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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 inclu... | 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 \( \{ i\} \notin \mathcal{F} \), wh... | 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... | Yes |
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 \( A \) who... | 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\n... | 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... | Proof. Exercise. | 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 ... | 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}... | 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 t... | 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 \ri... | 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\[ \le... | 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 al... | 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\[ \left| {{s}_{n} - L}\right| < \varepsilon \simeq 0. \]\... | 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 ... | 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... | 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{... | 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... | 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) }\... | 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 \( \left( {c - \delta, c + \delta }\right) \) ... | Yes |
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,... | Yes |
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 \( \var... | 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)... | 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)... | 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\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... | 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 8.7.1 (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 \( \del... | 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(... | 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 \( {\De... | 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 \( {f}^{\prime }\left( x\right) \simeq {f}^{\prime }\left( c\right) \)... | 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\l... | 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\righ... | 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... | 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 \( \l... | 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 Robin... | 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... | 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... | 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 define... | 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 \... | 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... | 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}_{... | 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 hypernat... | 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 - \fr... | 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 s... | 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} ... | 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 + \fr... | 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 \( \ma... | 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( {\l... | 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}\r... | 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... | 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\{ {... | Yes |
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{... | 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 * \( T \in \) * \( \math... | 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 \( {}^{ * }A \) . T... | 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... | 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... | 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\... | 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 \( k \leq n \), the... | (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 ... | 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 \c... | 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) \... | 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... | 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... | 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{\varepsilo... | 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 \( A \in \mathcal{A} \) ... | 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\... | 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 appro... | 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 + \f... | 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: \( {p}_{1} > {p}_{2} > \cdots \; \) ... | 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 s... | 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 hypersequenc... | 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 approachabl... | 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} \... | 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 |... | 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\[ \... | 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... | 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 h... | 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}_{... | 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)... | 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}... | 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., ... | 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... | No |
Theorem 1.1.1 If \( a, b \) are relatively prime, then we can find integers \( x, y \) such that \( {ax} + {by} = 1 \) . | Proof. We write \( a = {bq} + r \) by the Euclidean algorithm, and since \( a, b \) are relatively prime we know \( r \neq 0 \) so \( 0 < r < \left| b\right| \) . We see that \( b, r \) are relatively prime, or their common factor would have to divide \( a \) as well. So, \( b = r{q}_{1} + {r}_{1} \) with \( 0 < {r}_{1... | Yes |
Theorem 1.1.2 Every positive integer greater than 1 has a prime divisor. | Proof. Suppose that there is a positive integer having no prime divisors. Since the set of positive integers with no prime divisors is nonempty, there is a least positive integer \( n \) with no prime divisors. Since \( n \) divides itself, \( n \) is not prime. Hence we can write \( n = {ab} \) with \( 1 < a < n \) an... | Yes |
Theorem 1.1.4 If \( p \) is prime and \( p \mid {ab} \), then \( p \mid a \) or \( p \mid b \) . | Proof. Suppose that \( p \) is prime and \( p \mid {ab} \) where \( a \) and \( b \) are integers. If \( p \) does not divide \( a \), then \( a \) and \( p \) are coprime. Then \( \exists x, y \in \mathbb{Z} \) such that \( {ax} + {py} = 1 \) . Then we have \( {abx} + {pby} = b \) and \( {pby} = b - {abx} \) . Hence \... | Yes |
Example 1.1.6 Show that\n\n\[ S = 1 + \frac{1}{2} + \cdots + \frac{1}{n} \]\n\nis not an integer for \( n > 1 \) . | Solution. Let \( k \in \mathbb{Z} \) be the highest power of 2 less than \( n \), so that \( {2}^{k} \leq \) \( n < {2}^{k + 1} \) . Let \( m \) be the least common multiple of \( 1,2,\ldots, n \) excepting \( {2}^{k} \) . Then\n\n\[ {mS} = m + \frac{m}{2} + \cdots + \frac{m}{n}. \]\n\nEach of the numbers on the right-... | Yes |
Theorem 1.1.14 Given \( a, n \in \mathbb{Z},{a}^{\phi \left( n\right) } \equiv 1\left( {\;\operatorname{mod}\;n}\right) \) when \( \gcd \left( {a, n}\right) = 1 \) . This is a theorem due to Euler. | Proof. The case where \( n \) is prime is clearly a special case of Fermat’s little Theorem. The argument is basically the same as that of the alternate solution to Exercise 1.1.13.\n\nConsider the ring \( \mathbb{Z}/n\mathbb{Z} \) . If \( a, n \) are coprime, then \( \bar{a} \) is a unit in this ring. The units form a... | Yes |
Example 2.1.1 Let \( R \) be an integral domain. Suppose there is a map \( n : R \rightarrow \mathbb{N} \) such that:\n\n(i) \( n\left( {ab}\right) = n\left( a\right) n\left( b\right) \forall a, b \in R \) ; and\n\n(ii) \( n\left( a\right) = 1 \) if and only if \( a \) is a unit.\n\nWe call such a map a norm map, with ... | Solution. Suppose \( b \) is an element of \( R \) . We proceed by induction on the norm of \( b \) . If \( b \) is irreducible, then we have nothing to prove, so assume that \( b \) is an element of \( R \) which is not irreducible. Then we can write \( b = {ac} \) where neither \( a \) nor \( c \) is a unit. By condi... | Yes |
Theorem 2.1.6 If \( R \) is a principal ideal domain, then \( R \) is a unique factorization domain. | Proof. Let \( S \) be the set of elements of \( R \) that cannot be written as a product of irreducibles. If \( S \) is nonempty, take \( {a}_{1} \in S \) . Then \( {a}_{1} \) is not irreducible, so we can write \( {a}_{1} = {a}_{2}{b}_{2} \) where \( {a}_{2},{b}_{2} \) are not units. Then \( \left( {a}_{1}\right) \sub... | No |
Theorem 2.1.7 If \( R \) is a domain with a map \( \phi : R \rightarrow \mathbb{N} \), and given \( a, b \in R,\exists q, r \in R \) such that \( a = {bq} + r \) with \( r = 0 \) or \( \phi \left( r\right) < \phi \left( b\right) \), we call \( R \) a Euclidean domain. If a ring \( R \) is Euclidean, it is a principal i... | Proof. Given an ideal \( I \subseteq R \), take an element \( a \) of \( I \) such that \( \phi \left( a\right) \) is minimal among elements of \( I \) . Then given \( b \in I \), we can find \( q, r \in R \) such that \( b = {qa} + r \) where \( r = 0 \) or \( \phi \left( r\right) < \phi \left( a\right) \) . But then ... | Yes |
If \( R \) is a unique factorization domain, and \( f\left( x\right) \in R\left\lbrack x\right\rbrack \) , define the content of \( f \) to be the gcd of the coefficients of \( f \), denoted by \( \mathcal{C}\left( f\right) \) . For \( f\left( x\right), g\left( x\right) \in R\left\lbrack x\right\rbrack ,\mathcal{C}\lef... | Consider two polynomials \( f, g \in R\left\lbrack x\right\rbrack \), with \( \mathcal{C}\left( f\right) = c \) and \( \mathcal{C}\left( g\right) = d \) . Then we can write\n\n\[ f\left( x\right) = c{a}_{0} + c{a}_{1}x + \cdots + c{a}_{n}{x}^{n} \]\n\nand\n\n\[ g\left( x\right) = d{b}_{0} + d{b}_{1}x + \cdots + d{b}_{m... | Yes |
Example 2.3.5 Let \( \lambda = 1 - \rho ,\theta \in \mathbb{Z}\left\lbrack \rho \right\rbrack \) . Show that if \( \lambda \) does not divide \( \theta \) , then \( {\theta }^{3} \equiv \pm 1\left( {\;\operatorname{mod}\;{\lambda }^{4}}\right) \) . Deduce that if \( \alpha ,\beta ,\gamma \) are coprime to \( \lambda \)... | Solution. From the previous problem, we know that if \( \lambda \) does not divide \( \theta \) then \( \theta \equiv \pm 1\left( {\;\operatorname{mod}\;\lambda }\right) \) . Set \( \xi = \theta \) or \( - \theta \) so that \( \xi \equiv 1\left( {\;\operatorname{mod}\;\lambda }\right) \) . We write \( \xi \) as \( 1 + ... | Yes |
Example 3.1.1 Show that \( \sqrt{2}/3 \) is an algebraic number but not an algebraic integer. | Solution. Consider the polynomial \( f\left( x\right) = 9{x}^{2} - 2 \), which is in \( \mathbb{Q}\left\lbrack x\right\rbrack \) . Since \( f\left( {\sqrt{2}/3}\right) = 0 \), we know that \( \sqrt{2}/3 \) is an algebraic number.\n\nAssume \( \sqrt{2}/3 \) is an algebraic integer. Then there exists a monic polynomial i... | Yes |
Theorem 3.1.4 Let \( \alpha \) be an algebraic number. There exists a unique polynomial \( p\left( x\right) \) in \( \mathbb{Q}\left\lbrack x\right\rbrack \) which is monic, irreducible and of smallest degree, such that \( p\left( \alpha \right) = 0 \) . Furthermore, if \( f\left( x\right) \in \mathbb{Q}\left\lbrack x\... | Proof. Consider the set of all polynomials in \( \mathbb{Q}\left\lbrack x\right\rbrack \) for which \( \alpha \) is a root and pick one of smallest degree, say \( p\left( x\right) \) . If \( p\left( x\right) \) is not irreducible, it can be written as a product of two lower degree polynomials in \( \mathbb{Q}\left\lbra... | Yes |
Example 3.1.5 Show that the set of algebraic numbers is countable (and hence the set of transcendental numbers is uncountable). | Solution. All polynomials in \( \mathbb{Q}\left\lbrack x\right\rbrack \) have a finite number of roots. The set of rational numbers, \( \mathbb{Q} \), is countable and so the set \( \mathbb{Q}\left\lbrack x\right\rbrack \) is also countable. The set of algebraic numbers is the set of all roots of a countable number of ... | Yes |
Theorem 3.2.1 (Liouville) Given \( \alpha \), a real algebraic number of degree \( n \neq 1 \), there is a positive constant \( c = c\left( \alpha \right) \) such that for all rational numbers \( p/q,\left( {p, q}\right) = 1 \) and \( q > 0 \), the inequality\n\n\[ \left| {\alpha - \frac{p}{q}}\right| > \frac{c\left( \... | Proof. Let \( f\left( x\right) = {a}_{n}{x}^{n} + {a}_{n - 1}{x}^{n - 1} + \cdots + {a}_{0} \) be \( \in \mathbb{Z}\left\lbrack x\right\rbrack \) whose degree equals that of \( \alpha \) and for which \( \alpha \) is a root. \( \left( {\operatorname{So}\deg \left( f\right) \geq 2}\right) \) . Notice that\n\n\[ \left| {... | Yes |
Show that\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{1}{{10}^{n!}} \]\n\nis transcendental. | Solution. Suppose not, and call the sum \( \alpha \) . Look at the partial sum\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{k}\frac{1}{{10}^{n!}} = \frac{{p}_{k}}{{q}_{k}} \]\n\nwith \( {q}_{k} = {10}^{k!} \) . Thus,\n\n\[ \left| {\alpha - \frac{{p}_{k}}{{q}_{k}}}\right| = \left| {\mathop{\sum }\limits_{{n = k + 1}}^{\infty ... | Yes |
Example 3.2.3 Let \( f\left( {x, y}\right) \) be an irreducible polynomial of binary form of degree \( n \geq 3 \) . Assuming Thue’s theorem, show that \( f\left( {x, y}\right) = m \) for any fixed \( m \in {\mathbb{Z}}^{ * } \) has only finitely many solutions. | Solution. Suppose \( f\left( {x, y}\right) = m \) has infinitely many solutions, and write it in the form\n\n\[ f\left( {x, y}\right) = \mathop{\prod }\limits_{{i = 1}}^{n}\left( {x - {\alpha }_{i}y}\right) = m \]\n\nwhere \( {\alpha }_{i} \) is an algebraic number of degree \( \geq 3\forall i = 1,\ldots, n \) .\n\nWit... | Yes |
Example 3.3.1 Let \( \alpha \) be an algebraic number and define\n\n\[ \mathbb{Q}\left\lbrack \alpha \right\rbrack = \{ f\left( \alpha \right) : f \in \mathbb{Q}\left\lbrack x\right\rbrack \} \]\n\n a subring of \( \mathbb{C} \) . Show that \( \mathbb{Q}\left\lbrack \alpha \right\rbrack \) is a field. | Solution. Let \( f \) be the minimal polynomial of \( \alpha \), and consider the map \( \phi : \mathbb{Q}\left\lbrack x\right\rbrack \rightarrow \mathbb{Q}\left\lbrack \alpha \right\rbrack \) such that\n\n\[ \mathop{\sum }\limits_{{i = 0}}^{n}{a}_{i}{x}^{i} \rightarrow \mathop{\sum }\limits_{{i = 0}}^{n}{a}_{i}{\alpha... | Yes |
Theorem 3.3.2 (Theorem of the Primitive Element) If \( \alpha \) and \( \beta \) are algebraic numbers, then \( \exists \theta \), an algebraic number, such that \( \mathbb{Q}\left( {\alpha ,\beta }\right) = \mathbb{Q}\left( \theta \right) \) . | Proof. Let \( f \) be the minimal polynomial of \( \alpha \) and let \( g \) be the minimal polynomial of \( \beta \) . We want to show that we can find \( \lambda \in \mathbb{Q} \) such that \( \theta = \alpha + {\lambda \beta } \) and \( \mathbb{Q}\left( {\alpha ,\beta }\right) = \mathbb{Q}\left( \theta \right) \) . ... | Yes |
Example 3.3.10 Let \( K \) be an algebraic number field. Let \( {\mathcal{O}}_{K} \) be the set of all algebraic integers in \( K \) . Show that \( {\mathcal{O}}_{K} \) is a ring. | Solution. From the above theorem, we know that for \( \alpha ,\beta \), algebraic integers, \( \mathbb{Z}\left\lbrack \alpha \right\rbrack ,\mathbb{Z}\left\lbrack \beta \right\rbrack \) are finitely generated \( \mathbb{Z} \) -modules. Thus \( M = \mathbb{Z}\left\lbrack {\alpha ,\beta }\right\rbrack \) is also a finite... | Yes |
Lemma 4.1.1 If \( K \) is an algebraic number field of degree \( n \) over \( \mathbb{Q} \), and \( \alpha \in {\mathcal{O}}_{K} \) its ring of integers, then \( {\operatorname{Tr}}_{K}\left( \alpha \right) \) and \( {\mathrm{N}}_{K}\left( \alpha \right) \) are in \( \mathbb{Z} \). | Proof. We begin by writing \( \alpha {\omega }_{i} = \mathop{\sum }\limits_{{j = 1}}^{n}{a}_{ij}{\omega }_{j}\forall i \) . Then we have\n\n\[ \n{\alpha }^{\left( k\right) }{\omega }_{i}^{\left( k\right) } = \mathop{\sum }\limits_{{j = 1}}^{n}{a}_{ij}{\omega }_{j}^{\left( k\right) }\;\forall i, k, \n\] \n\nwhere \( {\a... | Yes |
Lemma 4.1.4 The bilinear pairing given by \( B\left( {x, y}\right) : K \times K \rightarrow \mathbb{Q} \) such that \( \left( {x, y}\right) \rightarrow {\operatorname{Tr}}_{K}\left( {xy}\right) \) is nondegenerate. | Proof. We recall that if \( V \) is a finite-dimensional vector space over a field \( F \) with basis \( {e}_{1},{e}_{2},\ldots ,{e}_{n} \) and \( B : V \times V \rightarrow F \) is a bilinear map, we can associate a matrix to \( B \) as follows. Write\n\n\[ v = \sum {a}_{i}{e}_{i}\;\text{ with }\;{a}_{i} \in F \]\n\n\... | Yes |
Theorem 4.2.2 Let \( {\alpha }_{1},{\alpha }_{2},\ldots ,{\alpha }_{n} \) be a set of generators for a finitely generated \( \mathbb{Z} \) -module \( M \), and let \( N \) be a submodule.\n\n(a) \( \exists {\beta }_{1},{\beta }_{2},\ldots ,{\beta }_{m} \) in \( N \) with \( m \leq n \) such that\n\n\[ N = \mathbb{Z}{\b... | Proof. (a) We will proceed by induction on the number of generators of a \( \mathbb{Z} \) -module. This is trivial when \( n = 0 \) . We can assume that we have proved the above statement to be true for all \( \mathbb{Z} \) -modules with \( n - 1 \) or fewer generators, and proceed to prove it for \( n \) . We define \... | Yes |
Example 4.3.1 Suppose that the minimal polynomial of \( \\alpha \) is Eisensteinian with respect to a prime \( p \), i.e., \( \\alpha \) is a root of the polynomial\n\n\\[ \n{x}^{n} + {a}_{n - 1}{x}^{n - 1} + \\cdots + {a}_{1}x + {a}_{0},\n\\]\n\nwhere \( p \\mid {a}_{i},0 \\leq i \\leq n - 1 \) and \( {p}^{2} \\nmid {... | Solution. Let \( M = \\mathbb{Z} + \\mathbb{Z}\\alpha + \\cdots + \\mathbb{Z}{\\alpha }^{n - 1} \) . First observe that since\n\n\\[ \n{\\alpha }^{n} + {a}_{n - 1}{\\alpha }^{n - 1} + \\cdots + {a}_{1}\\alpha + {a}_{0} = 0,\n\\]\n\nthen \( {\\alpha }^{n}/p \\in M \\subseteq {\\mathcal{O}}_{K} \) . Also, \( \\left| {{\\... | Yes |
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