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Proposition 4.4 With the above assumptions,\n\n\[ \n{\begin{Vmatrix}{T}_{c}\end{Vmatrix}}_{{L}^{q}} \leq M\parallel f{\parallel }_{{L}^{p}} \n\]\n\nfor any \( c \) with \( a \leq c \leq b \), where \( c = \left( {1 - \theta }\right) a + {\theta b} \) and \( 0 \leq \theta \leq 1 \) ; and\n\n\[ \n\frac{1}{p} = \frac{1 - ... | Once we have formulated this result, we in fact observe that we can prove it by essentially the same argument as in Section 2 in Chapter 2.\n\nWe write \( s = a\left( {1 - z}\right) + {bz} \), so \( z = \frac{s - a}{b - a} \), and the strip \( S \) is thereby transformed into the strip \( 0 \leq \operatorname{Re}\left(... | Yes |
Proposition 4.5 The Fourier transform \( {\widehat{K}}_{s}\left( \xi \right) \) is analytically continuable into the half-plane \( - \frac{d - 1}{2} \leq \operatorname{Re}\left( s\right) \) and satisfies \[ \mathop{\sup }\limits_{{\xi \in {\mathbb{R}}^{d}}}\left| {{\widehat{K}}_{s}\left( \xi \right) }\right| \leq M\;\t... | Proof. Write \( s\left( {s + 1}\right) \cdots \left( {s + N}\right) {u}^{s - 1} = {\left( \frac{d}{du}\right) }^{N + 1}{u}^{s + N} \) . Then an \( \left( {N + 1}\right) \) -fold integration by parts yields \[ {I}_{s}\left( \rho \right) = {\left( -1\right) }^{N + 1}{\int }_{0}^{\infty }{u}^{s + N}{\left( \frac{d}{du}\ri... | Yes |
Lemma 4.6 \( {I}_{s}\left( \rho \right) \) initially given above for \( \operatorname{Re}\left( s\right) > 0 \), has an analytic continuation into the half-space \( \operatorname{Re}\left( s\right) > - N - 1 \) . | Proof. Write \( s\left( {s + 1}\right) \cdots \left( {s + N}\right) {u}^{s - 1} = {\left( \frac{d}{du}\right) }^{N + 1}{u}^{s + N} \) . Then an \( \left( {N + 1}\right) \) -fold integration by parts yields\n\n\[ \n{I}_{s}\left( \rho \right) = {\left( -1\right) }^{N + 1}{\int }_{0}^{\infty }{u}^{s + N}{\left( \frac{d}{d... | Yes |
Proposition 5.1 Suppose \( f \in {L}^{p}\left( {\mathbb{R}}^{d}\right) \) is a radial function. Then \( \widehat{f} \) is continuous for \( \xi \neq 0 \) whenever \( 1 \leq p < {2d}/\left( {d + 1}\right) \) . Note the sequence of exponents \( \frac{2d}{\left( d + 1\right) } : 1,\frac{4}{3},\frac{3}{2},\frac{8}{5},\ldot... | Proof. Suppose \( f\left( x\right) = {f}_{0}\left( \left| x\right| \right) \) . Then \( \widehat{f}\left( \xi \right) = F\left( \left| \xi \right| \right) \) with \( F \) defined by (4), namely,\n\n(29)\n\n\[ F\left( \rho \right) = {2\pi }{\rho }^{-d/2 + 1}{\int }_{0}^{\infty }{J}_{d/2 - 1}\left( {2\pi \rho r}\right) {... | Yes |
Theorem 5.2 Suppose \( M \) has non-zero Gauss curvature at each point of the support of \( {d\mu } \) . Then the restriction inequality (31) holds for \( q = 2 \) and \( p = \frac{{2d} + 2}{d + 3} \) . | The proof starts with several quick observations. Let \( \mathcal{R} \) denote the restriction operator\n\n\[ \mathcal{R}\left( f\right) = {\left. \widehat{f}\left( \xi \right) \right| }_{M} = {\left. {\int }_{{\mathbb{R}}^{d}}{e}^{-{2\pi ix} \cdot \xi }f\left( x\right) dx\right| }_{M}, \]\n\nwhich is initially defined... | No |
Corollary 5.4 Under the assumptions of the theorem, the restriction inequality (31) holds for \( 1 \leq p \leq \frac{{2d} + 2}{d + 3} \) and \( q \leq \left( \frac{d - 1}{d + 1}\right) {p}^{\prime } \) . | This follows by combining the critical case \( p = \frac{{2d} + 2}{d + 3}, q \leq 2 \) (a consequence of the theorem and Hölder's inequality) with the trivial case \( p = 1, q = \infty \) via the Riesz interpolation theorem. | Yes |
Proposition 6.1 For each \( t \) :\n\n(i) \( {e}^{{it}\bigtriangleup } \) maps \( \mathcal{S} \) to \( \mathcal{S} \) . | Proof. That \( {e}^{{it}\bigtriangleup } \) maps \( \mathcal{S} \) to \( \mathcal{S} \) is clear because the multiplier \( {e}^{-{it4}{\pi }^{2}{\left| \xi \right| }^{2}} \) has the property that each derivative in \( \xi \) is of at most polynomial increase. | Yes |
Proposition 6.2 For each \( t \) :\n\n(i) The operator \( {e}^{{it}\bigtriangleup } \) is unitary on \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) .\n\n(ii) For every \( f \), the mapping \( t \mapsto {e}^{{it}\bigtriangleup }\left( f\right) \) is continuous in the \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) norm.\n\n(ii... | Proof. Conclusion (i) is immediate from Plancherel's theorem, since the multiplier \( {e}^{-{it4}{\pi }^{2}{\left| \dot{\xi }\right| }^{2}} \) has absolute value one. Now if \( \widehat{f} \in {L}^{2}\left( {\mathbb{R}}^{d}\right) \) , then clearly \( {e}^{-{it4}{\pi }^{2}{\left| \xi \right| }^{2}}\widehat{f}\left( \xi... | Yes |
Theorem 6.4 The solution \( {e}^{t{\left( \frac{d}{dx}\right) }^{3}}\left( f\right) \) satisfies | The proof of this is result is parallel with that of the previous theorem and reduces to a restriction theorem on \( {\mathbb{R}}^{2} \) for the cubic curve\n\n\[ \Gamma = \left\{ {\left( {{\xi }_{1},{\xi }_{2}}\right) : {\xi }_{2} = - 4{\pi }^{2}{\xi }_{1}^{3}}\right\} \]\n\nAccording to Corollary 5.5, what is needed ... | Yes |
Lemma 6.5 Let \( I\left( \xi \right) = {\int }_{\mathbb{R}}{e}^{{2\pi i}\left( {{\xi }_{1}t + {\xi }_{2}{t}^{3}}\right) }\psi \left( t\right) {dt} \), where \( \psi \) is a \( {C}^{\infty } \) function of compact support. Then | \[ I\left( \xi \right) = O\left( {\left| \xi \right| }^{-1/3}\right) ,\;\text{ as }\left| \xi \right| \rightarrow \infty . \] Proof. First note that \( I\left( \xi \right) = O\left( {\left| {\xi }_{2}\right| }^{-1/3}\right) \) . In fact \[ I\left( \xi \right) = {\int }_{\left| t\right| \leq {\left| {\xi }_{2}\right| }^... | Yes |
Proposition 6.6 Suppose \( F \) is a \( {C}^{\infty } \) function on \( {\mathbb{R}}^{d} \times \mathbb{R} \) of compact support. Then \( S\left( F\right) \) is a \( {C}^{\infty } \) function that satisfies (43) and (44). | Proof. Write \( F = {e}^{{it}\bigtriangleup }G\left( {\cdot, t}\right) \) with \( G\left( {x, t}\right) = i{\int }_{0}^{t}{e}^{-{is}\bigtriangleup }F\left( {\cdot, s}\right) {ds} \) . Now \( F\left( {\cdot, s}\right) \) is in the Schwartz space \( \mathcal{S}\left( {\mathbb{R}}^{d}\right) \) for each \( s \) and depend... | Yes |
Proposition 6.8 If \( F \in {L}^{p}\left( {{\mathbb{R}}^{d} \times \mathbb{R}}\right) \) then \( S\left( F\right) \) can be corrected (that is, redefined on a set of measure zero) so that for each \( t, S\left( F\right) \left( {\cdot, t}\right) \) belongs to \( {L}^{2}\left( {\mathbb{R}}^{d}\right) \) and, moreover, th... | This is based on the inequality\n\n(52)\n\n\[ \n{\begin{Vmatrix}{\int }_{\alpha }^{\beta }{e}^{-{is}\bigtriangleup }F\left( \cdot, s\right) ds\end{Vmatrix}}_{{L}^{2}\left( {\mathbb{R}}^{d}\right) } \leq c\parallel F{\parallel }_{{L}^{p}\left( {{\mathbb{R}}^{d} \times \mathbb{R}}\right) }, \]\n\nwith \( c \) independent... | Yes |
Proposition 7.2 Under the above assumptions we have \( \\begin{Vmatrix}{T}_{\\lambda }\\end{Vmatrix} \\leq c{\\lambda }^{-d/2} \) , \( \\lambda > 0 \), with \( \\parallel \\cdot \\parallel \) denoting the norm of the operator acting on \( {L}^{2}\\left( {\\mathbb{R}}^{d}\\right) \) . | The proof of the proposition is in many ways like that of the scalar version, Proposition 2.5 in Section 2, so we will be brief. As before, we begin by taking the precaution that \( \\psi \) is supported in a small ball. Now if \( T \) is an operator on \( {L}^{2} \), then \( \\begin{Vmatrix}{{T}^{ * }T}\\end{Vmatrix} ... | Yes |
Proposition 7.4 Assume that\n\n\[ \n\begin{Vmatrix}{{T}_{k}{T}_{j}^{ * }}\end{Vmatrix} \leq {a}^{2}\left( {k - j}\right) \;\text{ and }\;\begin{Vmatrix}{{T}_{k}^{ * }{T}_{j}}\end{Vmatrix} \leq {a}^{2}\left( {k - j}\right) .\n\]\n\nThen for every \( r \) ,\n\n(72)\n\n\[ \n\begin{Vmatrix}{\mathop{\sum }\limits_{{k = 0}}^... | Proof. We write \( T = \mathop{\sum }\limits_{{k = 0}}^{r}{T}_{k} \) and recall that \( \parallel T{\parallel }^{2} = \begin{Vmatrix}{T{T}^{ * }}\end{Vmatrix} \) . Since \( T{T}^{ * } \) is self-adjoint we may use this identity repeatedly to obtain \( \parallel T{\parallel }^{2n} = \) \( \begin{Vmatrix}{\left( T{T}^{ *... | Yes |
Proposition 8.1 \( \mathop{\sum }\limits_{{k = 1}}^{\mu }{r}_{2}\left( k\right) = {\pi \mu } + O\left( {\mu }^{1/2}\right) \), as \( \mu \rightarrow \infty \) . | The proof depends on the realization that \( \mathop{\sum }\limits_{{k = 0}}^{\mu }{r}_{2}\left( k\right) \) represents the number of lattice points in the disc of radius \( R \) with \( {R}^{2} = \mu \) . In fact, with \( {\mathbb{Z}}^{2} \) denoting the lattice points in \( {\mathbb{R}}^{2} \), that is, the points in... | Yes |
Proposition 8.2 Suppose \( f \) belongs to the Schwartz space \( \mathcal{S}\left( {\mathbb{R}}^{d}\right) \). Then\n\n\[ \mathop{\sum }\limits_{{n \in {\mathbb{Z}}^{d}}}f\left( n\right) = \mathop{\sum }\limits_{{n \in {\mathbb{Z}}^{d}}}\widehat{f}\left( n\right) \]\n\nHere \( {\mathbb{Z}}^{d} \) denotes the collection... | For the proof consider two sums\n\n\[ \mathop{\sum }\limits_{{n \in {\mathbb{Z}}^{d}}}f\left( {x + n}\right) \;\text{ and }\;\mathop{\sum }\limits_{{n \in {\mathbb{Z}}^{d}}}\widehat{f}\left( n\right) {e}^{{2\pi in} \cdot x}. \]\n\nBoth are rapidly converging series (since \( f \) and \( \widehat{f} \) are in \( \mathca... | Yes |
Theorem 8.3 \( N\left( R\right) = \pi {R}^{2} + O\left( {R}^{2/3}\right) \), as \( R \rightarrow \infty \) . | Proof. We replace the characteristic function \( {\chi }_{R} \) by a regularized version as follows. We fix a non-negative \ | No |
Theorem 8.5\n\n\[ \mathop{\sum }\limits_{{k = 1}}^{\mu }d\left( k\right) = \mu \log \mu + \left( {{2\gamma } - 1}\right) \mu + O\left( {{\mu }^{1/3}\log \mu }\right) \;\text{ as }\mu \rightarrow \infty .{}^{16} \] | Now as much as we might wish to follow the lines of the proof of Theorem 8.3, there are serious obstacles that seem to stand in the way. In fact, if \( {\chi }_{\mu } \) is the characteristic function of the region\n\n(90)\n\n\[ \left\{ {\left( {{x}_{1},{x}_{2}}\right) \in {\mathbb{R}}^{2} : {x}_{1}{x}_{2} \leq \mu ,{x... | No |
Corollary 8.7 The conclusions for \( {\mathfrak{J}}_{a, b}^{ - } \) are the same as those for \( {\mathfrak{J}}_{a, b}^{ + } \) stated in Proposition 8.6, except that (i) should be modified to read that uniformly in \( a, b \) , \[ \left( {\mathrm{i}}^{\prime }\right) {\mathfrak{J}}_{a, b}^{ - } = O\left( {\left| \lamb... | The only change occurs in the treatment of \( {II} \), namely \( \int {e}^{{i\lambda \Phi }\left( u\right) }\alpha \left( u\right) \frac{du}{u} \) , where now \( \Phi \left( u\right) = u - 1/u \) . In this case \( {\Phi }^{\prime }\left( u\right) = 1 + 1/{u}^{2} > 1 \), and there is no critical point. So Proposition 2.... | Yes |
Theorem 8.9 Let \( \widehat{f} \) be the Fourier transform of \( f\left( {x, y}\right) = {f}_{0}\left( {xy}\right) \) . Then \( \widehat{f} \) is a continuous function where \( {\xi \eta } \neq 0 \) . It is given by\n\n\[ \widehat{f}\left( {\xi ,\eta }\right) = 2{\int }_{0}^{\infty }{\mathfrak{J}}^{ + }\left( {-{2\pi }... | Proof. We approximate \( f \) by \( {f}_{\epsilon } \), with \( {f}_{\epsilon }\left( {x, y}\right) = {f}_{0}\left( {xy}\right) {\eta }_{\epsilon }\left( x\right) {\eta }_{\epsilon }\left( y\right) \) . Then each \( {f}_{\epsilon } \) is a \( {C}^{\infty } \) function of compact support, and clearly \( {f}_{\epsilon } ... | Yes |
Corollary 8.10 The Fourier transforms \( {\widehat{f}}_{\epsilon } \) and \( \widehat{f} \) satisfy the following estimate, uniformly in \( \epsilon \) :\n\n\[ \left| {{\widehat{f}}_{\epsilon }\left( {\xi ,\eta }\right) }\right| \leq {A}_{N}{\left| \xi \eta \right| }^{-N}\;\text{ when }\left| {\xi \eta }\right| \geq 1/... | This is a consequence of the asymptotic behavior of \( {\mathfrak{J}}^{ \pm }\left( \lambda \right) \) for \( \lambda \) as given in Proposition 8.6 and its corollary together with the fact that \( {\int }_{0}^{\infty }{e}^{-{4\pi i\rho }{\left| \xi \eta \right| }^{1/2}}{f}_{0}\left( {\rho }^{2}\right) {\rho d\rho } \)... | Yes |
Theorem 8.11\n\n\[ \mathop{\sum }\limits_{{k = 1}}^{\infty }{f}_{0}\left( k\right) d\left( k\right) = {\int }_{0}^{\infty }\left( {\log \rho + {2\gamma }}\right) {f}_{0}\left( \rho \right) {d\rho } + \mathop{\sum }\limits_{{k = 1}}^{\infty }{F}_{0}\left( k\right) d\left( k\right) ,\] | Proof. We apply the Poisson summation formula\n\n\[ \mathop{\sum }\limits_{{\mathbb{Z}}^{2}}{f}_{\epsilon }\left( {m, n}\right) = \mathop{\sum }\limits_{{\mathbb{Z}}^{2}}{\widehat{f}}_{\epsilon }\left( {m, n}\right) \]\n\nto the approximating functions \( {f}_{\epsilon } \), and then pass to the limit as \( \epsilon \r... | 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 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 [0] and unity [1], 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} ... | 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 \( s \) , \n\n\[ \n\mathop{\limsup }\limits_{{n \rightarrow \infty }}{s}_{n} = \mathop{\lim }\limits_{{n \rightarrow \infty }}{S}_{n} = \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {\mathop{\sup }\limits_{{m \geq n}}{s}_{m}}\right) .\n\] | Proof. First we show that \n\n\[ \n\overline{\lim } \leq {S}_{m}\;\text{ for all }m \in \mathbb{N}.\n\] \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 ... | 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\[ \text{“}\mathop{\lim }\limits_{{h \rightarrow 0}}g\left( h\right) = L\text{”} \]\n\ngiven in Section 7.3.\n\nThus when \( f \) is differentiable (i.e., has a derivative) at \( x \), we have\n... | 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}... | This last equation elucidates the role of the derivative function \( {f}^{\prime } \) as the best linear approximation to the function \( f \) at \( x \) . For the graph of the linear function\n\n\[ \nl\left( {\Delta x}\right) = f\left( x\right) + {f}^{\prime }\left( x\right) {\Delta x}\n\]\n\ngives the tangent to \( f... | Yes |
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\[{\Delta f} = {df} + {\varepsilon \Delta x} + {\delta \Delta y}\]\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\[{\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... | 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. | (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 ... | Yes |
Theorem 10.2.2 For any real number \( r \) ,\n\n\[\n\operatorname{hal}\left( r\right) = \bigcap \{ {}^{ * }A : r \in A\\text{ and }A\\text{ is open }\} .\n\] | 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\[ \left( {\forall x \in B}\right) \left( {\left| x\right| \leq b}\right) \text{.} \]\n\nNow, to prov... | 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... | No |
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 \( n \in \mathbb{N} \), \[ \varnothing \neq {A}_{n} \subseteq \mathbb{N} \] and so \( {A}_{n} \) has a least member \( {r}_{n} \) . This defines a poi... | 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.,\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 \( {}^{ * }A \) . 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\[ \n\{ R\left\lbrack x\right\rbrack : x \in \operatorname{dom}R\}\n\]... | 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 |
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 \( \... | 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} > \\... | 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}... | 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 |
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