.8.5 For any bounded real-valued sequence \( s \) , | Proof. First we show that\n\n\[ \overline{\lim } \leq {S}_{m}\;\text{ for all }m \in \mathbb{N}. \]\n\n(ii)\n\nTo see this, take an extended term \( {s}_{N} \) whose shadow is infinitely close to the cluster point \( \overline{\lim } \). Then if \( m \in \mathbb{N} \), we have \( {s}_{n} \leq {S}_{m} \) for all limited \( n \geq m \), and hence for all hypernatural \( n \geq m \) by transfer. In particular, \( {s}_{N} \leq {S}_{m} \), so as \( \overline{\lim } \simeq {s}_{N} \), this forces \( \overline{\lim } \leq {S}_{m} \) as required for (ii).\n\nNow let \( L = \mathop{\lim }\limits_{{n \rightarrow \infty }}{S}_{n} \). Then \( L \) is the greatest lower bound of the sequence \( S \), and by (ii), \( \overline{\lim } \) is a lower bound for this sequence, so \( \overline{\lim } \leq L \). But if \( \overline{\lim } < L \), we can choose a positive real \( \varepsilon \) with \( \overline{\lim } + \varepsilon < L \), and then by Theorem 6.8.4(1) there is some \( n \in \mathbb{N} \) such that the standard tail \( {s}_{n},{s}_{n + 1},{s}_{n + 2},\ldots \) is bounded above by \( \overline{\lim } + \varepsilon \). This implies that the least upper bound \( {S}_{n} \) of this tail is no bigger than \( \overline{\lim } + \varepsilon \). However, that gives\n\n\[ {S}_{n} \leq \overline{\lim } + \varepsilon < L \]\n\nwhich contradicts the fact that \( L \) is a lower bound of \( S \), and so \( L \leq {S}_{n} \).\n\nWe are left with the conclusion that \( \overline{\lim } = L \), as desired. | Yes |
Theorem 7.1.1 \( f \) is continuous at the real point \( c \) if and only if \( f\left( x\right) \simeq \) \( f\left( c\right) \) for all \( x \in {}^{ * }\mathbb{R} \) such that \( x \simeq c \), i.e., iff\n\n\[ f\left( {\operatorname{hal}\left( c\right) }\right) \subseteq \operatorname{hal}\left( {f\left( c\right) }\right) . \] | Proof. The standard definition is that \( f \) is continuous at \( c \) iff for each open interval \( \left( {f\left( c\right) - \varepsilon, f\left( c\right) + \varepsilon }\right) \) around \( f\left( c\right) \) in \( \mathbb{R} \) there is a corresponding open interval \( \left( {c - \delta, c + \delta }\right) \) around \( c \) that is mapped into \( \left( {f\left( c\right) - \varepsilon, f\left( c\right) + \varepsilon }\right) \) by \( f \) . Since \( a < c < b \), the number \( \delta \) can be chosen small enough so that the interval \( \left( {c - \delta, c + \delta }\right) \) is contained with \( \left( {a, b}\right) \), ensuring that \( f \) is indeed defined at all points that are within \( \delta \) of \( c \) .\n\nContinuity at \( c \) is thus formally expressed by the sentence\n\n\[ \left( {\forall \varepsilon \in {\mathbb{R}}^{ + }}\right) \left( {\exists \delta \in {\mathbb{R}}^{ + }}\right) \left( {\forall x \in \mathbb{R}}\right) \left( {\left| {x - c}\right| < \delta \rightarrow \left| {f\left( x\right) - f\left( c\right) }\right| < \varepsilon }\right) .\n\n(i)\n\nNow suppose \( x \simeq c \) implies \( f\left( x\right) \simeq f\left( c\right) \) . To show that (i) holds, let \( \varepsilon \) be a positive real number. Then we have to find a real \( \delta \) small enough to fulfill (i). First we show that this can be achieved if \ | No |
Corollary 7.1.2 The following are equivalent.\n\n(1) \( f \) is continuous at \( c \in \mathbb{R} \) .\n\n(2) \( f\left( x\right) \simeq f\left( c\right) \) whenever \( x \simeq c \) .\n\n(3) There is some positive \( d \simeq 0 \) such that \( f\left( x\right) \simeq f\left( c\right) \) whenever \( \left| {x - c}\right| < \n\nd \) . | Null | No |
Theorem 7.1.3 The following are equivalent.\n\n(1) \( f \) is continuous at \( c \) in \( A \) .\n\n(2) \( f\\left( x\\right) \\simeq f\\left( c\\right) \) for all \( x \\in * A \) with \( x \\simeq c \) .\n\n(3) There is some positive \( d \\simeq 0 \) such that \( f\\left( x\\right) \\simeq f\\left( c\\right) \) for all \( x \\in {}^{ * }A \) with \( \\left| {x - c}\\right| < d \) . | Null | No |
Theorem 7.7.1 \( f \) is uniformly continuous on \( A \) if and only if \( x \simeq y \) implies \( f\left( x\right) \simeq f\left( y\right) \) for all hyperreals \( x, y \in {}^{ * }A \) . | Proof. Exercise. | No |
Theorem 7.7.2 If the real function \( f \) is continuous on the closed interval \( \left\lbrack {a, b}\right\rbrack \) in \( \mathbb{R} \), then \( f \) is uniformly continuous on \( \left\lbrack {a, b}\right\rbrack \) . | Proof. Take hyperreals \( x, y \in * \left\lbrack {a, b}\right\rbrack \) with \( x \simeq y \) . Let \( c = \operatorname{sh}\left( x\right) \) . Then since \( a \leq x \leq b \) and \( x \simeq c \), we have \( c \in \left\lbrack {a, b}\right\rbrack \), and so \( f \) is continuous at \( c \) . Applying Theorem 7.1.1, we get \( f\left( x\right) \simeq f\left( c\right) \) and \( f\left( y\right) \simeq f\left( c\right) \), whence \( f\left( x\right) \simeq f\left( y\right) \) . Hence \( f \) is uniformly continuous by Theorem 7.7.1. | Yes |
Theorem 7.12.1 The sequence \( \left\langle {{f}_{n} : n \in \mathbb{N}}\right\rangle \) of real-valued functions defined on \( A \subseteq \mathbb{R} \) converges pointwise to the function \( f : A \rightarrow \mathbb{R} \) if and only if for each \( x \in A \) and each unlimited \( n \in {}^{ * }\mathbb{N},{f}_{n}\left( x\right) \simeq f\left( x\right) \) . | Null | No |
Theorem 7.12.2 \( \left\langle {{f}_{n} : n \in \mathbb{N}}\right\rangle \) converges uniformly to the function \( f \) : \( A \rightarrow \mathbb{R} \) if and only if for each \( x \in {}^{ * }A \) and each unlimited \( n \in {}^{ * }\mathbb{N},{f}_{n}\left( x\right) \simeq \) \( f\left( x\right) \) . | Proof. Exercise. | No |
Theorem 7.13.1 If the functions \( \left\langle {{f}_{n} : n \in \mathbb{N}}\right\rangle \) are all continuous on \( A \subseteq \) \( \mathbb{R} \), and the sequence converges uniformly to the function \( f : A \rightarrow \mathbb{R} \), then \( f \) is continuous on \( A \) . | Proof. Let \( c \) belong to \( A \) . To prove that \( f \) is continuous at \( c \), we invoke Theorem 7.1.3(2). If \( x \in * A \) with \( x \simeq c \), we want \( f\left( x\right) \simeq f\left( c\right) \), i.e., \( \mid f\left( x\right) - \) \( f\left( c\right) \mid < \varepsilon \) for any positive real \( \varepsilon \) . The key to this is to analyse the inequality\n\n\[ \left| {f\left( x\right) - f\left( c\right) }\right| \leq \left| {f\left( x\right) - {f}_{n}\left( x\right) }\right| + \left| {{f}_{n}\left( x\right) - {f}_{n}\left( c\right) }\right| + \left| {{f}_{n}\left( c\right) - f\left( c\right) }\right| .\](x)\n\nOn the right side, the middle term \( \left| {{f}_{n}\left( x\right) - {f}_{n}\left( c\right) }\right| \) will be infinitesimal for any \( n \in \mathbb{N} \) because \( x \simeq c \) and \( {f}_{n} \) is continuous at \( c \) . By taking a large enough \( n \), the first and last terms on the right can be made small enough that the sum of the three terms is less than \( \varepsilon \) .\n\nTo see how this works in detail, for a given \( \varepsilon \in {\mathbb{R}}^{ + } \) we apply the definition of uniform convergence to the number \( \varepsilon /4 \) to get that there is some integer \( m \in \mathbb{N} \) such that\n\n\[ n > m\text{ implies }\left| {{f}_{n}\left( x\right) - f\left( x\right) }\right| < \varepsilon /4 \]\n\nfor all \( n \in \mathbb{N} \) and all \( x \in A \), and hence for all \( n \in {}^{ * }\mathbb{N} \) and all \( x \in {}^{ * }A \) by universal transfer.\n\nNow fix \( n \) as a standard integer, say by putting \( n = m + 1 \) . Then for any \( x \in {}^{ * }A \) with \( x \simeq c \) it follows, since \( x, c \in {}^{ * }A \), that\n\n\[ \left| {{f}_{n}\left( x\right) - f\left( x\right) }\right| ,\left| {{f}_{n}\left( c\right) - f\left( c\right) }\right| < \varepsilon /4 \]\n\nand so in \( \left( \mathrm{x}\right) \) we get\n\n\[ \left| {f\left( x\right) - f\left( c\right) }\right| < \varepsilon /4 + \text{ infinitesimal } + \varepsilon /4 < \varepsilon \]\n\nas desired.\n\nNote that this proof is a mixture of standard and nonstandard arguments: it uses the hyperreal characterisation of continuity of \( {f}_{n} \) and \( f \), but the standard definition of uniform convergence of \( \left\langle {{f}_{n} : n \in \mathbb{N}}\right\rangle \) rather than the characterisation given by Theorem 7.12.2. | Yes |
Theorem 7.14.1 If the functions \( \left\langle {{f}_{n} : n \in \mathbb{N}}\right\rangle \) are all continuous on \( A \subseteq \) \( \mathbb{R} \), then for any \( n \in * \mathbb{N} \) and any \( y \in * A \) there is a positive infinitesimal \( d \) such that \( {f}_{n}\left( x\right) \simeq {f}_{n}\left( y\right) \) for all \( x \in {}^{ * }A \) with \( \left| {x - y}\right| < d \) . | Proof. The fact that \( {f}_{n} \) is continuous on \( A \) for all \( n \in \mathbb{N} \) is expressed by the sentence\n\n\[ \left( {\forall n \in \mathbb{N}}\right) \left( {\forall y \in A}\right) \]\n\n\[ \left( {\forall \varepsilon \in {\mathbb{R}}^{ + }}\right) \left( {\exists \delta \in {\mathbb{R}}^{ + }}\right) \left( {\forall x \in A}\right) \left( {\left| {x - y}\right| < \delta \rightarrow \left| {{f}_{n}\left( x\right) - {f}_{n}\left( y\right) }\right| < \varepsilon }\right) ,\]\n\nwhich states that \ | Yes |
Theorem 8.1.1 If \( f \) is defined at \( x \in \mathbb{R} \), then the real number \( L \in \mathbb{R} \) is the derivative of \( f \) at \( x \) if and only if for every nonzero infinitesimal \( \varepsilon \) , \( f\left( {x + \varepsilon }\right) \) is defined and\n\n\[ \frac{f\left( {x + \varepsilon }\right) - f\left( x\right) }{\varepsilon } \simeq L \] | Proof. Let \( g\left( h\right) = \frac{f\left( {x + h}\right) - f\left( x\right) }{h} \) and apply the characterisation of\n\n\[ \text{ “ }\mathop{\lim }\limits_{{h \rightarrow 0}}g\left( h\right) = L\text{ ” } \]\ngiven in Section 7.3.\n\nThus when \( f \) is differentiable (i.e., has a derivative) at \( x \), we have\n\n\[ {f}^{\prime }\left( x\right) = \operatorname{sh}\left( \frac{f\left( {x + \varepsilon }\right) - f\left( x\right) }{\varepsilon }\right) \]\n\nfor all infinitesimal \( \varepsilon \neq 0 \) . | No |
Theorem 8.2.1 If \( f \) is differentiable at \( x \in \mathbb{R} \), then \( f \) is continuous at \( x \) . | The differential of \( f \) at \( x \) corresponding to \( {\Delta x} \) is defined to be\n\n\[ \n{df} = {f}^{\prime }\left( x\right) {\Delta x}.\n\]\n\nThus whereas \( {\Delta f} \) represents the increment of the \ | No |
Theorem 8.2.2 (Incremental Equation) If \( {f}^{\prime }\left( x\right) \) exists at real \( x \) and \( {\Delta x} = {dx} \) is infinitesimal, then \( {\Delta f} \) and \( {df} \) are infinitesimal, and there is an infinitesimal \( \varepsilon \), dependent on \( x \) and \( {\Delta x} \), such that\n\n\[ \n{\Delta f} = {f}^{\prime }\left( x\right) {\Delta x} + {\varepsilon \Delta x} = {df} + {\varepsilon dx} \n\]\n\nand so\n\n\[ \nf\left( {x + {\Delta x}}\right) = f\left( x\right) + {f}^{\prime }\left( x\right) {\Delta x} + {\varepsilon \Delta x}. \n\] | Null | No |
Theorem 8.7.1 (Incremental Equation for Two Variables) If \( f \) is smooth at the real point \( \left( {a, b}\right) \) and \( {\Delta x} \) and \( {\Delta y} \) are infinitesimal, then\n\n\[ \n{\Delta f} = {df} + {\varepsilon \Delta x} + {\delta \Delta y} \n\] \n\nfor some infinitesimals \( \varepsilon \) and \( \delta \) . | Proof. The increment of \( f \) at \( \left( {a, b}\right) \) corresponding to \( {\Delta x},{\Delta y} \) can be written as\n\n\[ \n{\Delta f} = \left\lbrack {f\left( {a + {\Delta x}, b + {\Delta y}}\right) - f\left( {a + {\Delta x}, b}\right) }\right\rbrack + \left\lbrack {f\left( {a + {\Delta x}, b}\right) - f\left( {a, b}\right) }\right\rbrack . \n\] \n\n(ii)\n\nThe second main summand of (ii) is the increment at \( a \) corresponding to \( {\Delta x} \) of the one-variable function \( x \mapsto f\left( {x, b}\right) \), whose derivative \( {f}_{x}\left( {a, b}\right) \) is assumed to exist. Applying the one-variable incremental equation (Theorem 8.2.2) thus gives\n\n\[ \nf\left( {a + {\Delta x}, b}\right) - f\left( {a, b}\right) = {f}_{x}\left( {a, b}\right) {\Delta x} + {\varepsilon \Delta x} \n\] \n\n(iii)\n\nfor some infinitesimal \( \varepsilon \) .\n\nSimilarly, for the first summand we need to show that\n\n\[ \nf\left( {a + {\Delta x}, b + {\Delta y}}\right) - f\left( {a + {\Delta x}, b}\right) = {f}_{y}\left( {a, b}\right) {\Delta y} + {\delta \Delta y} \n\] \n\n(iv)\n\nfor some infinitesimal \( \delta \) . Then combining (ii)-(iv) will give\n\n\[ \n{\Delta f} = {f}_{x}\left( {a, b}\right) {\Delta x} + {f}_{y}\left( {a, b}\right) {\Delta y} + {\varepsilon \Delta x} + {\delta \Delta y}, \n\] \n\nwhich is the desired result. | Yes |
Theorem 8.10.1 If the nth derivative \( {f}^{\left( n\right) } \) exists on an open interval containing the real number \( x \), and \( {f}^{\left( n\right) } \) is continuous at \( x \), then for any infinitesimal \( {\Delta x} \) , | \[ f\left( {x + {\Delta x}}\right) = f\left( x\right) + {f}^{\prime }\left( x\right) {\Delta x} + \frac{{f}^{\prime \prime }\left( x\right) }{2!}\Delta {x}^{2} + \cdots + \frac{{f}^{\left( n\right) }\left( x\right) }{n!}\Delta {x}^{n} + {\varepsilon \Delta }{x}^{n} \] for some infinitesimal \( \varepsilon \) . | Yes |
Theorem 8.11.1 Let \( f \) be differentiable on an interval \( \left( {a, b}\right) \) in \( \mathbb{R} \) . Then the derivative \( {f}^{\prime } \) is continuous on \( \left( {a, b}\right) \) if and only if for each hyperreal \( x \) that is well inside \( {}^{ * }\left( {a, b}\right) \) and each infinitesimal \( {\Delta x} \) , \n\n\[ \n f\left( {x + {\Delta x}}\right) = f\left( x\right) + {f}^{\prime }\left( x\right) {\Delta x} + {\varepsilon \Delta x} \n\] \n\nfor some infinitesimal \( \varepsilon \) . | Proof. Assume that the incremental equation holds at points well inside \( {}^{ * }\left( {a, b}\right) \) . To prove continuity of \( {f}^{\prime } \), let \( c \) be a real point in \( \left( {a, b}\right) \) and suppose \( x \simeq c \) . We want \( {f}^{\prime }\left( x\right) \simeq {f}^{\prime }\left( c\right) \) .\n\nNow, if \( \Delta = \left( {x - c}\right) \simeq 0 \), then using Theorem 8.2.2 we get \n\n\[ \n f\left( x\right) = f\left( {c + \Delta }\right) = f\left( c\right) + {f}^{\prime }\left( c\right) \Delta + {\varepsilon \Delta } \n\] \n\nfor some \( \varepsilon \simeq 0 \) . But \( x \) is well inside \( {}^{ * }\left( {a, b}\right) \), since \( a < c < b \) and \( x \simeq c \), so by the assumed incremental equation at \( x \), applied to the infinitesimal \( - \Delta \) , we have \n\n\[ \n f\left( c\right) = f\left( {x + \left( {-\Delta }\right) }\right) = f\left( x\right) + {f}^{\prime }\left( x\right) \left( {-\Delta }\right) + {\varepsilon }^{\prime }\left( {-\Delta }\right) \n\] \n\nfor some \( {\varepsilon }^{\prime } \simeq 0 \) . Combining these equations leads to \n\n\[ \n {f}^{\prime }\left( x\right) - {f}^{\prime }\left( c\right) = \varepsilon - {\varepsilon }^{\prime } \simeq 0 \n\] \n\ngiving our desired conclusion \( {f}^{\prime }\left( x\right) \simeq {f}^{\prime }\left( c\right) \) .\n\nThe proof that continuity of \( {f}^{\prime } \) implies the incremental equation at points well inside \( {}^{ * }\left( {a, b}\right) \) is indicated in the following exercises, which also give an example to show what can happen when continuity fails. | No |
Theorem 9.4.1 The function \( F\left( x\right) = {\int }_{a}^{x}f\left( t\right) {dt} \) is differentiable on \( \left\lbrack {a, b}\right\rbrack \) , and its derivative is \( f \) . | There is a very intuitive explanation of why this relationship should hold. The increment\n\n\[ \n{\Delta F} = F\left( {x + {\Delta x}}\right) - F\left( x\right) \n\]\n\nof \( F \) at \( x \) corresponding to a positive infinitesimal \( {\Delta x} \) is closely approximated by the area of the rectangle of height \( f\left( x\right) \) and width \( {\Delta x} \), i.e, by \( f\left( x\right) {\Delta x} \) . Thus the quotient \( \frac{\Delta F}{\Delta x} \) should be closely approximated by \( f\left( x\right) \) itself.\n\nDoes \ | No |
Theorem 9.4.2 Fundamental Theorem of Calculus. If a function \( G \) has a continuous derivative \( f \) on \( \left\lbrack {a, b}\right\rbrack \), then \( {\int }_{a}^{b}f\left( x\right) {dx} = G\left( b\right) - G\left( a\right) \) . | Proof. This follows from Theorem 9.4.1 by standard arguments that require no ideas of limits or infinitesimals. For if \( F\left( x\right) = {\int }_{a}^{x}f\left( t\right) {dt} \), then on \( \left\lbrack {a, b}\right\rbrack \) we have \( {\left( G\left( x\right) - F\left( x\right) \right) }^{\prime } = f\left( x\right) - f\left( x\right) = 0 \), so there is a constant \( c \) with \( G\left( x\right) - F\left( x\right) = c \) . This implies \( G\left( b\right) - G\left( a\right) = F\left( b\right) - F\left( a\right) \) . But \( F\left( b\right) - F\left( a\right) = {\int }_{a}^{b}f\left( t\right) {dt}. \) | Yes |
Theorem 10.1.1 If \( A \subseteq \mathbb{R} \) and \( r \in \mathbb{R} \), (1) \( r \) is interior to \( A \) if and only if \( r \simeq x \) implies \( x \in * A \), i.e., iff \( \operatorname{hal}\left( r\right) \subseteq \) *A. | ## Proof. (1) Let \( r \in {A}^{ \circ } \). Then \( \left( {r - \varepsilon, r + \varepsilon }\right) \subseteq A \) for some real \( \varepsilon > 0 \). Then the sentence \[ \left( {\forall x \in \mathbb{R}}\right) \left( {\left| {r - x}\right| < \varepsilon \rightarrow x \in A}\right) \] (i) is true. But now if \( r \simeq x \) in \( {}^{ * }\mathbb{R} \), then \( \left| {r - x}\right| < \varepsilon \), so by universal transfer of (i), \( x \in {}^{ * }A \). This shows that \( \operatorname{hal}\left( r\right) \subseteq {}^{ * }A \). (An alternative way of putting this is to observe that \( \operatorname{hal}\left( r\right) \subseteq {}^{ * }\left( {r - \varepsilon, r + \varepsilon }\right) \subseteq {}^{ * }A \).) Conversely, if \( \operatorname{hal}\left( r\right) \subseteq {}^{ * }A \), then the sentence \[ \left( {\exists \varepsilon \in {}^{ * }{\mathbb{R}}^{ + }}\right) \left( {\forall x \in {}^{ * }\mathbb{R}}\right) \left( {\left| {r - x}\right| < \varepsilon \rightarrow x \in {}^{ * }A}\right) \] is seen to be true by interpreting \( \varepsilon \) as any positive infinitesimal. But then by existential transfer there is some real \( \varepsilon > 0 \) for which (i) holds, so \( \left( {r - \varepsilon, r + \varepsilon }\right) \subseteq A \) and hence \( r \in {A}^{ \circ } \). | Yes |
Theorem 10.2.2 For any real number \( r \) , \[ \operatorname{hal}\left( r\right) = \bigcap \{ {}^{ * }A : r \in A\text{ and }A\text{ is open }\} . \] | Proof. We have already observed that if \( r \in A \subseteq \mathbb{R} \) and \( A \) is open, then \( \operatorname{hal}\left( r\right) \subseteq {}^{ * }A \) . On the other hand, if \( x \notin \operatorname{hal}\left( r\right) \), then \( x ≄ r \), so there must exist some real \( \varepsilon > 0 \) such that \( \left| {r - x}\right| > \varepsilon \) . Put \( A = \left( {r - \varepsilon, r + \varepsilon }\right) \subseteq \mathbb{R} \) . Then \( r \in A \) and \( A \) is open, but \( x \notin {}^{ * }A = \{ y \in {}^{ * }\mathbb{R} : \left| {r - y}\right| < \varepsilon \} \) . | Yes |
Theorem 10.3.1 (Heine-Borel) A set \( B \subseteq \mathbb{R} \) is compact if and only if it is closed and bounded. | Proof. We have already seen that if \( B \) satisfies Robinson’s criterion, then it is closed and bounded (above and below).\n\nConversely, if \( B \) is closed and bounded, then there is some real \( b \) such that\n\n\[ \left( {\forall x \in B}\right) \left( {\left| x\right| \leq b}\right) . \]\n\nNow, to prove Robinson’s criterion, suppose \( x \in {}^{ * }B \) . Then by transfer, \( \left| x\right| \leq b \in \mathbb{R} \) . Hence \( x \) is limited, and so has a shadow \( r \in \mathbb{R} \) . Then \( r \simeq \) \( x \in {}^{ * }B \), and so \( r \in B \) because \( B \) is closed. Thus we have shown that \( x \) is infinitely close to the member \( r \) of \( B \), proving that \( B \) is compact. | Yes |
Theorem 10.4.1 The continuous image of a compact set is compact. | Proof. Let \( f \) be a continuous real function, and \( B \) a compact subset of \( \mathbb{R} \) included in the domain of \( f \) . Now, it is true, by definition of \( f\left( B\right) \), that\n\n\[ \left( {\forall y \in f\left( B\right) }\right) \left( {\exists x \in B}\right) \left( {y = f\left( x\right) }\right) .\n\]\n\nThus by transfer, if \( y \in \mathcal{N}\left( {f\left( B\right) }\right) \), then \( y = f\left( x\right) \) for some \( x \in \mathcal{B} \) . Since \( B \) is compact, \( x \simeq r \) for some \( r \in B \) . Then by continuity of \( f, f\left( x\right) \simeq f\left( r\right) \) , i.e., \( y \) is infinitely close to \( f\left( r\right) \in f\left( B\right) \) . This shows by Robinson’s criterion that \( f\left( B\right) \) is compact. | Yes |
Theorem 10.4.2 If \( f \) is continuous on a compact set \( B \subseteq \mathbb{R} \), then \( f \) is uniformly continuous on \( B \) . | Proof. By Theorem 7.7.1 we have to show that for all \( x, y \in {}^{ * }B \) ,\n\n\[ x \simeq y\;\text{ implies }\;f\left( x\right) \simeq f\left( y\right) . \]\n\nBut if \( x, y \in {}^{ * }B \), then by compactness \( x \simeq r \in B \) and \( y \simeq s \in B \) for some \( r, s \) . Thus if \( x \simeq y \), then \( r \simeq s \), and so \( r = s \), as both are real. Hence by continuity of \( f \) at \( r \in B, f\left( x\right) \simeq f\left( r\right) \) and \( f\left( y\right) \simeq f\left( r\right) \), whence \( f\left( x\right) \simeq f\left( y\right) \) as desired. | Yes |
Theorem 11.3.1 Any nonempty internal subset of \( {}^{ * }\mathbb{N} \) has a least member. | Proof. Let \( \left\lbrack {A}_{n}\right\rbrack \) be a nonempty internal subset of \( {}^{ * }\mathbb{N} \) . Then by the observations above we can assume that for each \( n \in \mathbb{N} \) ,\n\n\[ \varnothing \neq {A}_{n} \subseteq \mathbb{N} \]\n\nand so \( {A}_{n} \) has a least member \( {r}_{n} \) . This defines a point \( \left\lbrack {r}_{n}\right\rbrack \in {}^{ * }\mathbb{R} \) with\n\n\[ \left\{ {n \in \mathbb{N} : {r}_{n} \in {A}_{n}}\right\} = \mathbb{N} \in \mathcal{F} \]\n\nso \( \left\lbrack {r}_{n}\right\rbrack \in \left\lbrack {A}_{n}\right\rbrack \) . Moreover, if \( \left\lbrack {s}_{n}\right\rbrack \in \left\lbrack {A}_{n}\right\rbrack \), then\n\n\[ \left\{ {n \in \mathbb{N} : {s}_{n} \in {A}_{n}}\right\} \in \mathcal{F}\;\text{and}\;\left\{ {n \in \mathbb{N} : {s}_{n} \in {A}_{n}}\right\} \subseteq \left\{ {n \in \mathbb{N} : {r}_{n} \leq {s}_{n}}\right\} ,\]\n\nleading to the conclusion \( \left\lbrack {r}_{n}\right\rbrack \leq \left\lbrack {s}_{n}\right\rbrack \) in * \( \mathbb{R} \) . Hence \( \left\lbrack {A}_{n}\right\rbrack \) indeed has a least member, namely the hyperreal number \( \left\lbrack {r}_{n}\right\rbrack \) determined by the sequence of least members of the sets \( {A}_{n} \) .\n\nWriting \ | Yes |
Theorem 11.3.2 (Internal Induction) If \( X \) is an internal subset of \( {}^{ * }\mathbb{N} \) that contains 1 and is closed under the successor function \( n \mapsto n + 1 \) , then \( X = {}^{ * }\mathbb{N} \) . | Proof. Let \( Y = {}^{ * }\mathbb{N} - X \) . Then \( Y \) is internal \( \left( {{11.2}\left( 1\right) }\right) \), so if it is nonempty, it has a least element \( n \) . Then \( n \neq 1 \), as \( 1 \in X \), so \( n - 1 \in * \mathbb{N} \) . But now \( n - 1 \notin Y \), as \( n \) is least in \( Y \), so \( n - 1 \in X \), and therefore \( n = \left( {n - 1}\right) + 1 \) is in \( X \) by closure under successor. This contradiction forces us to conclude that \( Y = \varnothing \), and so \( X = {}^{ * }\mathbb{N} \) . | Yes |
Theorem 11.4.1 Let \( X \) be an internal subset of \( {}^{ * }\mathbb{N} \) and \( k \in \mathbb{N} \) . If \( n \in X \) for all \( n \in \mathbb{N} \) with \( k \leq n \), then there is an unlimited \( K \in * \mathbb{N} \) with \( n \in X \) for all \( n \in {}^{ * }\mathbb{N} \) with \( k \leq n \leq K \) . | Proof. If all unlimited hypernaturals are in \( X \), then any unlimited \( K \in {}^{ * }\mathbb{N} \) will do. Otherwise there are unlimited hypernaturals not in \( X \) . If we can show that there is a least such unlimited number \( H \), then all unlimited numbers smaller than \( H \) will be in \( X \), giving the desired result.\n\nTo spell this out: if \( {}^{ * }\mathbb{N} - X \) has unlimited members, then these must be greater than \( k \), and so the set\n\n\[ Y = \left\{ {n \in {}^{ * }\mathbb{N} : k < n \in {}^{ * }\mathbb{N} - X}\right\} \]\n\nis nonempty. But \( Y \) is internal, by the algebra of internal sets, since it is equal to\n\n\[ \left( {*\mathbb{N}-\{ 1,\ldots, k\} }\right) \cap \left( {*\mathbb{N} - X}\right) \]\n\nHence \( Y \) has a least element \( H \) by the internal least number principle. Then \( H \) is a hypernatural that is greater than \( k \) but not in \( X \), so it must be the case that \( H \notin \mathbb{N} \), because of our hypothesis that all limited \( n \geq k \) are in \( X \) . Thus \( H \) is unlimited. Then \( K = H - 1 \) is unlimited and meets the requirements of the theorem: \( H \) is the least hypernatural greater than \( k \) that is not in \( X \), so every \( n \in {}^{ * }\mathbb{N} \) with \( k \leq n \leq H - 1 \) does belong to \( X \) . | Yes |
Theorem 11.5.1 If a nonempty internal subset of \( {}^{ * }\mathbb{R} \) is bounded above/ below, then it has a least upper/greatest lower bound in \( {}^{ * }\mathbb{R} \) . | Proof. We treat the case of upper bounds. In effect, the point of the proof is to show that the least upper bound of a bounded internal set \( \left\lbrack {A}_{n}\right\rbrack \) is the hyperreal number determined by the sequence of least upper bounds of the \( {A}_{n} \) ’s:\n\n\[ \operatorname{lub}\left\lbrack {A}_{n}\right\rbrack = \left\lbrack {\operatorname{lub}{A}_{n}}\right\rbrack \]\n\nMore precisely, it is enough to require that \( \mathcal{F} \) -almost all \( {A}_{n} \) ’s have least upper bounds to make this work.\n\nSuppose then that a nonempty internal set \( \left\lbrack {A}_{n}\right\rbrack \) has an upper bound \( \left\lbrack {r}_{n}\right\rbrack . \) Write \( {A}_{n} \leq x \) to mean that \( x \) is an upper bound of \( {A}_{n} \) in \( \mathbb{R} \), and put\n\n\[ J = \left\{ {n \in \mathbb{N} : {A}_{n} \leq {r}_{n}}\right\} \]\n\nWe want \( J \in \mathcal{F} \) . If not, then \( {J}^{c} \in \mathcal{F} \) . But if \( n \in {J}^{c} \), there exists some \( {a}_{n} \) with \( {r}_{n} < {a}_{n} \in {A}_{n} \) . This leads to the conclusion \( \left\lbrack {r}_{n}\right\rbrack < \left\lbrack {a}_{n}\right\rbrack \in \left\lbrack {A}_{n}\right\rbrack \) , contradicting the fact that \( \left\lbrack {r}_{n}\right\rbrack \) is an upper bound of \( \left\lbrack {A}_{n}\right\rbrack \) .\n\nIt follows that \( J \in \mathcal{F} \) . Since \( \left\lbrack {A}_{n}\right\rbrack \neq \varnothing \), this then implies\n\n\[ {J}^{\prime } = \left\{ {n \in \mathbb{N} : \varnothing \neq {A}_{n} \leq {r}_{n}}\right\} \in \mathcal{F}. \]\n\nNow, if \( n \in {J}^{\prime } \), then \( {A}_{n} \) is a nonempty subset of \( \mathbb{R} \) bounded above (by \( {r}_{n} \) ), and so by the order-completeness of \( \mathbb{R},{A}_{n} \) has a least upper bound \( {s}_{n} \in \mathbb{R} \) . Then if \( \left\lbrack {b}_{n}\right\rbrack \in \left\lbrack {A}_{n}\right\rbrack \) ,\n\n\[ \left\{ {n \in \mathbb{N} : {b}_{n} \in {A}_{n}}\right\} \cap {J}^{\prime } \subseteq \left\{ {n \in \mathbb{N} : {b}_{n} \leq {s}_{n}}\right\} ,\]\n\nleading to \( \left\lbrack {b}_{n}\right\rbrack \leq \left\lbrack {s}_{n}\right\rbrack \), and showing that \( \left\lbrack {s}_{n}\right\rbrack \) is an upper bound of \( \left\lbrack {A}_{n}\right\rbrack \) . Finally, if \( \left\lbrack {t}_{n}\right\rbrack \) is any other upper bound of \( \left\lbrack {A}_{n}\right\rbrack \), then \( \left\{ {n : {A}_{n} \leq {t}_{n}}\right\} \in \mathcal{F} \) by the same argument as for \( \left\lbrack {r}_{n}\right\rbrack \), and\n\n\[ \left\{ {n \in \mathbb{N} : {A}_{n} \leq {t}_{n}}\right\} \cap {J}^{\prime } \subseteq \left\{ {n \in \mathbb{N} : {s}_{n} \leq {t}_{n}}\right\} \]\n\nso we get \( \left\lbrack {s}_{n}\right\rbrack \leq \left\lbrack {t}_{n}\right\rbrack \) . This shows that \( \left\lbrack {s}_{n}\right\rbrack \) is indeed the least upper bound of \( \left\lbrack {A}_{n}\right\rbrack \) in \( {}^{ * }\mathbb{R} \) . | Yes |
Theorem 11.8.1 Let \( X \) be an internal subset of \( {}^{ * }\mathbb{N} \), and let \( K \in {}^{ * }\mathbb{N} \) be unlimited. If every unlimited hypernatural \( H \leq K \) belongs to \( X \), then there is some \( k \in \mathbb{N} \) such that every limited \( n \) with \( k \leq n \) belongs to \( X \) . | Proof. For \( M, N \in * \mathbb{N} \) with \( M \leq N \), let\n\n\[ \lfloor M, N\rfloor = \left\{ {z \in {}^{ * }\mathbb{N} : M \leq z \leq N}\right\} \]\n\nbe the interval in \( {}^{ * }\mathbb{N} \) between \( M \) and \( N \) . Our hypothesis is that \( \lfloor H, K\rfloor \subseteq X \) for all unlimited hypernatural \( H \leq K \) . What we want to show is that \( \lfloor k, K\rfloor \subseteq X \) for some \( k \in \mathbb{N} \) . To put this more symbolically, we want to show that the set\n\n\[ Y = \left\{ {k \in {}^{ * }\mathbb{N} : \lfloor k, K\rfloor \subseteq X}\right\} \]\n\nhas a limited member.\n\nNow, if \( Y \) is internal, then by the internal least number principle it has a least element \( k \), and such a \( k \) must belong to \( \mathbb{N} \), because if it were unlimited, then \( k - 1 \) would be unlimited, so by our hypothesis \( k - 1 \) would also be in \( Y \) but less than \( k \) .\n\nIt thus suffices to show that \( Y \) is internal. But if \( \varphi \left( {x, y, A}\right) \) is the formula\n\n\[ x \in \mathbb{N} \land x \leq y \land \forall z \in \mathbb{N}\left( {x \leq z \leq y \rightarrow z \in A}\right) ,\]\nexpressing \ | Yes |
Theorem 11.9.1 If \( X \) is an internal subset of \( {}^{ * }\mathbb{R} \) that contains all points that are infinitely close to \( b \in {}^{ * }\mathbb{R} \), then there is a positive real \( \varepsilon \) such that \( X \) contains all points that are within \( \varepsilon \) of \( b \) . | Proof. Our hypothesis is that \( \operatorname{hal}\left( b\right) \subseteq X \) . For \( k \in * \mathbb{N} \), let \( \left( {b - \frac{1}{k}, b + \frac{1}{k}}\right) \) be the hyperreal interval\n\n\[ \left\{ {z \in {}^{ * }\mathbb{R} : \left| {z - b}\right| < \frac{1}{k}}\right\} . \]\n\nNow,\n\n\[ \left( {b - \frac{1}{k}, b + \frac{1}{k}}\right) \subseteq X \]\n\nwhenever \( k \) is unlimited, because in this case \( \frac{1}{k} \) is infinitesimal, and so\n\n\[ \left( {b - \frac{1}{k}, b + \frac{1}{k}}\right) \subseteq \operatorname{hal}\left( b\right) \subseteq X \]\n\nby our hypothesis. Thus the set\n\n\[ Y = \left\{ {k \in {}^{ * }\mathbb{N} : \left( {b - \frac{1}{k}, b + \frac{1}{k}}\right) \subseteq X}\right\} \]\n\ncontains all unlimited members of \( {}^{ * }\mathbb{N} \) . Hence by underflow we could conclude that \( \left( {b - \frac{1}{k}, b + \frac{1}{k}}\right) \subseteq X \) for some \( k \in \mathbb{N} \), and thereby complete the proof by putting \( \varepsilon = \frac{1}{k} \), provided that \( Y \) is internal. But applying internal set definition with \( \varphi \left( {x, y, A}\right) \) as the formula\n\n\[ x \in \mathbb{N} \land \left( {\forall z \in \mathbb{R}}\right) \left( {\left| {z - y}\right| < \frac{1}{x} \rightarrow z \in A}\right) \]\n\n(expressing \ | Yes |
Theorem 11.10.1 The intersection of a decreasing sequence\n\n\\[ \n{X}^{1} \supseteq {X}^{2} \supseteq \cdots \supseteq {X}^{k} \supseteq \cdots\n\\]\n\nof nonempty internal sets is always nonempty :\n\n\\[ \n\mathop{\\bigcap }\\limits_{{k \\in \\mathbb{N}}}{X}^{k} \\neq \\varnothing .\n\\] | Proof. This is a delicate analysis of the ultrapower construction, involving a kind of diagonalisation argument, that is not easy to motivate intuitively.\n\nFor each \\( k \\in \\mathbb{N} \\), let \\( {X}^{k} = \\left\\lbrack {A}_{n}^{k}\\right\\rbrack \\), so that \\( {X}^{k} \\) is the internal set defined by the sequence \\( \\left\\langle {{A}_{n}^{k} : n \\in \\mathbb{N}}\\right\\rangle \\) of subsets of \\( \\mathbb{R} \\) . Then by Section 11.2 the sets \\( \\left\{ {n \\in \\mathbb{N} : {A}_{n}^{k} \\neq \\varnothing }\\right\} \\) and \\( \\left\{ {n \\in \\mathbb{N} : {A}_{n}^{k} \\supseteq {A}_{n}^{k + 1}}\\right\} \\) belong to \\( \\mathcal{F} \\) . Hence if\n\n\\[ \n{J}^{k} = \\left\{ {n \\in \\mathbb{N} : {A}_{n}^{1} \\supseteq \\cdots \\supseteq {A}_{n}^{k} \\neq \\varnothing }\\right\}\n\\]\nthen by closure of \\( \\mathcal{F} \\) under finite intersections it follows for each \\( k \\in \\mathbb{N} \\) that \\( {J}^{k} \\in \\mathcal{F} \\) . Note that \\( {J}^{1} \\supseteq {J}^{2} \\supseteq \\cdots \\) .\n\nWe want to define a hyperreal \\( \\left\\lbrack {s}_{n}\\right\\rbrack \\) that belongs to every \\( {X}^{k} \\) . This will require that for each \\( k \\) we have \\( {s}_{n} \\in {A}_{n}^{k} \\) for \\( \\mathcal{F} \\) -almost all \\( n \\) . We will arrange this to work for almost all \\( n \\geq k \\), in the sense that\n\n\\[ \n\\{ n \\in \\mathbb{N} : k \\leq n\\} \\cap {J}^{k} \\subseteq \\left\{ {n \\in \\mathbb{N} : {s}_{n} \\in {A}_{n}^{k}}\\right\} .\n\\]\n\n(iii)\n\nBut the set \\( \\{ n \\in \\mathbb{N} : k \\leq n\\} \\) is cofinite in \\( \\mathbb{N} \\), and so belongs to the nonprincipal ultrafilter \\( \\mathcal{F} \\) . Since also \\( {J}^{k} \\in \\mathcal{F} \\) ,(iii) then yields \\( \\{ n \\in \\mathbb{N} \\) : \\( \\left. {{s}_{n} \\in {A}_{n}^{k}}\\right\\} \\in \\mathcal{F} \\), and therefore \\( \\left\\lbrack {s}_{n}\\right\\rbrack \\in {X}^{k} \\) as desired.\n\nIt thus remains to define \\( {s}_{n} \\) fulfilling (iii). For \\( n \\in {J}^{1} \\) let\n\n\\[ \n{k}_{n} = \\max \\left\{ {k : k \\leq n\\text{ and }n \\in {J}^{k}}\\right\}\n\\]\n\n(iv)\n\nThen \\( n \\in {J}^{{k}_{n}} \\), so by the definition of \\( {J}^{{k}_{n}} \\) we can choose some \\( {s}_{n} \\in {A}_{n}^{{k}_{n}} \\) , and hence\n\n\\[ \n{s}_{n} \\in {A}_{n}^{1} \\cap \\cdots \\cap {A}_{n}^{{k}_{n}}\n\\]\n\n(v)\n\nFor \\( n \\notin {J}^{1} \\) let \\( {s}_{n} \\) be arbitrary. Now, to prove (iii), observe that if \\( k \\leq n \\) and \\( n \\in {J}^{k} \\), then by (iv), \\( k \\leq {k}_{n} \\), and so by (v), \\( {s}_{n} \\in {A}_{n}^{k} \\) . | Yes |
Corollary 11.10.2 If \( \left\{ {{X}_{n} : n \in \mathbb{N}}\right\} \) is a collection of internal sets and \( X \) is internal, then:\n\n(1) \( \mathop{\bigcap }\limits_{{n \in \mathbb{N}}}{X}_{n} \neq \varnothing \) if \( \left\{ {{X}_{n} : n \in \mathbb{N}}\right\} \) has the finite intersection property. | (1) Let \( {Y}^{k} = {X}_{1} \cap \cdots \cap {X}_{k} \) . Then \( {Y}^{1} \supseteq {Y}^{2} \supseteq \cdots \), and each \( {Y}^{k} \) is internal by 11.2(1). The finite intersection property implies that \( {Y}^{k} \neq \varnothing \), so by the above theorem there is some hyperreal that belongs to every \( {Y}^{k} \), and hence to every \( {X}_{k} \) . | Yes |
Theorem 11.13.1 If \( X \) is internal, then \( \operatorname{sh}\left( X\right) \) is closed. | Proof. Let \( r \in \mathbb{R} \) be a closure point of \( \operatorname{sh}\left( X\right) \) . We need to show that \( r \in \operatorname{sh}\left( X\right) \), i.e., \( r \) is the shadow of some \( y \in X \) .\n\nNow, for each \( n \in \mathbb{N} \), the hyperreal open interval \( \left( {r - \frac{1}{n}, r + \frac{1}{n}}\right) \) meets \( \operatorname{sh}\left( X\right) \) in some real point \( {s}_{n} \) that must be the shadow of some \( {x}_{n} \in X \) . Hence \( {x}_{n} \simeq {s}_{n} \in \left( {r - \frac{1}{n}, r + \frac{1}{n}}\right) \), so the internal set\n\n\[ \begin{array}{l} {X}_{n} = X \cap \left( {r - \frac{1}{n}, r + \frac{1}{n}}\right) \end{array} \]\n\ncontains \( {x}_{n} \) and is thereby nonempty . The \( {X}_{n} \) ’s form a decreasing sequence, so by countable saturation there is a point \( y \) in their intersection. Then \( y \in X \) and \( \left| {y - r}\right| < \frac{1}{n} \) for all \( n \in \mathbb{N} \), so \( y \simeq r \) . Hence \( r = \operatorname{sh}\left( y\right) \in \operatorname{sh}\left( X\right) . \)\n\nThis shows that \( \operatorname{sh}\left( X\right) \) contains all its closure points and so is closed. | Yes |
Lemma 11.14.1 Every hyper-open set is a union of hyperreal open intervals. | Proof. Let \( A = \left\lbrack {A}_{n}\right\rbrack \) be hyper-open. Take a point \( r = \left\lbrack {r}_{n}\right\rbrack \) in \( A \) . Then we find that the set\n\n\[ J = \left\{ {n \in \mathbb{N} : {r}_{n} \in {A}_{n}\text{ and }{A}_{n}\text{ is open in }\mathbb{R}}\right\} \]\n\nbelongs to the ultrafilter \( \mathcal{F} \) . Our task is to show that \( r \) belongs to some hyperreal interval \( \left( {a, b}\right) \) that is included in \( A \) .\n\nNow, if \( n \in J \), then there is some real interval \( \left( {{a}_{n},{b}_{n}}\right) \subseteq \mathbb{R} \) with\n\n\[ {r}_{n} \in \left( {{a}_{n},{b}_{n}}\right) \subseteq {A}_{n} \]\n\nSince \( J \in \mathcal{F} \), this is enough to specify \( a \) as the hyperreal number \( \left\lbrack {a}_{n}\right\rbrack \) and \( b \) as \( \left\lbrack {b}_{n}\right\rbrack \) . Working with the properties of \( \mathcal{F} \), in a now familiar way, we can then show that \( \left\lbrack {a}_{n}\right\rbrack < \left\lbrack {r}_{n}\right\rbrack < \left\lbrack {b}_{n}\right\rbrack \), and also that \( \left\lbrack {s}_{n}\right\rbrack \in \left\lbrack {A}_{n}\right\rbrack \) whenever \( \left\lbrack {a}_{n}\right\rbrack < \left\lbrack {s}_{n}\right\rbrack < \left\lbrack {b}_{n}\right\rbrack \), so that\n\n\[ r \in \left( {a, b}\right) \subseteq A \]\n\nas desired. | Yes |
Theorem 11.14.4 If \( B \) is an internal set, then \( B \) is \( S \) -open if and only if it contains the halo of each of its points. | Proof. We have already observed that an S-open set is a union of halos.\n\nConversely, assume that \( \operatorname{hal}\left( r\right) \subseteq B \) whenever \( r \in B \) . For such an \( r \) , consider the set\n\n\[ X = \left\{ {n \in {}^{ * }\mathbb{N} : \left( {\forall x \in {}^{ * }\mathbb{R}}\right) \left( {\left| {r - x}\right| < \frac{1}{n} \rightarrow x \in B}\right) }\right\} . \]\n\nSince \( B \) is internal, it follows by the internal set definition principle that \( X \) is internal. Moreover, since \( \operatorname{hal}\left( r\right) \subseteq B \), it follows that \( \mathrm{X} \) contains every unlimited hypernatural \( n \), because for such an \( n,\left| {r - x}\right| < \frac{1}{n} \) implies \( x \in \) hal \( \left( r\right) \) . Hence by underflow, \( X \) must contain some standard \( n \in \mathbb{N} \), so \( B \) includes the real-radius interval \( \left( {r - \frac{1}{n}, r + \frac{1}{n}}\right) \) . But then since \( \frac{1}{n} \) is real,\n\n\[ r \in \left( \left( {r - \frac{1}{n}, r + \frac{1}{n}}\right) \right) \subseteq \left( {r - \frac{1}{n}, r + \frac{1}{n}}\right) \subseteq B. \]\n\nThis shows that \( B \) is the union of S-neighbourhoods, and is thereby S-open. | Yes |
Theorem 12.1.1 Let \( \left\langle {{f}_{n} : n \in \mathbb{N}}\right\rangle \) and \( \left\langle {{g}_{n} : n \in \mathbb{N}}\right\rangle \) be sequences of partial functions from \( \mathbb{R} \) to \( \mathbb{R} \) . Then the internal functions \( \left\lbrack {f}_{n}\right\rbrack \) and \( \left\lbrack {g}_{n}\right\rbrack \) are equal if and only if\n\n\[ \left\{ {n \in \mathbb{N} : {f}_{n} = {g}_{n}}\right\} \in \mathcal{F}. \]\n | Proof. Let \( {J}_{fg} = \left\{ {n \in \mathbb{N} : {f}_{n} = {g}_{n}}\right\} \), and suppose \( {J}_{fg} \in \mathcal{F} \) . Now in general, two functions are equal precisely when they have the same domain and assign the same values to all members of that domain. Thus\n\n\[ {J}_{fg} \subseteq \left\{ {n \in \mathbb{N} : \operatorname{dom}{f}_{n} = \operatorname{dom}{g}_{n}}\right\} \]\n\nleading by \( {11.2}\left( 3\right) \) to the conclusion that the internal sets \( \left\lbrack {\operatorname{dom}{f}_{n}}\right\rbrack \) and \( \left\lbrack {\operatorname{dom}{g}_{n}}\right\rbrack \) are equal, i.e., \( \operatorname{dom}\left\lbrack {f}_{n}\right\rbrack = \operatorname{dom}\left\lbrack {g}_{n}\right\rbrack \) . But for \( \left\lbrack {r}_{n}\right\rbrack \in \operatorname{dom}\left\lbrack {f}_{n}\right\rbrack \) ,\n\n\[ {J}_{fg} \cap \left\{ {n \in \mathbb{N} : {r}_{n} \in \operatorname{dom}{f}_{n}}\right\} \subseteq \left\{ {n \in \mathbb{N} : {f}_{n}\left( {r}_{n}\right) = {g}_{n}\left( {r}_{n}\right) }\right\} \]\n\nwhich leads to \( \left\lbrack {f}_{n}\right\rbrack \left( \left\lbrack {r}_{n}\right\rbrack \right) = \left\lbrack {g}_{n}\right\rbrack \left( \left\lbrack {r}_{n}\right\rbrack \right) \) . Hence \( \left\lbrack {f}_{n}\right\rbrack = \left\lbrack {g}_{n}\right\rbrack \) .\n\nFor the converse, suppose that \( {J}_{fg} \notin \mathcal{F} \) . Now, \( {J}_{fg}^{c} \) is a subset of the union\n\n\[ \left\{ {n \in \mathbb{N} : \operatorname{dom}{f}_{n} \neq \operatorname{dom}{g}_{n}}\right\} \cup \left\{ {n \in \mathbb{N} : \operatorname{dom}{f}_{n} = \operatorname{dom}{g}_{n}\text{ but }{f}_{n} \neq {g}_{n}}\right\} \]\n\nso either \( \left\{ {n : \operatorname{dom}{f}_{n} \neq \operatorname{dom}{g}_{n}}\right\} \in \mathcal{F} \), whence \( \operatorname{dom}\left\lbrack {f}_{n}\right\rbrack \neq \operatorname{dom}\left\lbrack {g}_{n}\right\rbrack \) and so \( \left\lbrack {f}_{n}\right\rbrack \neq \left\lbrack {g}_{n}\right\rbrack \), or else\n\n\[ J = \left\{ {n \in \mathbb{N} : \operatorname{dom}{f}_{n} = \operatorname{dom}{g}_{n}\text{ but }{f}_{n} \neq {g}_{n}}\right\} \in \mathcal{F}. \]\n\nBut for \( n \in J \) there exists some \( {r}_{n} \) with \( {f}_{n}\left( {r}_{n}\right) \neq {g}_{n}\left( {r}_{n}\right) \) . This leads to \( \left\lbrack {f}_{n}\right\rbrack \left( \left\lbrack {r}_{n}\right\rbrack \right) \neq \left\lbrack {g}_{n}\right\rbrack \left( \left\lbrack {r}_{n}\right\rbrack \right) \), and so \( \left\lbrack {f}_{n}\right\rbrack \neq \left\lbrack {g}_{n}\right\rbrack . | Yes |
Theorem 12.5.1 An internal set \( A \) is hyperfinite with internal cardinality \( N \) if and only if there is an internal bijection \( f : \{ 1,\ldots, N\} \rightarrow A \) . | Proof. Let \( A = \left\lbrack {A}_{n}\right\rbrack \) . If \( A \) is hyperfinite with internal cardinality \( N = \) \( \left\lbrack {N}_{n}\right\rbrack \), then we may suppose that for each \( n \in \mathbb{N},{A}_{n} \) is a finite set of cardinality \( {N}_{n} \) . Thus there is a bijection \( {f}_{n} : \left\{ {1,\ldots ,{N}_{n}}\right\} \rightarrow {A}_{n} \) . Let \( f = \) \( \left\lbrack {f}_{n}\right\rbrack \) . Then \( f \) is an internal function with domain \( \{ 1,\ldots, N\} \) that is injective \( \left( {{12.2}\left( 4\right) }\right) \) and has range \( A\left( {{12.2}\left( 1\right) }\right) \) .\n\nConversely, suppose that \( f = \left\lbrack {f}_{n}\right\rbrack \) is an internal bijection from \( \{ 1,\ldots, N\} \) onto \( A \) . Then\n\n\[ \left\lbrack {\operatorname{dom}{f}_{n}}\right\rbrack = \operatorname{dom}\left\lbrack {f}_{n}\right\rbrack = \{ 1,\ldots, N\} = \left\lbrack \left\{ {1,\ldots ,{N}_{n}}\right\} \right\rbrack ,\]\n\nso for \( \mathcal{F} \) -almost all \( n \) ,\n\n\[ \operatorname{dom}{f}_{n} = \left\{ {1,\ldots ,{N}_{n}}\right\} \]\n\n(i)\n\nAlso, as \( A \) is the image of \( \{ 1,\ldots, N\} \) under \( \left\lbrack {f}_{n}\right\rbrack \), Exercise 12.2(1) implies that \( A = \left\lbrack {{f}_{n}\left( \left\{ {1,\ldots ,{N}_{n}}\right\} \right) }\right\rbrack \), so\n\n\[ {f}_{n}\left( \left\{ {1,\ldots ,{N}_{n}}\right\} \right) = {A}_{n} \]\n\n(ii)\n\nfor \( \mathcal{F} \) -almost all \( n \) . Finally, by \( {12.2}\left( 4\right) \),\n\n\[ {f}_{n}\text{is injective} \]\n\n(iii)\n\nfor \( \mathcal{F} \) -almost all \( n \) . Then the set \( J \) of those \( n \in \mathbb{N} \) satisfying (i)-(iii) must belong to \( \mathcal{F} \) . But for \( n \in J,{A}_{n} \) is finite of cardinality \( {N}_{n} \) . Hence \( A \) is hyperfinite of internal cardinality \( N \) . | No |
Theorem 12.6.1 An internal set \( A = \left\lbrack {A}_{n}\right\rbrack \) is hyperfinite if and only if every injective internal function \( f \) whose domain includes \( A \) and has \( f\left( A\right) \subseteq A \) must in fact have \( f\left( A\right) = A \) . | Proof. Suppose \( A \) is hyperfinite. Let \( f = \left\lbrack {f}_{n}\right\rbrack \) be an internal injective function with \( A \subseteq \operatorname{dom}f \) and \( f\left( A\right) \subseteq A \) . Then each of the following is true for \( \mathcal{F} \) -almost all \( n \in \mathbb{N} \) :\n\n\[ \n{A}_{n}\text{is finite,}\n\]\n\n\[ \n{A}_{n} \subseteq \operatorname{dom}{f}_{n}\n\]\n\n\[ \n{f}_{n}\left( {A}_{n}\right) \subseteq {A}_{n},\;\text{ cf. }{12.2}\left( 1\right) ,{11.2}\left( 2\right)\n\]\n\n\( {f}_{n} \) is injective.\n\nThus the set \( J \) of those \( n \in \mathbb{N} \) satisfying all of these conditions must belong to \( \mathcal{F} \) . But\n\n\[ \nJ \subseteq \left\{ {n \in \mathbb{N} : {f}_{n}\left( {A}_{n}\right) = {A}_{n}}\right\}\n\]\n\nby the standard pigeonhole principle, and so \( f\left( A\right) = \left\lbrack {{f}_{n}\left( {A}_{n}\right) }\right\rbrack = \left\lbrack {A}_{n}\right\rbrack = A \) .\n\nFor the converse, suppose \( A \) is not hyperfinite. It follows that\n\n\[ \n{J}^{\prime } = \left\{ {n \in \mathbb{N} : {A}_{n}\text{ is infinite }}\right\} \in \mathcal{F}.\n\]\n\nBut for each \( n \in {J}^{\prime } \) there is an injective function \( {f}_{n} : {A}_{n} \rightarrow {A}_{n} \) and some \( {r}_{n} \in {A}_{n} - {f}_{n}\left( {A}_{n}\right) \) . Let \( f = \left\lbrack {f}_{n}\right\rbrack \) . This makes \( f \) an internal function with domain \( A \) that is injective \( \left( {{12.2}\left( 4\right) }\right) \) and has \( f\left( A\right) = \left\lbrack {{f}_{n}\left( {A}_{n}\right) }\right\rbrack \subseteq A \), while \( \left\lbrack {r}_{n}\right\rbrack \in A - f\left( A\right) \) and so \( f\left( A\right) \neq A \) . | Yes |
Any internal set belongs to a standard set that is transitive. Hence * \( \mathbb{U} \) is strongly transitive, and in particular, every member of an internal set is internal. | Proof. Let \( A \) be internal, with \( A \in {}^{ * }B \) for some \( B \in \mathbb{U} \) . By strong transitivity of \( \mathbb{U} \) there is a transitive \( T \in \mathbb{U} \) with \( B \subseteq T \) . But as we have seen, transitivity is preserved by the transfer map, so the standard set * \( T \in \) * \( \mathbb{U} \) is transitive. Also, \( {}^{ * }B \subseteq {}^{ * }T \), so \( A \in {}^{ * }T \), establishing the first part of the theorem. But then \( A \subseteq {}^{ * }T \), and every member of \( {}^{ * }T \) is internal by definition, so belongs to * \( \mathbb{U} \) . Thus \( A \subseteq {}^{ * }T \subseteq {}^{ * }\mathbb{U} \), completing the proof that * \( \mathbb{U} \) is strongly transitive.\n\nThe assertion that every member of an internal set is internal is now just the statement that \( {}^{ * }\mathbb{U} \) is transitive. | Yes |
Theorem 13.12.1 * \( \mathcal{P}\left( A\right) \) is the set of all internal subsets of \( {}^{ * }A \), i.e., \[ {}^{ * }\mathcal{P}\left( A\right) = \mathcal{P}\left( {{}^{ * }A}\right) \cap {}^{ * }\mathbb{U} = \{ B \subseteq {}^{ * }A : B\text{ is internal }\} . \] | Proof. We have \( {}^{ * }\mathcal{P}\left( A\right) \subseteq \mathcal{P}\left( {{}^{ * }A}\right) \) in general. Moreover, if \( B \in {}^{ * }\mathcal{P}\left( A\right) \), then \( B \) belongs to a standard entity, so \( B \) is internal.\n\nFor the converse, let \( B \) be an internal subset of \( {}^{ * }A \) . Then \( B \in {}^{ * }C \) for some \( C \in \mathbb{U} \) . Now, the sentence \[ \left( {\forall x \in C}\right) \left\lbrack {\left( {\forall y \in x}\right) \left( {y \in A}\right) \rightarrow x \in \mathcal{P}\left( A\right) }\right\rbrack \] is true, since it asserts of any \( x \in C \) that if \( x \subseteq A \), then \( x \) belongs to \( \mathcal{P}\left( A\right) \) . By transfer it follows that for any \( x \in {}^{ * }C \), if \( x \subseteq {}^{ * }A \), then \( x \) belongs to * \( \mathcal{P}\left( A\right) \) . But \( B \in {}^{ * }C \) and \( B \subseteq {}^{ * }A \), i.e., \[ \left( {\forall y \in B}\right) \left( {y \in {}^{ * }A}\right) \] so we get \( B \in {}^{ * }\mathcal{P}\left( A\right) \) as desired. | Yes |
Theorem 13.14.1 The image \( {}^{\mathrm{{im}}}A \) of any infinite set \( A \in \mathbb{U} \) is external. | Proof. The method of proof was hinted at in Exercise 12.2(8), which is itself the special case in which \( A \) is an infinite subset of \( \mathbb{R} \) (cf. also Section 11.7).\n\nIn general, if \( A \) is infinite, then there is an injection \( f : \mathbb{N} \rightarrow A \) . Put \( X = \{ f\left( n\right) : n \in \mathbb{N}\} \subseteq A \) . Then * \( X \) is internal (indeed standard), and so if \( {}^{\mathrm{{im}}}A \) were internal, then so too would be \( {}^{\mathrm{{im}}}A \cap {}^{ * }X \) (Section 13.11). Observe that\n\n\[
{}^{\mathrm{{im}}}A \cap {}^{ * }X = \left\{ {{}^{ * }a : a \in A\text{ and }{}^{ * }a \in {}^{ * }X}\right\} = \left\{ {{}^{ * }a : a \in X}\right\} = {}^{\mathrm{{im}}}X.
\]\n\nSince the transform \( {}^{ * }f : {}^{ * }\mathbb{N} \rightarrow {}^{ * }A \) is internal, this would then imply that the inverse image of \( {}^{\mathrm{{im}}}X \) under \( {}^{ * }f \) is internal (Section 13.11). But by transfer \( {}^{ * }f \) is injective, and from this it can be shown that \( {}^{ * }{f}^{-1}\left( {{}^{\mathrm{{im}}}X}\right) \) is equal to the external set \( \mathbb{N} \) .\n\nTherefore \( {}^{\mathrm{{im}}}A \) cannot be internal. | Yes |
Theorem 14.2.1 If \( \mathbb{U}\overset{ * }{ \rightarrow }{\mathbb{U}}^{\prime } \) is a universe embedding, then the following are equivalent.\n\n(1) \( {\mathbb{U}}^{\prime } \) is an enlargement of \( \mathbb{U} \) relative to \( \overset{ * }{ \rightarrow } \).\n\n(2) For any concurrent relation \( R \in \mathbb{U} \) there exists an entity \( b \in {\mathbb{U}}^{\prime } \) such that \( {}^{ * }x\left( {{}^{ * }R}\right) b \) for all \( x \in \operatorname{dom}R \).\n\n(3) For each set \( A \in \mathbb{U} \) there exists a hyperfinite subset \( B \) of \( {}^{ * }A \) that contains all the standard entities of \( {}^{ * }A \):\n\n\[ {}^{\mathrm{{im}}}A = \left\{ {{}^{ * }a : a \in A}\right\} \subseteq B \in {}^{ * }{\mathcal{P}}_{F}\left( A\right) . \] | Proof. First assume (1). If \( R \) is a binary relation in \( \mathbb{U} \), for each \( x \in \operatorname{dom}R \) let \( R\left\lbrack x\right\rbrack = \{ y \in \operatorname{ran}R : {xRy}\} \) . Then if \( R \) is concurrent, the collection\n\n\[ \{ R\left\lbrack x\right\rbrack : x \in \operatorname{dom}R\} \]\n\nhas the finite intersection property. Also, this collection is a subset of the power set \( \mathcal{P}\left( {\operatorname{ran}R}\right) \), so belongs to \( \mathbb{U} \) . Hence by (1) there is a \( b \in {\mathbb{U}}^{\prime } \) that belongs to every \( {}^{ * }\left( {R\left\lbrack x\right\rbrack }\right) \) . By transfer of \( \left( {\forall y \in R\left\lbrack x\right\rbrack }\right) \left( {xRy}\right) \) we then have \( {}^{ * }x\left( {{}^{ * }R}\right) b \) for all \( x \in \operatorname{dom}R \), establishing (2).\n\nThe proof that (2) implies (3) was indicated in the discussion of the last example above.\n\nTo show that (3) implies (1), let \( A \in \mathbb{U} \) be a collection of sets with the finite intersection property. Take a transitive \( T \in \mathbb{U} \) with \( A \subseteq T \) . Then the sentence\n\n\[ \left( {\forall Z \in {\mathcal{P}}_{F}\left( A\right) }\right) \left( {\exists y \in T}\right) \left( {\forall z \in Z}\right) \left( {y \in z}\right) \]\n\nis true. But assuming (3), there exists a hyperfinite \( B \subseteq {}^{ * }A \) containing \( {}^{\mathrm{{im}}}A \) . Then by transfer of this sentence, since \( B \in {}^{ * }{\mathcal{P}}_{F}\left( A\right) \), there exists some \( b \in * T \) with \( b \in z \) for all \( z \in B \), and hence \( b \in * C \) for all \( C \in A \) . Therefore (1) holds. | Yes |
Theorem 15.1.1 Let \( \varphi \left( x\right) \) be an internal \( {\mathcal{L}}_{{\mathbb{U}}^{\prime }} \) -formula with only the variable \( x \) free. Then\n\n(1) (Overflow) If there exists \( k \in \mathbb{N} \) such that \( \varphi \left( n\right) \) is true for all \( n \in \mathbb{N} \) with \( k \leq n \), then there exists \( K \in * \mathbb{N} - \mathbb{N} \) such that \( \varphi \left( n\right) \) is true for all \( n \in * \mathbb{N} \) with \( k \leq n \leq K \) . | (1) We adapt the proof of Theorem 11.4.1. Formula \( \left( {k < x}\right) \land \neg \varphi \left( x\right) \) is internal, and \( {}^{ * }\mathbb{N} \) is internal, so by the version in Section 13.15 of the internal set definition principle,\n\n\[ Y = \left\{ {n \in {}^{ * }\mathbb{N} : k < n\text{ and not }\varphi \left( n\right) }\right\} \]\n\nis an internal set. Now, if \( \varphi \left( n\right) \) is true for all \( * \mathbb{N} - \mathbb{N} \), then any \( K \in {}^{ * }\mathbb{N} \) – \( \mathbb{N} \) gives the desired result (remember we assume * \( \mathbb{N} \) – \( \mathbb{N} \neq \varnothing \) ). Otherwise \( Y \) is nonempty, and so by the internal least number principle has a least element \( H \) . Then \( H \) is unlimited and \( K = H - 1 \) gives the result. | Yes |
Lemma 15.2.1 If \( s : {}^{ * }\mathbb{N} \rightarrow {}^{ * }\mathbb{R} \) is an internal hypersequence such that \( {s}_{n} \) is infinitesimal for all standard \( n \in \mathbb{N} \), then there is an unlimited \( K \in {}^{ * }\mathbb{N} \) such that \( {s}_{n} \) is infinitesimal for all hypernatural \( n \leq K \) . | Proof. We cannot just apply overflow to \( \left\{ {n \in {}^{ * }\mathbb{N} : {s}_{n} \simeq 0}\right\} \), since we do not know that this set is internal. Instead we use \( \left\{ {n \in {}^{ * }\mathbb{N} : \left| {s}_{n}\right| < \frac{1}{n}}\right\} \) . Because \( s \) is internal, the formula\n\n\[ \left| {x \cdot s\left( x\right) }\right| < 1 \]\n\nis internal, and for each \( n \in \mathbb{N} \), since \( {s}_{n} \simeq 0 \) we have \( n \cdot {s}_{n} \simeq 0 \), and so \( \left| {n \cdot {s}_{n}}\right| < 1 \) . Hence by overflow there is a \( K \in {\mathbb{N}}_{\infty } \) such that if \( n \leq K \), then \( \left| {n \cdot {s}_{n}}\right| < 1 \) and so \( \left| {s}_{n}\right| < \frac{1}{n} \) . But when such an \( n \) is unlimited, \( \frac{1}{n} \simeq 0 \) and so \( {s}_{n} \simeq 0 \) . | Yes |
Theorem 15.2.3 If \( f \) is an internal \( {}^{ * }\mathbb{R} \) -valued function and \( f\left( x\right) \) is infinitesimal for all limited hyperreals \( x \), then there is an unlimited \( b \) such that \( f\left( x\right) \) is infinitesimal for all \( x \in \left\lbrack {-b, b}\right\rbrack \subseteq * \mathbb{R} \) . | Proof. As for the proof of Lemma 15.2.1, but using the internal formula \( \left| {x \cdot f\left( x\right) }\right| < 1 \) and the fact that if an internal set includes \( \mathbb{L} \) then it includes \( \left\lbrack {-b, b}\right\rbrack \) for some unlimited \( b \) (Section 11.6). | No |
Theorem 15.4.3 In a sequentially comprehensive enlargement, if \( X \) is any countable set of unlimited hypernaturals, then there is an unlimited hypernatural \( K \) less than every member of \( X \) . | Proof. Write \( X = \left\langle {{s}_{n} : n \in \mathbb{N}}\right\rangle \) and extend this to an internal hyperse-quence \( \left\langle {{s}_{n} : n \in * \mathbb{N}}\right\rangle \) by sequential comprehensiveness. Put\n\n\[ Y = \left\{ {k \in {}^{ * }\mathbb{N} : \left( {\forall n \in {}^{ * }\mathbb{N}}\right) \left( {n \leq k \rightarrow k < {s}_{n}}\right) }\right\} .\n\]\n\nThen \( Y \) is internal, as its defining formula is internal. Moreover, \( Y \) contains each standard \( k \in \mathbb{N} \), since in that case \( {s}_{1},\ldots ,{s}_{k} \) are all in \( X \), hence unlimited and so greater than \( k \) . By overflow, then, there is some unlimited \( K \) in \( Y \) . Then \( K < {s}_{n} \) whenever \( n \leq K \), which includes all \( n \in \mathbb{N} \) . Hence \( K < x \) whenever \( x \in X \) . | Yes |
Lemma 16.5.1 If \( B \) is Loeb measurable with respect to \( \mu \), then\n\n\[ \n{\mu }_{L}\left( B\right) = \inf \left\{ {{\mu }_{L}\left( A\right) : B \subseteq A \in \mathcal{A}}\right\} \n\] | Proof. By monotonicity, \( {\mu }_{L}\left( B\right) \) is a lower bound of the \( {\mu }_{L}\left( A\right) \) ’s for \( B \subseteq \) \( A \in \mathcal{A} \) . If \( {\mu }_{L}\left( B\right) = \infty \), then the result follows. If, however, \( {\mu }_{L}\left( B\right) < \infty \) , to show that it is the greatest lower bound it suffices to show that for any \( \varepsilon \in {\mathbb{R}}^{ + } \) there is some set \( {A}_{\varepsilon } \in \mathcal{A} \) with \( B \subseteq {A}_{\varepsilon } \) and \( {\mu }_{L}\left( {A}_{\varepsilon }\right) \leq {\mu }_{L}\left( B\right) + \varepsilon \) .\n\nNow, for such an \( \varepsilon \), by properties of outer measure there is an increasing sequence \( {A}_{1} \subseteq {A}_{2} \subseteq \cdots \) of \( \mathcal{A} \) -elements whose union includes \( B \) and has\n\n\[ \n{\mu }_{L}^{ + }\left( {\mathop{\bigcup }\limits_{{n \in \mathbb{N}}}{A}_{n}}\right) < {\mu }_{L}\left( B\right) + \varepsilon .\n\]\n\nThe sequence \( \left\langle {{A}_{n} : n \in \mathbb{N}}\right\rangle \) extends by sequential comprehensiveness to an internal sequence \( \left\langle {{A}_{n} : n \in * \mathbb{N}}\right\rangle \) of elements of \( \mathcal{A} \) . Then for each \( k \in \mathbb{N} \) we have\n\n\[ \n\left( {\forall n \in \mathbb{{N}^{d}}}\right) \left( {n \leq k\text{ implies }{A}_{n} \subseteq {A}_{k}\text{ and }\mu \left( {A}_{n}\right) < {\mu }_{L}\left( B\right) + \varepsilon }\right)\n\]\n\n(i)\n\n\( \left( {\mathrm{{since}}\mu \left( {A}_{n}\right) \simeq {\mu }_{L}\left( {A}_{n}\right) \leq {\mu }_{L}^{ + }\left( {\mathop{\bigcup }\limits_{{n \in \mathbb{N}}}{A}_{n}}\right) }\right) . \) But (i) is an internal assertion, since \( \mu \) and the extended sequence are internal, while \( k,\varepsilon \), and \( {\mu }_{L}\left( B\right) \) are fixed internal entities (real numbers). Therefore by overflow (i) must be true with some unlimited \( K \in {}^{ * }\mathbb{N} \) in place of \( k \) . For such a \( K \) we have \( {A}_{K} \in \mathcal{A} \) and \( {A}_{n} \subseteq {A}_{K} \) for all \( n \in \mathbb{N} \), so that\n\n\[ \nB \subseteq \mathop{\bigcup }\limits_{{n \in \mathbb{N}}}{A}_{n} \subseteq {A}_{K}\n\]\n\nwhile \( \mu \left( {A}_{K}\right) < {\mu }_{L}\left( B\right) + \varepsilon \) . Hence as \( \mu \left( {A}_{K}\right) \simeq {\mu }_{L}\left( {A}_{K}\right) \),\n\n\[ \n{\mu }_{L}\left( {A}_{K}\right) \leq {\mu }_{L}\left( B\right) + \varepsilon\n\]\n\nestablishing that \( {A}_{K} \) is the set \( {A}_{\varepsilon } \) we are looking for. | Yes |
Lemma 16.5.2 If \( B \) is Loeb measurable and \( {\mu }_{L} \) -finite, then\n\n\[{\mu }_{L}\left( B\right) = \sup \left\{ {{\mu }_{L}\left( A\right) : A \subseteq B}\right. \text{and}\left. {A \in \mathcal{A}}\right\} \text{.} | Proof. Given any \( \varepsilon \in {\mathbb{R}}^{ + } \), we will show that there is some set \( {A}_{\varepsilon } \in \mathcal{A} \) such that \( {A}_{\varepsilon } \subseteq B \) and \( {\mu }_{L}\left( B\right) - \varepsilon < \mu \left( {A}_{\varepsilon }\right) \). \n\nSince \( {\mu }_{L}\left( B\right) < \infty \), we know from Lemma 16.5.1 that there is some \( D \in \mathcal{A} \) with \( B \subseteq D \) and \( {\mu }_{L}\left( D\right) < \infty \). The desired result is obtained by using complementation relative to \( D \). Firstly, \( D - B \) is Loeb measurable and \( {\mu }_{L} \) -finite, so by Lemma 16.5.1 there is a set \( C \) with \( D - B \subseteq C \in \mathcal{A} \) and\n\n\[{\mu }_{L}\left( C\right) < {\mu }_{L}\left( {D - B}\right) + \varepsilon .\n\nWe may assume \( C \subseteq D \) (since we could replace \( C \) by \( C \cap D \) here). Let \( {A}_{\varepsilon } = D - C \in \mathcal{A} \). Then \( {A}_{\varepsilon } \subseteq B \), and \( C \) is the disjoint union of \( D - B \) and \( B - {A}_{\varepsilon } \), so\n\n\[{\mu }_{L}\left( {D - B}\right) + {\mu }_{L}\left( {B - {A}_{\varepsilon }}\right) = {\mu }_{L}\left( C\right) < {\mu }_{L}\left( {D - B}\right) + \varepsilon ,\n\nimplying that \( {\mu }_{L}\left( {B - {A}_{\varepsilon }}\right) < \varepsilon \). Therefore\n\n\[{\mu }_{L}\left( B\right) = {\mu }_{L}\left( {A}_{\varepsilon }\right) + {\mu }_{L}\left( {B - {A}_{\varepsilon }}\right) < {\mu }_{L}\left( {A}_{\varepsilon }\right) + \varepsilon ,\n\nso \( {\mu }_{L}\left( B\right) - \varepsilon < \mu \left( {A}_{\varepsilon }\right) \) as desired. | Yes |
Lemma 16.6.1 If \( B \) is Loeb measurable with \( {\mu }_{L}\left( B\right) < \infty \), then \( B \) is \( \mu \) - approximable. | Proof. Given \( \varepsilon \in {\mathbb{R}}^{ + } \), by Lemmas 16.5.1 and 16.5.2 there are \( {C}_{\varepsilon },{D}_{\varepsilon } \in \mathcal{A} \) with \( {C}_{\varepsilon } \subseteq B \subseteq {D}_{\varepsilon } \) and \( {\mu }_{L}\left( {D}_{\varepsilon }\right) < {\mu }_{L}\left( B\right) + \frac{\varepsilon }{2} \) while \( {\mu }_{L}\left( B\right) < {\mu }_{L}\left( {C}_{\varepsilon }\right) + \frac{\varepsilon }{2} \) . Then\n\n\[ \n{\mu }_{L}\left( {{D}_{\varepsilon } - {C}_{\varepsilon }}\right) = {\mu }_{L}\left( {D}_{\varepsilon }\right) - {\mu }_{L}\left( B\right) + {\mu }_{L}\left( B\right) - {\mu }_{L}\left( {C}_{\varepsilon }\right) < \varepsilon .\n\] | Yes |
Lemma 16.6.3 If \( B \) is \( \mu \) -approximable, then \( B \) is Loeb measurable. | Proof. We have to show that any \( E \subseteq S \) is split \( {\mu }_{L}^{ + } \) -additively by \( B \), for which it suffices that\n\n\[{\mu }_{L}^{ + }\left( E\right) \geq {\mu }_{L}^{ + }\left( {E \cap B}\right) + {\mu }_{L}^{ + }\left( {E - B}\right)\]\n\nNow, by Lemma 16.6.2 there is an \( A \in \mathcal{A} \) that is almost equal to \( B \). The desired inequality holds with \( A \) in place of \( B \), since \( A \) is Loeb measurable, and so as \( {\mu }_{L}^{ + }\left( {B\Delta A}\right) = 0 \), we can show it holds for \( B \) as well. Formally, let\n\n\[C = \left( {E \cap B}\right) - A,\]\n\n\[D = \left( {E \cap A}\right) - B\n\n\[G = E - \left( {A \cup B}\right)\n\n\[H = E \cap A \cap B\n\n(draw a Venn diagram!). Then \( C \) and \( D \) are included in \( {B\Delta A} \), so \( {\mu }_{L}^{ + }\left( C\right) = \) \( {\mu }_{L}^{ + }\left( D\right) = 0 \). Thus\n\n\[{\mu }_{L}^{ + }\left( {E \cap B}\right) = {\mu }_{L}^{ + }\left( {C \cup H}\right) \leq {\mu }_{L}^{ + }\left( C\right) + {\mu }_{L}^{ + }\left( H\right) = {\mu }_{L}^{ + }\left( H\right) ,\]\n\nand similarly\n\n\[{\mu }_{L}^{ + }\left( {E - B}\right) = {\mu }_{L}^{ + }\left( {D \cup G}\right) \leq {\mu }_{L}^{ + }\left( D\right) + {\mu }_{L}^{ + }\left( G\right) = {\mu }_{L}^{ + }\left( G\right) .\n\nHence\n\n\[{\mu }_{L}^{ + }\left( {E \cap B}\right) + {\mu }_{L}^{ + }\left( {E - B}\right) \leq {\mu }_{L}^{ + }\left( H\right) + {\mu }_{L}^{ + }\left( G\right)\]\n\n\( \leq {\mu }_{L}^{ + }\left( {E \cap A}\right) + {\mu }_{L}^{ + }\left( {E - A}\right) \) by monotonicity\n\n\[= {\mu }_{L}^{ + }\left( E\right)\]\n\nwith the last step given by the \( {\mu }_{L}^{ + } \) -measurability of \( A \) . | Yes |
Lemma 16.6.4 If \( {\mu }_{L}^{ + }\left( B\right) < \infty \), then \( B \) is Loeb measurable with respect to \( \mu \) if and only if it is \( \mu \) -approximable. | Proof. By Lemmas 16.6.1 and 16.6.3. | No |
An arbitrary subset \( B \) of \( S \) is Loeb measurable with respect to \( \mu \) if and only if \( B \cap A \) is \( \mu \) -approximable for all \( \mu \) -finite \( A \in \mathcal{A} \) . | Proof. Let \( B \) be Loeb measurable. If \( A \in \mathcal{A} \) is \( \mu \) -finite, then \( B \cap A \) is Loeb measurable, since the Loeb measurable sets are \( \cap \) -closed, and \( {\mu }_{L}^{ + } \) -finite since \( B \cap A \subseteq A \), so by Corollary 16.6.4 \( B \cap A \) is \( \mu \) -approximable.\n\nConversely, let \( B \cap A \) be \( \mu \) -approximable for all \( \mu \) -finite \( A \in \mathcal{A} \) . To show that \( B \) is Loeb measurable, we have to show that for any \( {\mu }_{L}^{ + } \) -finite \( E \subseteq S \), \[ {\mu }_{L}^{ + }\left( E\right) \geq {\mu }_{L}^{ + }\left( {E \cap B}\right) + {\mu }_{L}^{ + }\left( {E - B}\right) \] (iii)\n\nBut in such a case \( {\mu }_{L}^{ + }\left( {E \cap B}\right) < \infty \), so there must be a sequence \( \langle {A}_{n} : n \in \mathbb{N}\rangle \) of \( \mathcal{A} \) -elements that covers \( E \cap B \) with each \( {A}_{n} \) being \( \mu \) -finite. Each \( B \cap {A}_{n} \) is then \( \mu \) -approximable, by hypothesis, and so is Loeb measurable by Lemma 16.6.3. Hence if \[ A = \mathop{\bigcup }\limits_{{n \in \mathbb{N}}}\left( {B \cap {A}_{n}}\right) \] it follows that \( A \) is Loeb measurable, and so \[ {\mu }_{L}^{ + }\left( E\right) \geq {\mu }_{L}^{ + }\left( {E \cap A}\right) + {\mu }_{L}^{ + }\left( {E - A}\right) \] (iv)\n\nBut \[ E \cap A = E \cap \left( {\mathop{\bigcup }\limits_{{n \in \mathbb{N}}}\left( {B \cap {A}_{n}}\right) }\right) = E \cap B \] as the \( {A}_{n} \) ’s cover \( E \cap B \) . Then \( E - A = E - B \), and hence (iii) follows from (iv). | Yes |
Theorem 16.8.1 For any \( a, b \in \mathbb{R} \) with \( a < b \) , \[ {\mu }_{L}\left( {\{ s \in S : a < s < b\} }\right) = b - a. \] | Proof. Let \( A = \{ s \in S : a < s < b\} \) . Then \( A = S \cap {}^{ * }\left( {a, b}\right) \), so \( A \) is internal and belongs to \( \mathcal{A} \), hence is Loeb measurable. Moreover, \( A \) is hyperfinite, so has smallest and greatest elements, say \( s \) and \( t \) . Since \( a \) and \( b \) can be approximated infinitely closely by members of \( S \), we must then have \( a \simeq s \) and \( b \simeq t \) . Also, we can put \( s = \frac{K + 1}{N} \) and \( t = \frac{L}{N} \) for some \( K, L \in * \mathbb{Z} \) . Thus \[ \left. {A = \left\{ {\frac{K + 1}{N},\frac{K + 2}{N},\ldots ,\frac{L}{N}}\right. }\right\} = \left\{ {\frac{M}{N} : K < M \leq L}\right\} , \] which is hyperfinite of cardinality \( L - K \), since the internal function \( f\left( x\right) = \) \( \frac{K + x}{N} \) is a bijection from \( \{ 1,\ldots, L - K\} \) onto \( A \) . It follows that \[ \frac{\left| A\right| }{N} = \frac{L - K}{N} = \frac{L}{N} - \frac{K}{N} \simeq b - a, \] and so \( {\mu }_{L}\left( A\right) = b - a \) as desired. | Yes |
Lemma 16.8.3 \( \mathcal{M} \) includes the Borel algebra \( {\mathcal{B}}_{\mathbb{R}} \), and \( \nu \) agrees with Lebesgue measure on all Borel sets. | Proof. Each open interval \( \left( {a, b}\right) \subseteq \mathbb{R} \) belongs to \( \mathcal{M} \), since \( {\operatorname{sh}}^{-1}\left( \left( {a, b}\right) \right) \) is the union of the sequence \( \left\langle {{A}_{n} : n \in \mathbb{N}}\right\rangle \), where\n\n\[ \n{A}_{n} = S \cap {}^{ * }\left( {a + \frac{1}{n}, b - \frac{1}{n}}\right) \in \mathcal{A}.\n\]\n\nBut \( {\mathcal{B}}_{\mathbb{R}} \) is the smallest \( \sigma \) -algebra containing all open intervals \( \left( {a, b}\right) \), so this implies that \( {\mathcal{B}}_{\mathbb{R}} \subseteq \mathcal{M} \) . Also, by Theorem 16.8.1,\n\n\[ \n{\mu }_{L}\left( {A}_{n}\right) = \left( {b - \frac{1}{n}}\right) - \left( {a + \frac{1}{n}}\right) = b - a - \frac{2}{n},\n\]\n\nand hence as the \( {A}_{n} \) ’s form an increasing sequence,\n\n\[ \n\nu \left( \left( {a, b}\right) \right) = {\mu }_{L}\left( {{\operatorname{sh}}^{-1}\left( \left( {a, b}\right) \right) = \mathop{\lim }\limits_{{n \rightarrow \infty }}{\mu }_{L}\left( {A}_{n}\right) = b - a.}\right.\n\]\n\nThus \( \nu \) is a measure on \( {\mathcal{B}}_{\mathbb{R}} \) that agrees with \( \lambda \) on all open intervals. But any such measure must agree with \( \lambda \) on all Borel sets (16.4(1)). | Yes |
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