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Corollary 1. Let \( E \) and the functions \( {F}_{n} : {E}^{1} \rightarrow E \) be as in Theorem 1. Suppose the series\n\n\[ \sum {F}_{n}\left( p\right) \]\n\nconverges for some \( p \in I \), and\n\n\[ \sum {F}_{n}^{\prime } \]\n\nconverges uniformly on \( J - Q \), for each finite subinterval \( J \subseteq I \). \n\nThen \( \sum {F}_{n} \) converges uniformly on each such \( J \), and\n\n\[ F = \mathop{\sum }\limits_{{n = 1}}^{\infty }{F}_{n} \]\n\nis differentiable on \( I - Q \), with\n\n\[ {F}^{\prime } = {\left( \mathop{\sum }\limits_{{n = 1}}^{\infty }{F}_{n}\right) }^{\prime } = \mathop{\sum }\limits_{{n - 1}}^{\infty }{F}_{n}^{\prime }\text{ there. } \]\n\n(7)\n\nIn other words, the series can be differentiated termwise.
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Proof. Let\n\n\[ {s}_{n} = \mathop{\sum }\limits_{{k = 1}}^{n}{F}_{k},\;n = 1,2,\ldots, \]\n\nbe the partial sums of \( \sum {F}_{n} \) . From our assumptions, it then follows that the \( {s}_{n} \) satisfy all conditions of Theorem 1. (Verify!) Thus the conclusions of Theorem 1 hold, with \( {F}_{n} \) replaced by \( {s}_{n} \) .\n\nWe have \( F = \lim {s}_{n} \) and \( {F}^{\prime } = {\left( \lim {s}_{n}\right) }^{\prime } = \lim {s}_{n}^{\prime } \), whence (7) follows.
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No
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Corollary 2. If \( E \) and the \( {f}_{n} \) are as in Theorem 2 and if \( \sum {f}_{n} \) converges uniformly to \( f \) on each finite interval \( J \subseteq I \), then \( \int f \) exists on \( I \), and
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\[ {\int }_{p}^{x}f = {\int }_{p}^{x}\mathop{\sum }\limits_{{n = 1}}^{\infty }{f}_{n} = \mathop{\sum }\limits_{{n = 1}}^{\infty }{\int }_{p}^{x}{f}_{n}\text{ for any }p, x \in I. \]
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Yes
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Corollary 3. Let a function \( f : {E}^{1} \rightarrow {E}^{m}\left( {{}^{ * }{C}^{m}}\right) \) be relatively continuous on \( \left\lbrack {p,{x}_{0}}\right\rbrack \left( {\text{or}\left\lbrack {{x}_{0}, p}\right\rbrack }\right) ,{x}_{0} \neq p{.}^{4} \) If\n\n\[ f\left( x\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{\left( x - p\right) }^{n}\text{ for }p \leq x < {x}_{0}\text{ (respectively,}{x}_{0} < x \leq p\text{),} \]\n\nand if \( \sum {a}_{n}{\left( {x}_{0} - p\right) }^{n} \) converges, then necessarily\n\n\[ f\left( {x}_{0}\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{\left( {x}_{0} - p\right) }^{n}. \]
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The proof is sketched in Problems 4 and 5.
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No
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Theorem 1. Let the functions \( f, g, h \) be real or complex (or let \( f, g \) be vector valued and \( h \) scalar valued).\n\nIf they are regulated on \( I \), so are \( f \pm g,{fh} \), and \( \left| f\right| \) ; so also is \( f/h \) if \( h \) is bounded away from 0 on \( I \), i.e., \( \left( {\exists \varepsilon > 0}\right) \left| h\right| \geq \varepsilon \) on \( I \) .
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The proof, based on the usual limit properties, is left to the reader.
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No
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Lemma 1 (Heine-Borel). If a closed interval \( A = \left\lbrack {a, b}\right\rbrack \) in \( {E}^{1} \) (or \( {E}^{n} \) ) is covered by open sets \( {G}_{i}\left( {i \in I}\right) \), i.e.,
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The proof was sketched in Problem 10 of Chapter 4, §6.
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No
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Lemma 2. If a function \( f : {E}^{1} \rightarrow T \) is regulated on \( I = \left\lbrack {a, b}\right\rbrack \), then \( f \) can be uniformly approximated by simple step functions on \( I \) .
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Proof. By assumption, \( f\left( {p}^{ - }\right) \) exists for each \( p \in (a, b\rbrack \), and \( f\left( {p}^{ + }\right) \) exists for \( p \in \lbrack a, b) \), all finite.\n\nThus, given \( \varepsilon > 0 \) and any \( p \in I \), there is \( {G}_{p}\left( \delta \right) \) ( \( \delta \) depending on \( p \) ) such that \( \rho \left( {f\left( x\right), r}\right) < \varepsilon \) whenever \( r = f\left( {p}^{ - }\right) \) and \( x \in \left( {p - \delta, p}\right) \), and \( \rho \left( {f\left( x\right), s}\right) < \varepsilon \) whenever \( s = f\left( {p}^{ + }\right) \) and \( x \in \left( {p, p + \delta }\right) ;x \in I \) .\n\nWe choose such a \( {G}_{p}\left( \delta \right) \) for every \( p \in I \) . Then the open globes \( {G}_{p} = {G}_{p}\left( \delta \right) \) cover the closed interval \( I = \left\lbrack {a, b}\right\rbrack \), so by Lemma \( 1, I \) is covered by a finite number of such globes, say,\n\n\[ I \subseteq \mathop{\bigcup }\limits_{{k = 1}}^{n}{G}_{{p}_{k}}\left( {\delta }_{k}\right) ,\;a \in {G}_{{p}_{1}}, a \leq {p}_{1} < {p}_{2} < \cdots < {p}_{n} \leq b. \]\n\nWe now define the step function \( g \) on \( I \) as follows.\n\nIf \( x = {p}_{k} \), we put\n\n\[ g\left( x\right) = f\left( {p}_{k}\right) ,\;k = 1,2,\ldots, n. \]\n\nIf \( x \in \left\lbrack {a,{p}_{1}}\right) \), then\n\n\[ g\left( x\right) = f\left( {p}_{1}^{ - }\right) . \]\n\nIf \( x \in \left( {{p}_{1},{p}_{1} + {\delta }_{1}}\right) \), then\n\n\[ g\left( x\right) = f\left( {p}_{1}^{ + }\right) \]\n\nMore generally, if \( x \) is in \( {G}_{\neg {p}_{k}}\left( {\delta }_{k}\right) \) but in none of the \( {G}_{{p}_{i}}\left( {\delta }_{i}\right), i < k \), we put\n\n\[ g\left( x\right) = f\left( {p}_{k}^{ - }\right) \;\text{if}x < {p}_{k} \]\n\nand\n\n\[ g\left( x\right) = f\left( {p}_{k}^{ + }\right) \;\text{ if }x > {p}_{k} \]\n\nThen by construction, \( \rho \left( {f, g}\right) < \varepsilon \) on each \( {G}_{{p}_{k}} \), hence on \( I \) .
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Yes
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Theorem 2. If \( f : {E}^{1} \rightarrow E \) is regulated on an interval \( I \subseteq {E}^{1} \) and if \( E \) is complete, then \( \int f \) exists on \( I \), exact at every continuity point of \( f \) in \( {I}^{0} \) .
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Proof. In view of Problem 14 of \( §5 \), it suffices to consider closed intervals.\n\nThus let \( I = \left\lbrack {a, b}\right\rbrack, a < b \), in \( {E}^{1} \) . Suppose first that \( f \) is the char-\n\n\n\nFIGURE 25\n\nacteristic function \( {C}_{J} \) of a subinterval \( J \subseteq I \) with endpoints \( c \) and \( d \) \( \left( {a \leq c \leq d \leq b}\right) \), so \( f = 1 \) on \( J \) , and \( f = 0 \) on \( I - J \) . We then define \( F\left( x\right) = x \) on \( J, F = c \) on \( \left\lbrack {a, c}\right\rbrack \), and \( F = d \) on \( \left\lbrack {d, b}\right\rbrack \) (see Figure 25). Thus \( F \) is continuous (why?), and \( {F}^{\prime } = f \) on \( I - \{ a, b, c, d\} \) (why?). Hence \( F = \int f \) on \( I \) ; i.e., characteristic functions are integrable.\n\nThen, however, so is any simple step function\n\n\[ f = \mathop{\sum }\limits_{{k = 1}}^{m}{a}_{k}{C}_{{I}_{k}} \]\n\nby repeated use of Corollary 1 in \( §5.{}^{4} \)\n\nFinally, let \( f \) be any regulated function on \( I \) . Then by Lemma 2, for any \( {\varepsilon }_{n} = \frac{1}{n} \), there is a simple step function \( {g}_{n} \) such that\n\n\[ \mathop{\sup }\limits_{{x \in I}}\left| {{g}_{n}\left( x\right) - f\left( x\right) }\right| \leq \frac{1}{n},\;n = 1,2,\ldots \]\n\nAs \( \frac{1}{n} \rightarrow 0 \), this implies that \( {g}_{n} \rightarrow f \) (uniformly) on \( I \) (see Chapter 4, \( §{12} \) , Theorem 1). Also, by what was proved above, the step functions \( {g}_{n} \) have\n\n--- \n\n\( {}^{4} \) The corollary applies here also if the \( {a}_{k} \) are vectors \( \left( {C}_{{I}_{k}}\right. \) is scalar valued \( ) \).\n\n---\n\nantiderivatives, hence so has \( f \) (Theorem 2 in \( §9 \) ); i.e., \( F = \int f \) exists on \( I \), as claimed. Moreover, \( \int f \) is exact at continuity points of \( f \) in \( {I}^{0} \) (Problem 10 in §5).
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Yes
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Theorem 4 (second law of the mean). Suppose \( f \) and \( g \) are real, \( f \) is monotone with \( f = \int {f}^{\prime } \) on \( I \), and \( g \) is regulated on \( I;I = \left\lbrack {a, b}\right\rbrack \) . Then\n\n\[{\int }_{a}^{b}{fg} = f\left( a\right) {\int }_{a}^{q}g + f\left( b\right) {\int }_{q}^{b}g\text{ for some }q \in I.\]
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Proof. To fix ideas, let \( f \uparrow \) ; i.e., \( {f}^{\prime } \geq 0 \) on \( I \) .\n\nThe formula \( f = \int {f}^{\prime } \) means that \( f \) is relatively continuous (hence regulated) on \( I \) and differentiable on \( I - Q \) ( \( Q \) countable). As \( g \) is regulated,\n\n\[{\int }_{a}^{x}g = G\left( x\right)\]\ndoes exist on \( I \), so \( G \) has similar properties, with \( G\left( a\right) = {\int }_{a}^{a}g = 0 \).\n\nBy Theorems 1 and 2, \( \int f{G}^{\prime } = \int {fg} \) exists on \( I \) . (Why?) Hence by Corollary 5 in \( §5 \), so does \( \int G{f}^{\prime } \), and we have\n\n\[{\int }_{a}^{b}{fg} = {\int }_{a}^{b}f{G}^{\prime } = {\left. f\left( x\right) G\left( x\right) \right| }_{a}^{b} - {\int }_{a}^{b}G{f}^{\prime } = f\left( b\right) G\left( b\right) - {\int }_{a}^{b}G{f}^{\prime }.\]\n\nNow \( G \) has the Darboux property on \( I \) (being relatively continuous), and\n\n\( {}^{5} \) One can also assume \( g \leq 0 \) on \( I \) ; in this case, simply apply the theorem to \( - g \) .\n\n\( {f}^{\prime } \geq 0 \) . Also, \( \int G \) and \( \int G{f}^{\prime } \) exist on \( I \) . Thus by Problems 7 and 8 in \( §5 \),\n\n\[{\int }_{a}^{b}G{f}^{\prime } = {\left. G\left( q\right) {\int }_{a}^{b}{f}^{\prime } = G\left( q\right) f\left( x\right) \right| }_{a}^{b},\;q \in I.\]\n\nCombining all, we obtain the required result (1) since\n\n\[\int {fg} = f\left( b\right) G\left( b\right) - {\int }_{a}^{b}G{f}^{\prime }\]\n\n\[= f\left( b\right) G\left( b\right) - f\left( b\right) G\left( q\right) + f\left( a\right) G\left( q\right)\]\n\n\[= f\left( b\right) {\int }_{q}^{b}g + f\left( a\right) {\int }_{a}^{q}g.\]
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Yes
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Theorem 5 (integral test of convergence). If \( f : {E}^{1} \rightarrow {E}^{1} \) is nonnegative and nonincreasing on \( I = \lbrack a, + \infty ) \), then\n\n\[ \n{\int }_{a}^{\infty }f\text{ converges iff }\mathop{\sum }\limits_{{n = 1}}^{\infty }f\left( n\right) \text{ does. } \n\]
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Proof. As \( f \downarrow, f \) is regulated, so \( \int f \) exists on \( I = \lbrack a, + \infty ) \) . We fix some natural \( k \geq a \) and define\n\n\[ \nF\left( x\right) = {\int }_{k}^{x}f\text{ for }x \in I. \n\]\n\nBy Theorem 3(iii) in \( §5, F \uparrow \) on \( I \) . Thus by monotonicity,\n\n\[ \n\mathop{\lim }\limits_{{x \rightarrow + \infty }}F\left( x\right) = \mathop{\lim }\limits_{{x \rightarrow + \infty }}{\int }_{k}^{x}f = {\int }_{k}^{\infty }f \n\]\n\nexists in \( {E}^{ * } \) ; so does \( {\int }_{a}^{k}f \) . Since\n\n\[ \n{\int }_{a}^{x}f = {\int }_{a}^{k}f + {\int }_{k}^{x}f \n\]\n\nwhere \( {\int }_{a}^{k}f \) is finite by definition, we have\n\n\[ \n{\int }_{a}^{\infty }f < + \infty \;\text{ iff }\;{\int }_{k}^{\infty }f < + \infty . \n\]\n\nSimilarly,\n\n\[ \n\mathop{\sum }\limits_{{n = 1}}^{\infty }f\left( n\right) < + \infty \;\text{ iff }\;\mathop{\sum }\limits_{{n = k}}^{\infty }f\left( n\right) < + \infty . \n\]\n\nThus we may replace \
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Yes
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Theorem 2. F has the limits\n\n\\[ \nF\\left( {1}^{ - }\\right) = \\frac{\\pi }{2}\\text{ and }F\\left( {-{1}^{ + }}\\right) = - \\frac{\\pi }{2}.\n\\]\n\nThus \\( F \\) becomes relatively continuous on \\( \\bar{I} = \\left\\lbrack {-1,1}\\right\\rbrack \\) if one sets\n\n\\[ \nF\\left( 1\\right) = \\frac{\\pi }{2}\\text{ and }F\\left( {-1}\\right) = - \\frac{\\pi }{2}\n\\]\n\ni.e.,\n\n\\[ \n\\arcsin 1 = \\frac{\\pi }{2}\\text{ and }\\arcsin \\left( {-1}\\right) = - \\frac{\\pi }{2}.\n\\]
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Proof. We have\n\n\\[ \nF\\left( x\\right) = {\\int }_{0}^{x}f = {\\int }_{0}^{c}f + {\\int }_{c}^{x}f,\\;c = \\sqrt{\\frac{1}{2}}.\n\\]\n\nBy substituting \\( s = \\sqrt{1 - {t}^{2}} \\) in the last integral and setting, for brevity, \\( y = \\) \\( \\sqrt{1 - {x}^{2}} \\), we obtain\n\n\\[ \n{\\int }_{c}^{x}f = {\\int }_{c}^{x}\\frac{1}{\\sqrt{1 - {t}^{2}}}{dt} = {\\int }_{y}^{c}\\frac{1}{\\sqrt{1 - {s}^{2}}}{ds} = F\\left( c\\right) - F\\left( y\\right) .\\;\\text{ (Verify!) }\n\\]\n\nNow as \\( x \\rightarrow {1}^{ - } \\), we have \\( y = \\sqrt{1 - {x}^{2}} \\rightarrow 0 \\), and hence \\( F\\left( y\\right) \\rightarrow F\\left( 0\\right) = 0 \\) (for \\( F \\) is continuous at 0 ). Thus\n\n\\[ \nF\\left( {1}^{ - }\\right) = \\mathop{\\lim }\\limits_{{x \\rightarrow {1}^{ - }}}F\\left( x\\right) = \\mathop{\\lim }\\limits_{{y \\rightarrow 0}}\\left( {{\\int }_{0}^{c}f + {\\int }_{y}^{c}f}\\right) = {\\int }_{0}^{c}f + F\\left( c\\right) = 2{\\int }_{0}^{c}f = \\frac{\\pi }{2}.\n\\]\n\nSimilarly, one gets \\( F\\left( {-{1}^{ + }}\\right) = - \\pi /2 \\).\n\nThe function \\( F \\) as redefined in Theorem 2 will be denoted by \\( {F}_{0} \\). It is a primitive of \\( f \\) on the closed interval \\( \\bar{I} \\) (exact on \\( I \\) ). Thus \\( {F}_{0}\\left( x\\right) = {\\int }_{0}^{x}f \\) , \\( - 1 \\leq x \\leq 1 \\), and we may now write\n\n\\[ \n\\frac{\\pi }{2} = {\\int }_{0}^{1}f\\text{ and }\\pi = {\\int }_{-1}^{0}f + {\\int }_{0}^{1}f = {\\int }_{-1}^{1}f.\n\\]
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No
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Theorem 3.\n\n(i) Each \( {F}_{n} \) is differentiable on \( I = \left( {-1,1}\right) \) and relatively continuous on \( \bar{I} = \left\lbrack {-1,1}\right\rbrack \) .\n\n(ii) \( {F}_{n} \) is increasing on \( \bar{I} \) if \( n \) is even, and decreasing if \( n \) is odd.\n\n(iii) \( {F}_{n}^{\prime }\left( x\right) = \frac{{\left( -1\right) }^{n}}{\sqrt{1 - {x}^{2}}} \) on \( I \) .\n\n(iv) \( {F}_{n}\left( {-1}\right) = {F}_{n - 1}\left( {-1}\right) = {n\pi } - {\left( -1\right) }^{n}\frac{\pi }{2};{F}_{n}\left( 1\right) = {F}_{n - 1}\left( 1\right) = {n\pi } + {\left( -1\right) }^{n}\frac{\pi }{2} \) .
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The proof is obvious from (12) and the properties of \( {F}_{0} \) . Assertion (iv) ensures\n\n\n\nFIGURE 26\n\nthat the graphs of the \( {F}_{n} \) add up to one curve. By (ii), each \( {F}_{n} \) is one to one (strictly monotone) on \( \bar{I} \) . Thus it has a strictly monotone inverse on the interval \( \overline{{J}_{n}} = {F}_{n}\left\lbrack \left\lbrack {-1,1}\right\rbrack \right\rbrack \), i.e., on the \( {F}_{n} \) -image of \( \bar{I} \) .
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No
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Theorem 4. The sine and cosine functions (s and c) are differentiable, hence continuous, on all of \( {E}^{1} \), with derivatives \( {s}^{\prime } = c \) and \( {c}^{\prime } = - s \) ; that is,\n\n\[{\left( \sin x\right) }^{\prime } = \cos x\\text{ and }{\left( \cos x\right) }^{\prime } = - \sin x.\]
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Proof. It suffices to consider the intervals \( \\overline{{J}_{0}} \) and \( \\overline{{J}_{1}} \), for, by (15), all properties of \( s \) and \( c \) repeat themselves, with period \( {2\\pi } \), on the rest of \( {E}^{1} \). By (13),\n\n\[s = {F}_{0}^{-1}\\text{ on }\\overline{{J}_{0}} = \\left\\lbrack {-\\frac{\\pi }{2},\\frac{\\pi }{2}}\\right\\rbrack\]\n\nwhere \( {F}_{0} \) is differentiable on \( I = \\left( {-1,1}\\right) \). Thus Theorem 3 of \( §2 \) shows that \( s \) is differentiable on \( {J}_{0} = \\left( {-\\pi /2,\\pi /2}\\right) \) and that\n\n\[{s}^{\\prime }\\left( q\\right) = \\frac{1}{{F}_{0}^{\\prime }\\left( p\\right) }\\text{ whenever }p \\in I\\text{ and }q = {F}_{0}\\left( p\\right) ;\]\n\ni.e., \( q \\in J \) and \( p = s\\left( q\\right) \). However, by Theorem 3(iii),\n\n\[{F}_{0}^{\\prime }\\left( p\\right) = \\frac{1}{\\sqrt{1 - {p}^{2}}}.\n\]\n\nHence\n\n\[{s}^{\\prime }\\left( q\\right) = \\sqrt{1 - {\\sin }^{2}q} = \\cos q = c\\left( q\\right) ,\\;q \\in J.\]\n\nThis proves the theorem for interior points of \( \\overline{{J}_{0}} \) as far as \( s \) is concerned.\n\nAs\n\n\[c = \\sqrt{1 - {s}^{2}} = {\\left( 1 - {s}^{2}\\right) }^{\\frac{1}{2}}\\text{ on }{J}_{0}\\text{ (by (13)),}\]\n\nwe can use the chain rule (Theorem 3 in \( §1 \)) to obtain\n\n\[{c}^{\\prime } = \\frac{1}{2}{\\left( 1 - {s}^{2}\\right) }^{-\\frac{1}{2}}\\left( {-{2s}}\\right) {s}^{\\prime } = - s\]\n\non noting that \( {s}^{\\prime } = c = {\\left( 1 - {s}^{2}\\right) }^{\\frac{1}{2}} \) on \( {J}_{0} \). Similarly, using (14), one proves that \( {s}^{\\prime } = c \) and \( {c}^{\\prime } = - s \) on \( {J}_{1} \) (interior of \( \\overline{{J}_{1}} \) ).\n\nNext, let \( q \) be an endpoint, say, \( q = \\pi /2 \). We take the left derivative\n\n\[{s}_{ - }^{\\prime }\\left( q\\right) = \\mathop{\\lim }\\limits_{{x \\rightarrow {q}^{ - }}}\\frac{s\\left( x\\right) - s\\left( q\\right) }{x - q},\\;x \\in {J}_{0}.\n\]\n\nBy L'Hôpital's rule, we get\n\n\[{s}_{ - }^{\\prime }\\left( q\\right) = \\mathop{\\lim }\\limits_{{x \\rightarrow {q}^{ - }}}\\frac{{s}^{\\prime }\\left( x\\right) }{1} = \\mathop{\\lim }\\limits_{{x \\rightarrow {q}^{ - }}}c\\left( x\\right)\]\n\nsince \( {s}^{\\prime } = c \) on \( {J}_{0} \). However, \( s = {F}_{0}^{-1} \) is left continuous at \( q \) (why?); hence so is \( c = \\sqrt{1 - {s}^{2}} \). (Why?) Therefore,\n\n\[{s}_{ - }^{\\prime }\\left( q\\right) = \\mathop{\\lim }\\limits_{{x \\rightarrow {q}^{ - }}}c\\left( x\\right) = c\\left( q\\right) ,\\;\\text{ as required. }\n\]\n\nSimilarly, one shows that \( {s}_{ + }^{\\prime }\\left( q\\right) = c\\left( q\\right) \). Hence \( {s}^{\\prime }\\left( q\\right) = c\\left( q\\right) \) and \( {c}^{\\prime }\\left( q\\right) = - s\\left( q\\right) \), as before.
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Yes
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Problem 33. There is yet another bijection that lets us prove that a set of size \( n \) has \( {2}^{n} \) subsets. Namely, for each subset \( S \) of \( \left\lbrack n\right\rbrack = \{ 1,2,\ldots, n\} \), define a function (traditionally denoted by \( {\chi }_{S} \) ) as follows. \( {}^{a} \n\n\[ \n{\chi }_{S}\left( i\right) = \left\{ \begin{array}{ll} 1 & \text{ if }i \in S \\ 0 & \text{ if }i \notin S \end{array}\right. \n\] \n\nThe function \( {\chi }_{S} \) is called the characteristic function of \( S \) . Notice that the characteristic function is a function from \( \left\lbrack n\right\rbrack \) to \( \{ 0,1\} \) .
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(b) We define a function \( f \) from the set of subsets of \( \left\lbrack n\right\rbrack = \{ 1,2,\ldots, n\} \) to the set of functions from \( \left\lbrack n\right\rbrack \) to \( \{ 0,1\} \) by \( f\left( S\right) = {\chi }_{S} \) . Explain why \( f \) is a bijection.\n\n(c) Why does the fact that \( f \) is a bijection prove that \( \left\lbrack n\right\rbrack \) has \( {2}^{n} \) subsets?
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No
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Theorem 1.2.7. The number of subsets of an n-element set is \( {2}^{n} \) .
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The proofs in Problem 28 and Problem 33 use essentially the same bijection, but they interpret sequences of zeros and ones differently, and so end up being different proofs. We will give yet another proof, using bijections similar to those we used in proving the Pascal Equation, at the beginning of Chapter 2.
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No
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Explain why it is that the number of bijections from an \( n \) - element set to an \( n \) -element set is equal to \( n \) times the number of bijections from an \( \left( {n - 1}\right) \) -element subset to an \( \left( {n - 1}\right) \) -element set. What does this have to do with Problem 27?
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and if \( {b}_{n} \) stands for the number of bijections from an \( n \) -element set to an \( n \) -element set, then\n\n\[ {b}_{n} = n{b}_{n - 1}\text{.} \]\n\n(2.2)
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Yes
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Problem 120. Show that the following algorithm (known as Dijkstra's algorithm) applied to a weighted graph whose vertices are labelled 1 to \( n \) gives, for each \( i \), the distance from vertex 1 to \( \mathrm{i} \) as \( d\left( i\right) \) .
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1. Let \( d\left( 1\right) = 0 \) . Let \( d\left( i\right) = \infty \) for all other \( i \) . Let \( v\left( 1\right) = 1 \) . Let \( v\left( j\right) = 0 \) for all other \( j \) . For each \( i \) and \( j \), let \( w\left( {i, j}\right) \) be the minimum weight of an\n\nedge between \( i \) and \( j \), or \( \infty \) if there are no such edges. Let \( k = 1 \) . Let \( t = 1 \) .\n\n2. For each \( i \), if \( d\left( i\right) > d\left( k\right) + w\left( {k, i}\right) \) let \( d\left( i\right) = d\left( k\right) + w\left( {k, i}\right) \) .\n\n3. Among those \( i \) with \( v\left( i\right) = 0 \), choose one with \( d\left( i\right) \) a minimum, and let \( k = i \) . Increase \( t \) by 1 . Let \( v\left( i\right) = 1 \) .\n\n4. Repeat the previous two steps until \( t = n \
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Yes
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Theorem 3.1.4. If \( 0 \leq k \leq n \), the number of \( k \) -element subsets of an \( n \) -element set is given by\n\n\[ \left( \begin{array}{l} n \\ k \end{array}\right) = \frac{{n}^{\underline{k}}}{k!} = \frac{n!}{k!\left( {n - k}\right) !}. \]
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We define \( \left( \begin{array}{l} n \\ k \end{array}\right) \) to be 0 if \( k > n \), because then there are no \( k \) -element subsets of an \( n \) -element set. Notice that this is what the middle term of the formula in the theorem gives us. This explains the entries of row 8 of our table. For now we jump over row 9.
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No
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Problem 123. Suppose we wish to place the books in Problem 122 (satisfying the assumptions we made there) so that each shelf gets at least one book. Now in how many ways may we place the books? (Hint: how can you make sure that each shelf gets at least one book before you start the process described in Problem 122?) (h)
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The assignment of which books go to which shelves of a bookcase is simply a function from the books to the shelves. But a function doesn't determine which book sits to the left of which others on the shelf, and this information is part of how the books are arranged on the shelves. In other words, the order in which the shelves receive their books matters. Our function must thus assign an ordered list of books to each shelf. We will call such a function an ordered function. More precisely, an ordered function from a set \( S \) to a set \( T \) is a function that assigns an (ordered) list of elements of \( S \) to some, but not necessarily all, elements of \( T \) in such a way that each element of \( S \) appears on one and only one of the lists. \( {}^{1} \) (Notice that although it is not the usual definition of a function from \( S \) to \( T \), a function can be described as an assignment of subsets of \( S \) to some, but not necessarily all, elements of \( T \) so that each element of \( S \) is in one and only one of these subsets.) Thus the number of ways to place the books into the bookcase is the entry in the middle column of row 5 of our table. If in addition we require each shelf to get at least one book, we are discussing the entry in the middle column of row 6 of our table. An ordered onto function is one which assigns a list to each element of \( T \) . In Problem 122 you showed that the number of ordered functions from a \( k \) -element set to an \( n \) -element set is \( \mathop{\prod }\limits_{{i = 1}}^{k}\left( {n + i - 1}\right) \) . This product occurs frequently enough that it has a name; it is called the \( k \) th rising factorial power of \( n \) and is denoted by \( {n}^{\bar{k}} \) . It is read as \
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Yes
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Problem 155. Write down the rows of Stirling's triangle of the first kind for \( k = 0 \) to 6 .
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By definition, the Stirling numbers of the first kind are also change of basis coefficients. The Stirling numbers of the first and second kind are change of basis coefficients from the falling factorial powers of \( x \) to the ordinary factorial powers, and vice versa.
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No
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Problem 233. We get a more elegant proof if we ask for a picture enumerator for \( {A}_{1} \cup {A}_{2} \cup \cdots \cup {A}_{n} \) . so let us assume \( A \) is a set with a picture function \( P \) defined on it and that each set \( {A}_{i} \) is a subset of \( A \) .
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(a) By thinking about how we got the formula for the size of a union, write down instead a conjecture for the picture enumerator of a union. You could use notation like \( {E}_{P}\left( {\mathop{\bigcap }\limits_{{i : i \in S}}{A}_{i}}\right) \) for the picture enumerator of the intersection of the sets \( {A}_{i} \) for \( i \) in a subset of \( S \) of \( \left\lbrack n\right\rbrack \) .\n\n(b) If \( x \in \mathop{\bigcup }\limits_{{i = 1}}^{n}{A}_{i} \), what is the coefficient for \( P\left( x\right) \) in (the inclusion-exclusion side of) your formula for \( {E}_{P}\left( {\mathop{\bigcup }\limits_{{i = 1}}^{n}{A}_{i}}\right) \) ? (h)\n\n(c) If \( x \notin \mathop{\bigcup }\limits_{{i = 1}}^{n}{A}_{i} \), what is the coefficient of \( P\left( x\right) \) in (the inclusion-exclusion side of) your formula for \( {E}_{P}\left( {\mathop{\bigcup }\limits_{{i = 1}}^{n}{A}_{i}}\right) \) ?\n\n(d) How have you proved your conjecture for the picture enumerator of the union of the sets \( {A}_{i} \) ?\n\n(e) How can you get the formula for the principle of inclusion and exclusion from your formula for the picture enumerator of the union?
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No
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Problem 234. Frequently when we apply the principle of inclusion and exclusion, we will have a situation like that of part (d) of Problem 231.d. That is, we will have a set \( A \) and subsets \( {A}_{1},{A}_{2},\ldots ,{A}_{n} \) and we will want the size or the probability of the set of elements in \( A \) that are not in the union. This set is known as the complement of the union of the \( {A}_{i}\mathrm{\;s} \) in \( A \) , and is denoted by \( A \smallsetminus \mathop{\bigcup }\limits_{{i = 1}}^{n}{A}_{i} \), or if \( A \) is clear from context, by \( \overline{\mathop{\bigcup }\limits_{{i = 1}}^{n}{A}_{i}} \) . Give the fomula for \( \overline{\mathop{\bigcup }\limits_{{i = 1}}^{n}{A}_{i}} \) .
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We can find a very elegant way of writing the formula in Problem 234 if we let \( \mathop{\bigcap }\limits_{{i : i \in \varnothing }}{A}_{i} = A \) . for this reason, if we have a family of subsets \( {A}_{i} \) of a set \( A \), we define \( {}^{1}\mathop{\bigcap }\limits_{{i : i \in \varnothing }}{A}_{i} = A \) .
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No
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Problem 272. If the list \( \left( {{i\sigma }\left( i\right) {\sigma }^{2}\left( i\right) \ldots {\sigma }^{n}\left( i\right) }\right) \) does not have repeated elements but the list \( \left( {{i\sigma }\left( i\right) {\sigma }^{2}\left( i\right) \ldots {\sigma }^{n}\left( i\right) {\sigma }^{n + 1}\left( i\right) }\right) \) does have repeated elements, then what is \( {\sigma }^{n + 1}\left( i\right) \) ? \( {}_{\left( \mathrm{h}\right) } \)
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We say \( {\sigma }^{j}\left( i\right) \) is an element of the cycle \( \left( {{i\sigma }\left( i\right) {\sigma }^{2}\left( i\right) \ldots {\sigma }^{n}\left( i\right) }\right) \) . Notice that the case \( j = 0 \) means \( i \) is an element of the cycle. Notice also that if \( j > n,{\sigma }^{j}\left( i\right) = {\sigma }^{j - n - 1}\left( i\right) \) , so the distinct elements of the cycle are \( i,\sigma \left( i\right) ,{\sigma }^{2}\left( i\right) \), through \( {\sigma }^{n}\left( i\right) \) . We think of the cycle \( \left( {{i\sigma }\left( i\right) {\sigma }^{2}\left( i\right) }\right. \ldots \left. {{\sigma }^{n}\left( i\right) }\right) \) as representing the permutation \( \sigma \) restricted to the set of elements of the cycle. We say that the cycles \( \left( {{i\sigma }\left( i\right) {\sigma }^{2}\left( i\right) \ldots {\sigma }^{n}\left( i\right) }\right) \) and \( \left( {{j\sigma }\left( j\right) {\sigma }^{2}\left( j\right) \ldots {\sigma }^{n}\left( j\right) }\right) \) are equivalent if there is an integer \( k \) such that \( j = {\sigma }^{k}\left( i\right) \) .
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No
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Problem 275. If two cycles of \( \sigma \) have an element in common, what can we say about them?
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Problem 275 leads almost immediately to the following theorem.
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No
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Theorem 6.2.3. Suppose a group acts on a set \( S \) . The orbits of \( G \) form a partition of \( S \) .
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It is probably worth pointing out that this theorem tells us that the orbit \( {Gx} \) is also the orbit \( {Gy} \) for any element \( y \) of \( {Gx} \) .
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No
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Problem 290. Complete the proof of Theorem 6.2.3.
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Notice that thinking in terms of orbits actually hides some information about the action of our group. When we computed the multiset of all results of acting on \( \{ 1,2\} \) with the elements of \( {D}_{4} \), we got an eight-element multiset containing each side twice. When we computed the multiset of all results of acting on \( \{ 1,3\} \) with the elements of \( {D}_{4} \), we got an eight-element multiset containing each diagonal of the square four times. These multisets remind us that we are acting on our two-element sets with an eight-element group. The multiorbit of \( G \) determined by an element \( x \) of \( S \) is the multiset\n\n\[ \n\{ \bar{\sigma }\left( x\right) \mid \sigma \in G\} \n\]\n\nand is denoted by \( G{x}_{\text{multi }} \) .\n\nWhen we used the quotient principle to count circular seating arrangements or necklaces, we partitioned up a set of lists of people or beads into blocks of equivalent lists. In the case of seating \( n \) people around a round table, what made two lists equivalent was, in retrospect, the action of the rotation group \( {C}_{n} \) . In the case of stringing \( n \) beads on a string to make a necklace, what made two lists equivalent was the action of the dihedral group. Thus the blocks of our partitions were orbits of the rotation group or the dihedral group, and we were counting the number of orbits of the group action. In Problem 45, we were not able to apply the quotient principle because we had blocks of different sizes. However, these blocks were still orbits of the action of the group \( {D}_{4} \) . And, even though the orbits have different sizes, we expect that each orbit corresponds naturally to a multiorbit and that the multiorbits all have the same size. Thus if we had a version of the quotient rule for a union of multisets, we could hope to use it to count the number of multiorbits.
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No
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Problem 300. A second computation of the result of Problem 299 can be done as follows.\n\n(a) Let \( \widehat{\chi }\left( {\sigma, x}\right) = 1 \) if \( \sigma \left( x\right) = x \) and let \( \widehat{\chi }\left( {\sigma, x}\right) = 0 \) otherwise. Notice that \( \widehat{\chi } \) is different from the \( \chi \) in the previous problem, because it is a function of two variables. Use \( \widehat{\chi } \) to convert the single summation in Problem 298 into a double summation over elements \( x \) of \( S \) and elements \( \sigma \) of \( G \) .
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(b) Reverse the order of the previous summation in order to convert it into a single sum involving the function \( \chi \) given by\n\n\[ \chi \left( \sigma \right) = \text{the number of elements of}S\text{left fixed by}\sigma \text{.} \]
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No
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Problem 331. If \( f : S \rightarrow T \) is a function, we say that \( f \) maps \( x \) to \( y \) as another way to say that \( f\left( x\right) = y \) . Suppose \( S = T = \{ 1,2,3\} \) . Give a function from \( S \) to \( T \) that is not onto.
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Notice that two different members of \( S \) have mapped to the same element of \( T \) . Thus when we say that \( f \) associates one and only one element of \( T \) to each element of \( S \), it is quite possible that the one and only one element \( f\left( 1\right) \) that \( f \) maps 1 to is exactly the same as the one and only one element \( f\left( 2\right) \) that \( f \) maps 2 to.
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No
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If we reverse all the arrows in the digraph of a bijection \( f \), we get the digraph of another function \( g \) . Is \( g \) a bijection? What is \( f\left( {g\left( x\right) }\right) \) ? What is \( g\left( {f\left( x\right) }\right) \) ?
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If \( f \) is a function from \( S \) to \( T \), if \( g \) is a function from \( T \) to \( S \), and if \( f\left( {g\left( x\right) }\right) = x \) for each \( x \) in \( T \) and \( g\left( {f\left( x\right) }\right) = x \) for each \( x \) in \( S \), then we say that \( g \) is an inverse of \( f \) (and \( f \) is an inverse of \( g \) ).
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No
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Explain why a bijection must have an inverse.
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Since a function with an inverse has exactly one inverse \( g \), we call \( g \) the inverse of \( f \) . From now on, when \( f \) has an inverse, we shall denote its inverse by \( {f}^{-1} \) . Thus \( f\left( {{f}^{-1}\left( x\right) }\right) = x \) and \( {f}^{-1}\left( {f\left( x\right) }\right) = x \) . Equivalently \( f \circ {f}^{-1} = \iota \) and \( {f}^{-1} \circ f = \iota \) .
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No
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Suppose that \( R \) is an equivalence relation on a set \( X \) and for each \( x \in X \), let \( {C}_{x} = \{ y \mid y \in X \) and \( {yRx}\} \) . If \( {C}_{x} \) and \( {C}_{z} \) have an element \( y \) in common, what can you conclude about \( {C}_{x} \) and \( {C}_{z} \) (besides the fact that they have an element in common!)?
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In Problem 352 the sets \( {C}_{x} \) are called equivalence classes of the equivalence relation \( R \) . You have just proved that if \( R \) is an equivalence relation of the set \( X \), then each element of \( X \) is in exactly one equivalence class of \( R \) . Recall that a partition of a set \( X \) is a set of disjoint sets whose union is \( X \) . For example, \( \{ 1,3\} \) , \( \{ 2,4,6\} ,\{ 5\} \) is a partition of the set \( \{ 1,2,3,4,5,6\} \) . Thus another way to describe what you proved in Problem 352 is the following:\n\nTheorem A.2.1. If \( R \) is an equivalence relation on \( X \), then the set of equivalence classes of \( R \) is a partition of \( X \) .
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No
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What postage do you think we can make with five and six cent stamps? Is there a number \( N \) such that if \( n \geq N \), then we can make \( n \) cents worth of postage?
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You probably see that we can make \( n \) cents worth of postage as long as \( n \) is at least 20. However you didn't try to make 26 cents in postage by working with 25 cents; rather you saw that you could get 20 cents and then add six cents to that to get 26 cents. Thus if we want to prove by induction that we are right that if \( n \geq {20} \) , then we can make \( n \) cents worth of postage, we are going to have to use the strong version of the principle of mathematical induction.\n\nWe know that we can make 20 cents with four five-cent stamps. Now we let \( k \) be a number greater than 20 , and assume that it is possible to make any amount between 20 and \( k - 1 \) cents in postage with five and six cent stamps. Now if \( k \) is less than 25, it is 21, 22, 23, or 24 . We can make 21 with three fives and one six. We can make 22 with two fives and two sixes, 23 with one five and three sixes, and 24 with four sixes. Otherwise \( k - 5 \) is between 20 and \( k - 1 \) (inclusive) and so by our inductive hypothesis, we know that \( k - 5 \) cents can be made with five and six cent stamps, so with one more five cent stamp, so can \( k \) cents. Thus by the (strong) principle of mathematical induction, we can make \( n \) cents in stamps with five and six cent stamps for each \( n \geq {20} \).
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Yes
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If \( \mathcal{F} \) and \( \mathcal{G} \) are species of subsets of \( X \), how is the EGF for \( \mathcal{F} \cdot \mathcal{G} \) related to the EGFs for \( F \) and \( G \) ?
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Prove you are right. (h)
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No
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Problem 398. Without giving the proof, how can you compute the EGF \( f\left( x\right) \) for the number of structures using a set of size \( n \) in the species \( {\mathcal{F}}_{1} \cdot {\mathcal{F}}_{2}\cdots \cdot {\mathcal{F}}_{k} \) of structures on \( k \) -tuples of subsets of \( X \) from the EGFs \( {f}_{i}\left( x\right) \) for \( {\mathcal{F}}_{i} \) for each \( i \) from 1 to \( k \) ? (Here we are using the natural extension of the idea of the pair of structures to the idea of a \( k \) -tuple structure.)
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Theorem C.4.1. If \( {\mathcal{F}}_{1},{\mathcal{F}}_{2},\ldots ,{\mathcal{F}}_{n} \) are species of subsets of the set \( X \) and \( {\mathcal{F}}_{i} \) has EGF \( {f}_{i}\left( x\right) \) , then the family of \( k \) -tuple structures \( {\mathcal{F}}_{1} \cdot {\mathcal{F}}_{2}\cdots \cdot {\mathcal{F}}_{n} \) has EGF \( \mathop{\prod }\limits_{{i = 1}}^{n}{f}_{i}\left( x\right) \) .
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Yes
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Lemma 8.8. If \( \bigtriangleup {ABC} \) is nondegenerate and its angle bisector at \( A \) intersects \( \left\lbrack {BC}\right\rbrack \) at the point \( D \) . Then\n\n\[ \frac{AB}{AC} = \frac{DB}{DC}. \]
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Proof. Applying Claim 20.22, we get that\n\n\n\n\[ \frac{\operatorname{area}\left( {\bigtriangleup {ABD}}\right) }{\operatorname{area}\left( {\bigtriangleup {ACD}}\right) } = \frac{BD}{CD} \]\n\nBy Proposition 8.10 the triangles \( {ABD} \) and \( {ACD} \) have equal altitudes from \( D \) . Applying Claim 20.22 again, we get that\n\n\[ \frac{\operatorname{area}\left( {\bigtriangleup {ABD}}\right) }{\operatorname{area}\left( {\bigtriangleup {ACD}}\right) } = \frac{AB}{AC} \]\n\nand hence the result.
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Yes
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Theorem 1.1.2 Let \( A, B \), and \( C \) be subsets of a universal set \( X \) . Then the following hold:\n\n(a) \( A \cup {A}^{c} = X \) ;\n\n(b) \( A \cap {A}^{c} = \varnothing \) ;\n\n(c) \( {\left( {A}^{c}\right) }^{c} = A \) ;\n\n(d) (Distributive law) \( A \cap \left( {B \cup C}\right) = \left( {A \cap B}\right) \cup \left( {A \cap C}\right) \) ;\n\n(e) (Distributive law) \( A \cup \left( {B \cap C}\right) = \left( {A \cup B}\right) \cap \left( {A \cup C}\right) \) ;\n\n(f) (DeMorgan’s law) \( A \smallsetminus \left( {B \cup C}\right) = \left( {A \smallsetminus B}\right) \cap \left( {A \smallsetminus C}\right) \) ;\n\n(g) (DeMorgan’s law) \( A \smallsetminus \left( {B \cap C}\right) = \left( {A \smallsetminus B}\right) \cup \left( {A \smallsetminus C}\right) \) ;\n\n(h) \( A \smallsetminus B = A \cap {B}^{c} \) .
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Proof: We prove some of the results and leave the rest for the exercises.\n\n(a) Clearly, \( A \cup {A}^{c} \subset X \) since both \( A \) and \( {A}^{c} \) are subsets of \( X \) . Now let \( x \in X \) . Then either \( x \) is an element of \( A \) or it is not an element of \( A \) . In the first case, \( x \in A \) and, so, \( x \in A \cup {A}^{c} \) . In the second case, \( x \in {A}^{c} \) and, so, \( x \in A \cup {A}^{c} \) . Thus, \( X \subset A \cup {A}^{c} \) .\n\n(b) No element of \( x \) can be simultaneously in \( A \) and not in \( A \) . Thus, \( A \cap {A}^{c} = \varnothing \) .\n\n(d) Let \( x \in A \cap \left( {B \cup C}\right) \) . Then \( x \in A \) and \( x \in B \cup C \) . Therefore, \( x \in B \) or \( x \in C \) . In the first case, since \( x \) is also in \( A \) we get \( x \in A \cap B \) and, hence, \( x \in \left( {A \cap B}\right) \cup \left( {A \cap C}\right) \) . In the second case, \( x \in A \cap C \) and, hence, \( x \in \left( {A \cap B}\right) \cup \left( {A \cap C}\right) \) . Thus, in all cases, \( x \in \left( {A \cap B}\right) \cup \left( {A \cap C}\right) \) . This shows \( A \cap \left( {B \cup C}\right) \subset \left( {A \cap B}\right) \cup \left( {A \cap C}\right) .\n\nNow we prove the other inclusion. Let \( x \in \left( {A \cap B}\right) \cup \left( {A \cap C}\right) \) . Then \( x \in A \cap B \) or \( x \in A \cap C \) . In either case, \( x \in A \) . In the first case, \( x \in B \) and, hence, \( x \in B \cup C \) . It follows in this case that \( x \in A \cap \left( {B \cup C}\right) \) . In the second case, \( x \in C \) and, hence, \( x \in B \cup C \) . Again, we conclude \( x \in A \cap \left( {B \cup C}\right) \) . Therefore, \( \left( {A \cap B}\right) \cup \left( {A \cap C}\right) \subset A \cap \left( {B \cup C}\right) \) as desired. \( ▱ \)
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No
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Theorem 1.1.3 Let \( \left\{ {{A}_{i} : i \in I}\right\} \) be an indexed family of subsets of a universal set \( X \) and let \( B \) be a subset of \( X \) . Then the following hold:\n\n(a) \( B \cup \left( {\mathop{\bigcap }\limits_{{i \in I}}{A}_{i}}\right) = \mathop{\bigcap }\limits_{{i \in I}}B \cup {A}_{i} \);
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Proof of (a): Let \( x \in B \cup \left( {\mathop{\bigcap }\limits_{{i \in I}}{A}_{i}}\right) \) . Then \( x \in B \) or \( x \in \mathop{\bigcap }\limits_{{i \in I}}{A}_{i} \) . If \( x \in B \), then \( x \in B \cup {A}_{i} \) for all \( i \in I \) and, thus, \( x \in \mathop{\bigcap }\limits_{{i \in I}}B \cup {A}_{i} \) . If \( x \in \mathop{\bigcap }\limits_{{i \in I}}{A}_{i} \), then \( x \in {A}_{i} \) for all \( i \in I \) . Therefore, \( x \in B \cup {A}_{i} \) for all \( i \in I \) and, hence, \( x \in \mathop{\bigcap }\limits_{{i \in I}}B \cup {A}_{i} \) . We have thus showed \( B \cup \left( {\mathop{\bigcap }\limits_{{i \in I}}{A}_{i}}\right) \subset \mathop{\bigcap }\limits_{{i \in I}}B \cup {A}_{i} \) .\n\nNow let \( x \in \mathop{\bigcap }\limits_{{i \in I}}B \cup {A}_{i} \) . Then \( x \in B \cup {A}_{i} \) for all \( i \in I \) . If \( x \in B \), then \( x \in B \cup \left( {\mathop{\bigcap }\limits_{{i \in I}}{A}_{i}}\right) \) . If \( x \notin B \) , then we must have that \( x \in {A}_{i} \) for all \( i \in I \) . Therefore, \( x \in \mathop{\bigcap }\limits_{{i \in I}}{A}_{i} \) and, hence, \( x \in B \cup \left( {\mathop{\bigcap }\limits_{{i \in I}}{A}_{i}}\right) \) . This proves the other inclusion and, so, the equality. \( ▱ \)
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Yes
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Theorem 1.2.1 Let \( f : X \rightarrow Y \) . If there are two functions \( g : Y \rightarrow X \) and \( h : Y \rightarrow X \) such that \( g\left( {f\left( x\right) }\right) = x \) for every \( x \in X \) and \( f\left( {h\left( y\right) }\right) = y \) for every \( y \in Y \), then \( f \) is bijective and \( g = h = {f}^{-1} \) .
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Proof: First we prove that \( f \) is surjective. Let \( y \in Y \) and set \( x = h\left( y\right) \) . Then, from the assumption on \( h \), we have \( f\left( x\right) = f\left( {h\left( y\right) }\right) = y \) . This shows that \( f \) is surjective.\n\nNext we prove that \( f \) is injective. Let \( x,{x}^{\prime } \in X \) be such that \( f\left( x\right) = f\left( {x}^{\prime }\right) \) . Then \( x = g\left( {f\left( x\right) }\right) = \) \( g\left( {f\left( {x}^{\prime }\right) }\right) = {x}^{\prime } \) . Thus, \( f \) is injective.\n\nWe have shown that for each \( y \in Y \), there is a unique \( x \in X \), which we denote \( {f}^{-1}\left( y\right) \) such that \( f\left( x\right) = y \) . Since for such a \( y, g\left( y\right) = g\left( {f\left( x\right) }\right) = x \), we obtain \( g\left( y\right) = {f}^{-1}\left( y\right) \) . Since \( f\left( {h\left( y\right) }\right) = y \), we also conclude that \( h\left( y\right) = x = {f}^{-1}\left( y\right) \) . \( ▱ \)
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Yes
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Theorem 1.2.3 Let \( f : X \rightarrow Y \) be a function, let \( A \) be a subset of \( X \), and let \( B \) be a subset of \( Y \). The following hold:\n\n(a) \( A \subset {f}^{-1}\left( {f\left( A\right) }\right) \).\n\n(b) \( f\left( {{f}^{-1}\left( B\right) }\right) \subset B \).
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Proof: We prove (a) and leave (b) as an exercise.\n\n(a) Let \( x \in A \). By the definition of image, \( f\left( x\right) \in f\left( A\right) \). Now, by the definition of preimage, \( x \in {f}^{-1}\left( {f\left( A\right) }\right) \). \( ▱ \)
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No
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Theorem 1.2.4 Let \( f : X \rightarrow Y \) be a function, let \( A, B \subset X \), and let \( C, D \subset Y \). The following hold:\n\n(a) If \( C \subset D \), then \( {f}^{-1}\left( C\right) \subset {f}^{-1}\left( D\right) \) ;\n\n(b) \( {f}^{-1}\left( {D \smallsetminus C}\right) = {f}^{-1}\left( D\right) \smallsetminus {f}^{-1}\left( C\right) \) ;\n\n(c) If \( A \subset B \), then \( f\left( A\right) \subset f\left( B\right) \) ;\n\n(d) \( f\left( {A \smallsetminus B}\right) \supset f\left( A\right) \smallsetminus f\left( B\right) \) .
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Proof: We prove (b) and leave the other parts as an exercise.\n\n(b) We prove first that \( {f}^{-1}\left( {D \smallsetminus C}\right) \subset {f}^{-1}\left( D\right) \smallsetminus {f}^{-1}\left( C\right) \) . Let \( x \in {f}^{-1}\left( {D \smallsetminus C}\right) \) . Then, from the definition of inverse image, we get \( f\left( x\right) \in D \smallsetminus C \) . Thus, \( f\left( x\right) \in D \) and \( f\left( x\right) \notin C \) . Hence \( x \in {f}^{-1}\left( D\right) \) and \( x \notin {f}^{-1}\left( C\right) \) . We conclude that \( x \in {f}^{-1}\left( D\right) \smallsetminus {f}^{-1}\left( C\right) \) .\n\nNext we prove \( {f}^{-1}\left( D\right) \smallsetminus {f}^{-1}\left( C\right) \subset {f}^{-1}\left( {D \smallsetminus C}\right) \) . Let \( x \in {f}^{-1}\left( D\right) \smallsetminus {f}^{-1}\left( C\right) \) . Thus, \( x \in {f}^{-1}\left( D\right) \) and \( x \notin {f}^{-1}\left( C\right) \) . Therefore, \( f\left( x\right) \in D \) and \( f\left( x\right) \notin C \) . This means \( f\left( x\right) \in D \smallsetminus C \) and, so, \( x \in {f}^{-1}\left( {D \smallsetminus C}\right) \) . \( ▱ \)
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No
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Theorem 1.2.5 Let \( f : X \rightarrow Y \) be a function, let \( {\left\{ {A}_{\alpha }\right\} }_{\alpha \in I} \) be an indexed family of subsets of \( X \), and let \( {\left\{ {B}_{\beta }\right\} }_{\beta \in J} \) be an indexed family of subsets of \( Y \) . The following hold:\n\n(a) \( f\left( {\mathop{\bigcup }\limits_{{\alpha \in I}}{A}_{\alpha }}\right) = \mathop{\bigcup }\limits_{{\alpha \in I}}f\left( {A}_{\alpha }\right) \) ;
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(a) Let \( y \in f\left( {\mathop{\bigcup }\limits_{{\alpha \in I}}{A}_{\alpha }}\right) \) . From the definition of image of a set, there is \( x \in \mathop{\bigcup }\limits_{{\alpha \in I}}{A}_{\alpha } \) such that \( y = f\left( x\right) \) . From the definition of union of a family of sets, there is \( {\alpha }_{0} \in I \) such that \( x \in {A}_{{\alpha }_{0}} \) . Therefore, \( y = f\left( x\right) \in f\left( {A}_{{\alpha }_{0}}\right) \) and, so, \( y \in \mathop{\bigcup }\limits_{{\alpha \in I}}f\left( {A}_{\alpha }\right) \) . \( ▱ \) .
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Yes
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Theorem 1.2.6 Let \( f : X \rightarrow Y \) and \( g : Y \rightarrow Z \) be two functions and let \( B \subset Z \) . The following hold:\n\n(a) \( {\left( g \circ f\right) }^{-1}\left( B\right) = {f}^{-1}\left( {{g}^{-1}\left( B\right) }\right) \) ;\n\n(b) If \( f \) and \( g \) are injective, then \( g \circ f \) is injective;\n\n(c) If \( f \) and \( g \) are surjective, then \( g \circ f \) is surjective;\n\n(d) If \( g \circ f \) is injective, then \( f \) is injective;\n\n(e) If \( g \circ f \) is surjective, then \( g \) is surjective.
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Proof: We prove (d) and leave the other parts as an exercise.\n\n(d) Suppose \( g \circ f \) is injective and let \( x,{x}^{\prime } \in X \) be such that \( f\left( x\right) = f\left( {x}^{\prime }\right) \) . Then \( \left( {g \circ f}\right) \left( x\right) = \) \( g\left( {f\left( x\right) }\right) = g\left( {f\left( {x}^{\prime }\right) }\right) = \left( {g \circ f}\right) \left( {x}^{\prime }\right) \) . Since \( g \circ f \) is injective, it follows that \( x = {x}^{\prime } \) . We conclude that \( f \) is injective. \( ▱ \)
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No
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Theorem 1.2.7 Suppose \( A \) is an infinite set. Then there exists a one-to-one function \( f : \mathbb{N} \rightarrow A \) .
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Proof: Let \( A \) be an infinite set. We define \( f \) as follows. Choose any element \( {a}_{1} \in A \) and set \( f\left( 1\right) = {a}_{1} \) . Now the set \( A \smallsetminus \left\{ {a}_{1}\right\} \) is again infinite (otherwise \( A = \{ a\} \cup \left( {A \smallsetminus \left\{ {a}_{1}\right\} }\right) \) would be the union of two finite sets). So we may choose \( {a}_{2} \in A \) with \( {a}_{2} \neq {a}_{1} \) and we define \( f\left( 2\right) = {a}_{2}{}^{2} \) . Having defined \( f\left( 1\right) ,\ldots, f\left( k\right) \), we choose \( {a}_{k + 1} \in A \) such that \( {a}_{k + 1} \in A \smallsetminus \left\{ {{a}_{1},\ldots ,{a}_{k}}\right\} \) and define \( f\left( {k + 1}\right) = {a}_{k + 1} \) (such an \( {a}_{k + 1} \) exists because \( A \smallsetminus \left\{ {{a}_{1},\ldots ,{a}_{k}}\right\} \) is infinite and, so, nonempty). The function \( f \) so defined clearly has the desired properties. \( ▱ \)
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Yes
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Theorem 1.3.1 — Principle of Mathematical Induction. For each natural number \( n \in \mathbb{N} \) , suppose that \( P\left( n\right) \) denotes a proposition which is either true or false. Let \( A = \{ n \in \mathbb{N} : P\left( n\right) \) is true \( \} \) . Suppose the following conditions hold:\n\n(a) \( 1 \in A \) .\n\n(b) For each \( k \in \mathbb{N} \), if \( k \in A \), then \( k + 1 \in A \) .\n\nThen \( A = \mathbb{N} \) .
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Proof: Suppose conditions (a) and (b) hold. Assume, by way of contradiction, that \( A \neq \mathbb{N} \) . Set \( B = \mathbb{N} \smallsetminus A \), that is, \( B = \{ n \in \mathbb{N} : P\left( n\right) \) is false \( \} \) . Then \( B \) is a nonempty subset of \( \mathbb{N} \) . By the Well-Ordering Property of the natural numbers, there exists a smallest element \( \ell \in B \) . By condition (a), \( 1 \notin B \) . Hence, \( \ell \geq 2 \) . It follows that \( k = \ell - 1 \) is a natural number. Since \( k < \ell, k \notin B \) and, hence, we have that \( P\left( k\right) \) is true. By condition (b), we obtain that \( P\left( {k + 1}\right) \) is true. But \( k + 1 = \ell \), and \( P\left( \ell \right) \) is false, since \( \ell \in B \) . This is a contradiction, so the conclusion follows. \( ▱ \)
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Yes
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Theorem 1.3.2 — Generalized Principle of Mathematical Induction. Let \( {n}_{0} \in \mathbb{N} \) and for each natural \( n \geq {n}_{0} \), suppose that \( P\left( n\right) \) denotes a proposition which is either true or false. Let \( A = \{ n \in \) \( \mathbb{N} : P\left( n\right) \) is true \( \} \) . Suppose the following two conditions hold:\n\n(a) \( {n}_{0} \in A \) .\n\n(b) For each \( k \in \mathbb{N}, k \geq {n}_{0} \), if \( k \in A \), then \( k + 1 \in A \) .\n\nThen \( \left\{ {k \in \mathbb{N} : k \geq {n}_{0}}\right\} \subset A \) .
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Proof: Suppose conditions (a) and (b) hold. Assume, by way of contradiction, that \( A \nsupseteq \{ k \in \mathbb{N} \) : \( \left. {k \geq {n}_{0}}\right\} \) . Set \( B = \left\{ {n \in \mathbb{N} : n \geq {n}_{0}, P\left( n\right) \text{is false}}\right\} \) . Then \( B \) is a nonempty subset of \( \mathbb{N} \) . By the Well-Ordering Property of the natural numbers, there exists a smallest element \( \ell \in B \) . By condition (a), \( {n}_{0} \notin B \) . Hence, \( \ell \geq {n}_{0} + 1 \) . It follows that \( k = \ell - 1 \geq {n}_{0} \) . Since \( k < \ell, k \notin B \) and, so, we have that \( P\left( k\right) \) is true. By condition (b), we obtain that \( P\left( {k + 1}\right) \) is true. But \( k + 1 = \ell \), and \( P\left( \ell \right) \) is false, since \( \ell \in B \) . This is a contradiction, so the conclusion follows. \( ▱ \)
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Yes
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Theorem 1.3.3 — Principle of Strong Induction. For each natural \( n \in \mathbb{N} \) , suppose that \( P\left( n\right) \) denotes a proposition which is either true or false. Let \( A = \{ n \in \mathbb{N} : P\left( n\right) \) is true \( \} \) . Suppose the following two conditions hold:\n\n(a) \( 1 \in A \) .\n\n(b) For each \( k \in \mathbb{N} \), if \( 1,2,\ldots, k \in A \), then \( k + 1 \in A \) .\n\nThen \( A = \mathbb{N} \) .
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Remark 1.3.4 Note that the inductive step above says that, in order to prove \( P\left( {k + 1}\right) \) is true, we may assume not only that \( P\left( k\right) \) is true, but also that \( P\left( 1\right), P\left( 2\right) ,\ldots, P\left( {k - 1}\right) \) are true.\n\nThere is also a generalized version of this theorem where the base case is for some integer \( {n}_{0} > 1 \) .
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Yes
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Proposition 1.4.1 For \( x, y, z \in \mathbb{R} \), the following hold:\n\n(a) If \( x + y = x + z \), then \( y = z \) ;
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Proof: (a) Suppose \( x + y = x + z \) . Adding \( - x \) (which exists by axiom (2d)) to both sides, we have\n\n\[ \left( {-x}\right) + \left( {x + y}\right) = \left( {-x}\right) + \left( {x + z}\right) . \]\n\nThen axiom (1a) gives\n\n\[ \left\lbrack {\left( {-x}\right) + x}\right\rbrack + y = \left\lbrack {\left( {-x}\right) + x}\right\rbrack + z. \]\n\nThus, again by axiom (2d), \( 0 + y = 0 + z \) and, by axiom (1c), \( y = z \) .
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Yes
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Proposition 1.4.2 Let \( x, y, M \in \mathbb{R} \) and suppose \( M > 0 \) . The following properties hold:\n\n(a) \( \left| x\right| \geq 0 \) ;\n\n(b) \( \left| {-x}\right| = \left| x\right| \)\n\n(c) \( \left| {xy}\right| = \left| x\right| \left| y\right| \)\n\n(d) \( \left| x\right| < M \) if and only if \( - M < x < M \) . (The same holds if \( < \) is replaced with \( \leq \) .)
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Proof: We prove (d) and leave the other parts as an exercise.\n\n(d) Suppose \( \left| x\right| < M \) . In particular, this implies \( M > 0 \) . We consider the two cases separately: \( x \geq 0 \) and \( x < 0 \) . Suppose first \( x \geq 0 \) . Then \( \left| x\right| = x \) and, hence, \( - M < 0 \leq x = \left| x\right| < M \) . Now suppose \( x < 0 \) . Then \( \left| x\right| = - x \) . Therefore, \( - x < M \) and, so \( x > - M \) . It follows that \( - M < x < 0 < M \) .\n\nFor the converse, suppose \( - M < x < M \) . Again, we consider different cases. If \( x \geq 0 \), then \( \left| x\right| = x < M \) as desired. Next suppose \( x < 0 \) . Now, \( - M < x \) implies \( M > - x \) . Then \( \left| x\right| = - x < M \) . \( ▱ \)
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No
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Theorem 1.4.3 — Triangle Inequality. Given \( {x, y} \in \mathbb{R} \) , \n\n\[ \left| {x + y}\right| \leq \left| x\right| + \left| y\right| \]
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Proof: From the observation above, we have \n\n\[ - \left| x\right| \leq x \leq \left| x\right| \] \n\n\[ - \left| y\right| \leq y \leq \left| y\right| \] \n\nAdding up the inequalities gives \n\n\[ - \left| x\right| - \left| y\right| \leq x + y \leq \left| x\right| + \left| y\right| \] \n\nSince \( - \left| x\right| - \left| y\right| = - \left( {\left| x\right| + \left| y\right| }\right) \), the conclusion follows from Proposition 1.4.2 (d). \( ▱ \)
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Yes
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Proposition 1.5.1 Let \( A \) be a nonempty subset of \( \mathbb{R} \) that is bounded above. Then \( \alpha = \sup A \) if and only if\n\n(1’) \( x \leq \alpha \) for all \( x \in A \) ;\n\n(2’) For any \( \varepsilon > 0 \), there exists \( a \in A \) such that \( \alpha - \varepsilon < a \) .
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Proof: Suppose first that \( \alpha = \sup A \) . Then clearly (1’) holds (since this is identical to condition (1) in the definition of supremum). Now let \( \varepsilon > 0 \) . Since \( \alpha - \varepsilon < \alpha \), condition (2) in the definition of supremum implies that \( \alpha - \varepsilon \) is not an upper bound of \( A \) . Therefore, there must exist an element \( a \) of \( A \) such that \( \alpha - \varepsilon < a \) as desired.\n\nConversely, suppose conditions (1') and (2') hold. Then all we need to show is that condition (2) in the definition of supremum holds. Let \( M \) be an upper bound of \( A \) and assume, by way of contradiction, that \( M < \alpha \) . Set \( \varepsilon = \alpha - M \) . By condition (2) in the statement, there is \( a \in A \) such that \( a > \alpha - \varepsilon = M \) . This contradicts the fact that \( M \) is an upper bound. The conclusion now follows. \( ▱ \)
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Yes
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Theorem 1.5.3 Let \( A \) and \( B \) be nonempty sets and \( A \subset B \). Suppose \( B \) is bounded above. Then \( \sup A \leq \sup B \).
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Proof: Let \( M \) be an upper bound for \( B \), then for \( x \in B, x \leq M \). In particular, it is also true that \( x \leq M \) for \( x \in A \). Thus, \( A \) is also bounded above. Now, since \( \sup B \) is an upper bound for \( B \), it is also an upper bound for \( A \). Then, by the second condition in the definition of supremum, \( \sup A \leq \sup B \) as desired. \( ▱ \)
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Yes
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Theorem 1.6.1 — The Archimedean Property. The set of natural numbers is unbounded above.
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Proof: Let us assume by contradiction that \( \mathbb{N} \) is bounded above. Since \( \mathbb{N} \) is nonempty,\n\n\[ \alpha = \sup \mathbb{N} \]\n\nexists and is a real number. By Proposition 1.5.1 (with \( \varepsilon = 1 \) ), there exists \( n \in \mathbb{N} \) such that\n\n\[ \alpha - 1 < n \leq \alpha \text{.} \]\n\nBut then \( n + 1 > \alpha \) . This is a contradiction since \( n + 1 \) is a natural number.
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Yes
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Theorem 1.6.2 The following hold:\n\n(a) For any \( x \in \mathbb{R} \), there exists \( n \in \mathbb{N} \) such that \( x < n \) ;
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Proof: (a) Fix any \( x \in \mathbb{R} \) . Since \( \mathbb{N} \) is not bounded above, \( x \) cannot be an upper bound of \( \mathbb{N} \) . Thus, there exists \( n \in \mathbb{N} \) such that \( x < n \) .
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Yes
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Theorem 1.6.3 - The Density Property of \( \\mathbb{Q} \) . If \( x \) and \( y \) are two real numbers such that \( x < y \), then there exists a rational number \( r \) such that\n\n\[ x < r < y\\text{.} \]
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Proof: We are going to prove that there exist an integer \( m \) and a positive integer \( n \) such that\n\n\[ x < m/n < y \]\n\nor, equivalently,\n\n\[ {nx} < m < {ny} = {nx} + n\\left( {y - x}\\right) . \]\n\nSince \( y - x > 0 \), by Theorem 1.6.2 (3), there exists \( n \\in \\mathbb{N} \) such that \( 1 < n\\left( {y - x}\\right) \) . Then\n\n\[ {ny} = {nx} + n\\left( {y - x}\\right) > {nx} + 1. \]\n\nBy Theorem 1.6.2 (4), one can choose \( m \\in \\mathbb{Z} \) such that\n\n\[ m - 1 \\leq {nx} < m\\text{.} \]\n\nThen \( {nx} < m \\leq {nx} + 1 < {ny} \) . Therefore,\n\n\[ x < m/n < y\\text{.} \]\n\nThe proof is now complete.
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Yes
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Proposition 1.6.4 The number \( \sqrt{2} \) is irrational.
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Proof: Suppose, by way of contradiction, that \( \sqrt{2} \in \mathbb{Q} \) . Then there are integers \( r \) and \( s \) with \( s \neq 0 \) , such that\n\n\[ \sqrt{2} = \frac{r}{s} \]\n\nBy canceling out the common factors of \( r \) and \( s \), we may assume that \( r \) and \( s \) have no common factors.\n\nNow, by squaring both sides of the equation above, we get\n\n\[ 2 = \frac{{r}^{2}}{{s}^{2}} \] \nand, hence,\n\n\[ 2{s}^{2} = {r}^{2} \]\n\n(1.3)\n\nIt follows that \( {r}^{2} \) is an even integer. Therefore, \( r \) is an even integer (see Exercise 1.4.1). We can then write \( r = {2j} \) for some integer \( j \) . Hence \( {r}^{2} = 4{j}^{2} \) . Substituting in (1.3), we get \( {s}^{2} = 2{j}^{2} \) . Therefore, \( {s}^{2} \) is even. We conclude as before that \( s \) is even. Thus, both \( r \) and \( s \) have a common factor, which is a contradiction. \( ▱ \)
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Yes
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Theorem 1.6.5 Let \( x \) and \( y \) be two real numbers such that \( x < y \). Then there exists an irrational number \( t \) such that\n\n\[ x < t < y\text{.} \]
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Proof: Since \( x < y \), one has\n\n\[ x - \sqrt{2} < y - \sqrt{2} \]\n\nBy Theorem 1.6.3, there exists a rational number \( r \) such that\n\n\[ x - \sqrt{2} < r < y - \sqrt{2} \]\n\nThis implies\n\n\[ x < r + \sqrt{2} < y \]\n\nSince \( r \) is rational, the number \( t = r + \sqrt{2} \) is irrational (see Exercise 1.6.4) and \( x < t < y \).
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Yes
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Lemma 2.1.2 Let \( \ell \geq 0 \) . If \( \ell < \varepsilon \) for all \( \varepsilon > 0 \), then \( \ell = 0 \) .
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Proof: This is easily proved by contraposition. If \( \ell > 0 \), then there is a positive number, for example \( \varepsilon = \ell /2 \), such that \( \varepsilon < \ell \) . \( ▱ \)
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Yes
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Theorem 2.1.3 A convergent sequence \( \left\{ {a}_{n}\right\} \) has at most one limit.
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Proof: Suppose \( \left\{ {a}_{n}\right\} \) converges to \( a \) and \( b \) . Then given \( \varepsilon > 0 \), there exist positive integers \( {N}_{1} \) and \( {N}_{2} \) such that\n\n\[ \left| {{a}_{n} - a}\right| < \varepsilon /2\text{ for all }n \geq {N}_{1} \]\n\nand\n\n\[ \left| {{a}_{n} - b}\right| < \varepsilon /2\text{ for all }n \geq {N}_{2}. \]\n\nLet \( N = \max \left\{ {{N}_{1},{N}_{2}}\right\} \) . Then\n\n\[ \left| {a - b}\right| \leq \left| {a - {a}_{N}}\right| + \left| {{a}_{N} - b}\right| < \varepsilon /2 + \varepsilon /2 = \varepsilon . \]\n\nSince \( \varepsilon > 0 \) is arbitrary, by Lemma 2.1.2, \( \left| {a - b}\right| = 0 \) and, hence, \( a = b \) . \( ▱ \)
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Yes
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Lemma 2.1.4 Given real numbers \( a, b \), then \( a \leq b \) if and only if \( a < b + \varepsilon \) for all \( \varepsilon > 0 \) .
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Proof: Suppose \( a < b + \varepsilon \) for all \( \varepsilon > 0 \) . And suppose, by way of contradiction, that \( a > b \) . then set \( {\varepsilon }_{0} = a - b \) . Then \( {\varepsilon }_{0} > 0 \) . By assumption, we should have \( a < b + {\varepsilon }_{0} = b + a - b = a \), which is a contradiction. It follows that \( a \leq b \) .\n\nThe other direction follows immediately from the order axioms. \( ▱ \)
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Yes
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Theorem 2.1.5 — Comparison Theorem. Suppose \( \left\{ {a}_{n}\right\} \) and \( \left\{ {b}_{n}\right\} \) converge to \( a \) and \( b \) , respectively, and \( {a}_{n} \leq {b}_{n} \) for all \( n \in \mathbb{N} \) . Then \( a \leq b \) .
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Proof: For any \( \varepsilon > 0 \), there exist \( {N}_{1},{N}_{2} \in \mathbb{N} \) such that\n\n\[ a - \frac{\varepsilon }{2} < {a}_{n} < a + \frac{\varepsilon }{2},\;\text{ for }n \geq {N}_{1}, \]\n\n\[ b - \frac{\varepsilon }{2} < {b}_{n} < b + \frac{\varepsilon }{2},\;\text{ for }n \geq {N}_{2}. \]\n\nChoose \( N > \max \left\{ {{N}_{1},{N}_{2}}\right\} \) . Then\n\n\[ a - \frac{\varepsilon }{2} < {a}_{N} \leq {b}_{N} < b + \frac{\varepsilon }{2}. \]\n\nThus, \( a < b + \varepsilon \) for any \( \varepsilon > 0 \) . Using Lemma 2.1.4 we conclude \( a \leq b \) .
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Yes
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Theorem 2.1.6 - The Squeeze Theorem. Suppose the sequences \( \left\{ {a}_{n}\right\} ,\left\{ {b}_{n}\right\} \), and \( \left\{ {c}_{n}\right\} \) satisfy\n\n\[ \n{a}_{n} \leq {b}_{n} \leq {c}_{n}\text{ for all }n \in \mathbb{N},\n\]\n\nand \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = \mathop{\lim }\limits_{{n \rightarrow \infty }}{c}_{n} = \ell \) . Then \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{b}_{n} = \ell \) .
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Proof: Fix any \( \varepsilon > 0 \) . Since \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = \ell \), there exists \( {N}_{1} \in \mathbb{N} \) such that\n\n\[ \n\ell - \varepsilon < {a}_{n} < \ell + \varepsilon\n\]\n\nfor all \( n \geq {N}_{1} \) . Similarly, since \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{c}_{n} = \ell \), there exists \( {N}_{2} \in \mathbb{N} \) such that\n\n\[ \n\ell - \varepsilon < {c}_{n} < \ell + \varepsilon\n\]\n\nfor all \( n \geq {N}_{2} \) . Let \( N = \max \left\{ {{N}_{1},{N}_{2}}\right\} \) . Then, for \( n \geq N \), we have\n\n\[ \n\ell - \varepsilon < {a}_{n} \leq {b}_{n} \leq {c}_{n} < \ell + \varepsilon\n\]\n\nwhich implies \( \left| {{b}_{n} - \ell }\right| < \varepsilon \) . Therefore, \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{b}_{n} = \ell \) . \( ▱ \)
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Yes
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Theorem 2.1.7 A convergent sequence is bounded.
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Proof: Suppose the sequence \( \left\{ {a}_{n}\right\} \) converges to \( a \) . Then, for \( \varepsilon = 1 \), there exists \( N \in \mathbb{N} \) such that\n\n\[ \left| {{a}_{n} - a}\right| < 1\text{ for all }n \geq N. \]\n\nSince \( \left| {a}_{n}\right| - \left| a\right| \leq \left| \right| {a}_{n}\left| -\right| a\left| \right| \leq \left| {{a}_{n} - a}\right| \), this implies \( \left| {a}_{n}\right| < 1 + \left| a\right| \) for all \( n \geq N \) . Set\n\n\[ M = \max \left\{ {\left| {a}_{1}\right| ,\ldots ,\left| {a}_{N - 1}\right| ,\left| a\right| + 1}\right\} . \]\n\nThen \( \left| {a}_{n}\right| \leq M \) for all \( n \in \mathbb{N} \) . Therefore, \( \left\{ {a}_{n}\right\} \) is bounded. \( ▱ \)
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Yes
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Lemma 2.1.8 Let \( {\left\{ {n}_{k}\right\} }_{k} \) be a sequence of positive integers with\n\n\[ \n{n}_{1} < {n}_{2} < {n}_{3} < \cdots \n\]\n\nThen \( {n}_{k} \geq k \) for all \( k \in \mathbb{N} \) .
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Proof: We use mathematical induction. When \( k = 1 \), it is clear that \( {n}_{1} \geq 1 \) since \( {n}_{1} \) is a positive integer. Assume \( {n}_{k} \geq k \) for some \( k \) . Now \( {n}_{k + 1} > {n}_{k} \) and, since \( {n}_{k} \) and \( {n}_{k + 1} \) are integers, this implies, \( {n}_{k + 1} \geq {n}_{k} + 1 \) . Therefore, \( {n}_{k + 1} \geq k + 1 \) by the inductive hypothesis. The conclusion now follows by the principle of mathematical induction. \( ▱ \)
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Yes
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Theorem 2.1.9 If a sequence \( \left\{ {a}_{n}\right\} \) converges to \( a \), then any subsequence \( \left\{ {a}_{{n}_{k}}\right\} \) of \( \left\{ {a}_{n}\right\} \) also converges to \( a \) .
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Proof: Suppose \( \left\{ {a}_{n}\right\} \) converges to \( a \) and let \( \varepsilon > 0 \) be given. Then there exists \( N \) such that\n\n\[ \left| {{a}_{n} - a}\right| < \varepsilon \text{for all}n \geq N\text{.}\]\n\nFor any \( k \geq N \), since \( {n}_{k} \geq k \), we also have\n\n\[ \left| {{a}_{{n}_{k}} - a}\right| < \varepsilon \]\n\nThus, \( \left\{ {a}_{{n}_{k}}\right\} \) converges to \( a \) as \( k \rightarrow \infty \) . \( ▱ \)
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Yes
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Theorem 2.3.1 — Monotone Convergence Theorem. Let \( \{ {a}_{n}\} \) be a sequence of real numbers. The following hold:\n\n(a) If \( \left\{ {a}_{n}\right\} \) is increasing and bounded above, then it is convergent.\n\n(b) If \( \left\{ {a}_{n}\right\} \) is decreasing and bounded below, then it is convergent.
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Proof: (a) Let \( \left\{ {a}_{n}\right\} \) be an increasing sequence that is bounded above. Define\n\n\[ A = \left\{ {{a}_{n} : n \in \mathbb{N}}\right\} \]\n\nThen \( A \) is a subset of \( \mathbb{R} \) that is nonempty and bounded above and, hence, \( \sup A \) exists. Let \( \ell = \sup A \) and let \( \varepsilon > 0 \). By Proposition 1.5.1, there exists \( N \in \mathbb{N} \) such that\n\n\[ \ell - \varepsilon < {a}_{N} \leq \ell \]\n\nSince \( \left\{ {a}_{n}\right\} \) is increasing,\n\n\[ \ell - \varepsilon < {a}_{N} \leq {a}_{n}\text{ for all }n \geq N. \]\n\nOn the other hand, since \( \ell \) is an upper bound for \( A \), we have \( {a}_{n} \leq \ell \) for all \( n \). Thus,\n\n\[ \ell - \varepsilon < {a}_{n} < \ell + \varepsilon \text{ for all }n \geq N. \]\n\nTherefore, \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = \ell \).\n\n(b) Let \( \left\{ {a}_{n}\right\} \) be a decreasing sequence that is bounded below. Define\n\n\[ {b}_{n} = - {a}_{n} \]\n\nThen \( \left\{ {b}_{n}\right\} \) is increasing and bounded above (if \( M \) is a lower bound for \( \left\{ {a}_{n}\right\} \), then \( - M \) is an upper bound for \( \left\{ {b}_{n}\right\} \) ). Let\n\n\[ \ell = \mathop{\lim }\limits_{{n \rightarrow \infty }}{b}_{n} = \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {-{a}_{n}}\right) . \]\n\nThen \( \left\{ {a}_{n}\right\} \) converges to \( - \ell \) by Theorem 2.2.1. \( ▱ \)
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Yes
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Theorem 2.3.3 - Nested Intervals Theorem. Let \( {\left\{ {I}_{n}\right\} }_{n = 1}^{\infty } \) be a sequence of nonempty closed bounded intervals satisfying \( {I}_{n + 1} \subset {I}_{n} \) for all \( n \in \mathbb{N} \) . Then the following hold:\n\n(a) \( \mathop{\bigcap }\limits_{{n = 1}}^{\infty }{I}_{n} \neq \varnothing \) .\n\n(b) If, in addition, the lengths of the intervals \( {I}_{n} \) converge to zero, then \( \mathop{\bigcap }\limits_{{n = 1}}^{\infty }{I}_{n} \) consists of a single point.
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Proof: Let \( \left\{ {I}_{n}\right\} \) be as in the statement with \( {I}_{n} = \left\lbrack {{a}_{n},{b}_{n}}\right\rbrack \) . In particular, \( {a}_{n} \leq {b}_{n} \) for all \( n \in \mathbb{N} \) . Given that \( {I}_{n + 1} \subset {I}_{n} \), we have \( {a}_{n} \leq {a}_{n + 1} \) and \( {b}_{n + 1} \leq {b}_{n} \) for all \( n \in \mathbb{N} \) . This shows that \( \left\{ {a}_{n}\right\} \) is an increasing sequence bounded above by \( {b}_{1} \) and \( \left\{ {b}_{n}\right\} \) is a decreasing sequence bounded below by \( {a}_{1} \) . By the\n\nMonotone Convergence Theorem (Theorem 2.3.1), there exist \( a, b \in \mathbb{R} \) such that \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = a \) and \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{b}_{n} = b \) . Since \( {a}_{n} \leq {b}_{n} \) for all \( n \), by Theorem 2.1.5, we get \( a \leq b \) . Now, we also have \( {a}_{n} \leq a \) and \( b \leq {b}_{n} \) for all \( n \in \mathbb{N} \) (since \( \left\{ {a}_{n}\right\} \) is increasing and \( \left\{ {b}_{n}\right\} \) is decreasing). This shows that if \( a \leq x \leq b \), then \( x \in {I}_{n} \) for all \( n \in \mathbb{N} \) . Thus, \( \left\lbrack {a, b}\right\rbrack \subset \mathop{\bigcap }\limits_{{n = 1}}^{\infty }{I}_{n} \) . It follows that \( \mathop{\bigcap }\limits_{{n = 1}}^{\infty }{I}_{n} \neq \varnothing \) . This proves part (a).\n\nNow note also that \( \mathop{\bigcap }\limits_{{n = 1}}^{\infty }{I}_{n} \subset \left\lbrack {a, b}\right\rbrack \) . Indeed, if \( x \in \mathop{\bigcap }\limits_{{n = 1}}^{\infty }{I}_{n} \), then \( x \in {I}_{n} \) for all \( n \) . Therefore, \( {a}_{n} \leq x \leq {b}_{n} \) for all \( n \) . Using Theorem 2.1.5, we conclude \( a \leq x \leq b \) . Thus, \( x \in \left\lbrack {a, b}\right\rbrack \) . This proves the desired inclusion and, hence, \( \mathop{\bigcap }\limits_{{n = 1}}^{\infty }{I}_{n} = \left\lbrack {a, b}\right\rbrack \) .\n\nWe now prove part (b). Suppose the lengths of the intervals \( {I}_{n} \) converge to zero. This means \( {b}_{n} - {a}_{n} \rightarrow 0 \) as \( n \rightarrow \infty \) . Then \( b = \mathop{\lim }\limits_{{n \rightarrow \infty }}{b}_{n} = \mathop{\lim }\limits_{{n \rightarrow \infty }}\left\lbrack {\left( {{b}_{n} - {a}_{n}}\right) + {a}_{n}}\right\rbrack = a \) . It follows that \( \mathop{\bigcap }\limits_{{n = 1}}^{\infty }{I}_{n} = \) \( \{ a\} \) as desired. \( ▱ \)
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Yes
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Theorem 2.3.5 If a sequence \( \left\{ {a}_{n}\right\} \) is increasing and not bounded above, then\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = \infty \]
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Proof: Fix any real number \( M \) . Since \( \left\{ {a}_{n}\right\} \) is not bounded above, there exists \( N \in \mathbb{N} \) such that \( {a}_{N} \geq M \) . Then\n\n\[ {a}_{n} \geq {a}_{N} \geq M\text{ for all }n \geq N \]\n\nbecause \( \left\{ {a}_{n}\right\} \) is increasing. Therefore, \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = \infty \) . The proof for the second case is similar.
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No
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Theorem 2.3.6 Let \( \\left\\{ {a}_{n}\\right\\} ,\\left\\{ {b}_{n}\\right\\} \), and \( \\left\\{ {c}_{n}\\right\\} \) be sequences of real numbers and let \( k \) be a constant. Suppose\n\n\[ \n\\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}{a}_{n} = \\infty ,\\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}{b}_{n} = \\infty \\text{, and}\\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}{c}_{n} = - \\infty \n\]\n\nThen\n\n(a) \( \\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}\\left( {{a}_{n} + {b}_{n}}\\right) = \\infty \) ;\n\n(b) \( \\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}\\left( {{a}_{n}{b}_{n}}\\right) = \\infty \) ;\n\n(c) \( \\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}\\left( {{a}_{n}{c}_{n}}\\right) = - \\infty \) ;\n\n(d) \( \\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}k{a}_{n} = \\infty \) if \( k > 0 \), and \( \\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}k{a}_{n} = - \\infty \) if \( k < 0 \) ;\n\n(e) \( \\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}\\frac{1}{{a}_{n}} = 0 \) . (Here we assume \( {a}_{n} \\neq 0 \) for all \( n \) .)
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Proof: We provide proofs for (a) and (e) and leave the others as exercises.\n\n(a) Fix any \( M \\in \\mathbb{R} \) . Since \( \\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}{a}_{n} = \\infty \), there exists \( {N}_{1} \\in \\mathbb{N} \) such that\n\n\[ \n{a}_{n} \\geq \\frac{M}{2}\\text{ for all }n \\geq {N}_{1} \n\]\n\nSimilarly, there exists \( {N}_{2} \\in \\mathbb{N} \) such that\n\n\[ \n{b}_{n} \\geq \\frac{M}{2}\\text{ for all }n \\geq {N}_{1} \n\]\n\nLet \( N = \\max \\left\\{ {{N}_{1},{N}_{2}}\\right\\} \) . Then it is clear that\n\n\[ \n{a}_{n} + {b}_{n} \\geq M\\text{ for all }n \\geq N. \n\]\n\nThis implies (a).\n\n(e) For any \( \\varepsilon > 0 \), let \( M = \\frac{1}{\\varepsilon } \) . Since \( \\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}{a}_{n} = \\infty \), there exists \( N \\in \\mathbb{N} \) such that\n\n\[ \n{a}_{n} > \\frac{1}{\\varepsilon }\\text{ for all }n \\geq N \n\]\n\nThis implies that for \( n \\geq N \),\n\n\[ \n\\left| {\\frac{1}{{a}_{n}} - 0}\\right| = \\frac{1}{{a}_{n}} < \\varepsilon \n\]\n\nThus, (e) holds. \( ▱ \)
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No
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Theorem 2.4.2 A convergent sequence is a Cauchy sequence.
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Proof: Let \( \left\{ {a}_{n}\right\} \) be a convergent sequence and let\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = a \]\n\nThen for any \( \varepsilon > 0 \), there exists a positive integer \( N \) such that\n\n\[ \left| {{a}_{n} - a}\right| < \varepsilon /2\text{ for all }n \geq N. \]\n\nFor any \( m, n \geq N \), one has\n\n\[ \left| {{a}_{m} - {a}_{n}}\right| \leq \left| {{a}_{m} - a}\right| + \left| {{a}_{n} - a}\right| < \varepsilon /2 + \varepsilon /2 = \varepsilon . \]\n\nThus, \( \left\{ {a}_{n}\right\} \) is a Cauchy sequence.
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Yes
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Theorem 2.4.3 A Cauchy sequence is bounded.
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Proof: Let \( \left\{ {a}_{n}\right\} \) be a Cauchy sequence. Then for \( \varepsilon = 1 \), there exists a positive integer \( N \) such that\n\n\[ \left| {{a}_{m} - {a}_{n}}\right| < 1\text{for all}m, n \geq N\text{.} \]\n\nIn particular,\n\n\[ \left| {{a}_{n} - {a}_{N}}\right| < 1\text{ for all }n \geq N. \]\n\nLet \( M = \max \left\{ {\left| {a}_{1}\right| ,\ldots ,\left| {a}_{N - 1}\right| ,1 + \left| {a}_{N}\right| }\right\} \) . Then, for \( n = 1,\ldots, N - 1 \), we clearly have \( \left| {a}_{n}\right| \leq M \) . Moreover, for \( n \geq N \),\n\n\[ \left| {a}_{n}\right| = \left| {{a}_{n} - {a}_{N} + {a}_{N}}\right| \leq \left| {{a}_{n} - {a}_{N}}\right| + \left| {a}_{N}\right| \leq 1 + \left| {a}_{N}\right| \leq M. \]\n\nTherefore, \( \left| {a}_{n}\right| \leq M \) for all \( n \in \mathbb{N} \) and, thus, \( \left\{ {a}_{n}\right\} \) is bounded. \( ▱ \)
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Yes
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Lemma 2.4.4 A Cauchy sequence that has a convergent subsequence is convergent.
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Proof: Let \( \left\{ {a}_{n}\right\} \) be a Cauchy sequence that has a convergent subsequence. For any \( \varepsilon > 0 \), there exists a positive integer \( N \) such that\n\n\[ \left| {{a}_{m} - {a}_{n}}\right| \leq \varepsilon /2\text{for all}m, n \geq N\text{.}\]\n\nLet \( \left\{ {a}_{{n}_{k}}\right\} \) be a subsequence of \( \left\{ {a}_{n}\right\} \) that converges to some point \( a \) . For the above \( \varepsilon \), there exists a positive number \( K \) such that\n\n\[ \left| {{a}_{{n}_{k}} - a}\right| < \varepsilon /2\text{ for all }k \geq K.\]\n\nThus, we can find a positive integer \( {n}_{\ell } > N \) such that\n\n\[ \left| {{a}_{{n}_{\ell }} - a}\right| < \varepsilon /2 \]\n\nThen for any \( n \geq N \), we have\n\n\[ \left| {{a}_{n} - a}\right| \leq \left| {{a}_{n} - {a}_{{n}_{\ell }}}\right| + \left| {{a}_{{n}_{\ell }} - a}\right| < \varepsilon .\n\nTherefore, \( \left\{ {a}_{n}\right\} \) converges to \( a \) . \( ▱ \)
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Yes
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Theorem 2.4.5 Any Cauchy sequence of real numbers is convergent.
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Proof: Let \( \left\{ {a}_{n}\right\} \) be a Cauchy sequence. Then it is bounded by Theorem 2.4.3. By the Bolzano-Weierstrass theorem, \( \left\{ {a}_{n}\right\} \) has a convergent subsequence. Therefore, it is convergent by Lemma 2.4.4. \( ▱ \)
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Yes
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Theorem 2.4.7 Every contractive sequence is convergent.
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Proof: By induction, one has\n\n\[ \left| {{a}_{n + 1} - {a}_{n}}\right| \leq {k}^{n - 1}\left| {{a}_{2} - {a}_{1}}\right| \text{ for all }n \in \mathbb{N}. \]\n\nThus,\n\n\[ \left| {{a}_{n + p} - {a}_{n}}\right| \leq \left| {{a}_{n + 1} - {a}_{n}}\right| + \left| {{a}_{n + 2} - {a}_{n + 1}}\right| + \cdots + \left| {{a}_{n + p} - {a}_{n + p - 1}}\right| \]\n\n\[ \leq \left( {{k}^{n - 1} + {k}^{n} + \cdots + {k}^{n + p - 2}}\right) \left| {{a}_{2} - {a}_{1}}\right| \]\n\n\[ \leq {k}^{n - 1}\left( {1 + k + {k}^{2} + \cdots + {k}^{p - 1}}\right) \left| {{a}_{2} - {a}_{1}}\right| \]\n\n\[ \leq \frac{{k}^{n - 1}}{1 - k}\left| {{a}_{2} - {a}_{1}}\right| \]\n\nfor all \( n, p \in \mathbb{N} \). Since \( {k}^{n - 1} \rightarrow 0 \) as \( n \rightarrow \infty \) (independently of \( p \) ), this implies \( \left\{ {a}_{n}\right\} \) is a Cauchy sequence and, hence, it is convergent. \( ▱ \)
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Yes
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Proposition 2.5.1 Let \( \\left\\{ {a}_{n}\\right\\} \) be a bounded sequence. Define\n\n\[ \n{s}_{n} = \\sup \\left\\{ {{a}_{k} : k \\geq n}\\right\\} \n\]\n\n(2.8)\n\nand\n\n\[ \n{t}_{n} = \\inf \\left\\{ {{a}_{k} : k \\geq n}\\right\\} \n\]\n\n(2.9)\n\nThen \( \\left\\{ {s}_{n}\\right\\} \) and \( \\left\\{ {t}_{n}\\right\\} \) are convergent.
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Proof: If \( n \\leq m \), then \( \\left\\{ {{a}_{k} : k \\geq m}\\right\\} \\subset \\left\\{ {{a}_{k} : k \\geq n}\\right\\} \) . Therefore, it follows from Theorem 1.5.3 that \( {s}_{n} \\geq {s}_{m} \) and, so, the sequence \( \\left\\{ {s}_{n}\\right\\} \) is decreasing. Since \( \\left\\{ {a}_{n}\\right\\} \) is bounded, then so is \( \\left\\{ {s}_{n}\\right\\} \) . In particular, \( \\left\\{ {s}_{n}\\right\\} \) is bounded below. Similarly, \( \\left\\{ {t}_{n}\\right\\} \) is increasing and bounded above. Therefore, both sequences are convergent by Theorem 2.3.1. \( ▱ \)
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Yes
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Theorem 2.5.2 If \( \\left\\{ {a}_{n}\\right\\} \) is not bounded above, then\n\n\[ \n\\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}{s}_{n} = \\infty \n\]\n\nwhere \( \\left\\{ {s}_{n}\\right\\} \) is defined in (2.8).\n\nSimilarly, if \( \\left\\{ {a}_{n}\\right\\} \) is not bounded below, then\n\n\[ \n\\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}{t}_{n} = - \\infty \n\]\n\nwhere \( \\left\\{ {t}_{n}\\right\\} \) is defined in (2.9).
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Proof: Suppose \( \\left\\{ {a}_{n}\\right\\} \) is not bounded above. Then for any \( k \\in \\mathbb{N} \), the set \( \\left\\{ {{a}_{i} : i \\geq k}\\right\\} \) is also not bounded above. Thus, \( {s}_{k} = \\sup \\left\\{ {{a}_{i} : i \\geq k}\\right\\} = \\infty \) for all \( k \) . Therefore, \( \\mathop{\\lim }\\limits_{{k \\rightarrow \\infty }}{s}_{k} = \\infty \) . The proof for the second case is similar. \( ▱ \)
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Yes
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Theorem 2.5.4 Let \( \\left\\{ {a}_{n}\\right\\} \) be a sequence and \( \\ell \\in \\mathbb{R} \) . The following are equivalent:\n\n(a) \( \\lim \\mathop{\\sup }\\limits_{{n \\rightarrow \\infty }}{a}_{n} = \\ell \) .\n\n(b) For any \( \\varepsilon > 0 \), there exists \( N \\in \\mathbb{N} \) such that\n\n\[ \n{a}_{n} < \\ell + \\varepsilon \\text{ for all }n \\geq N,\n\]\n\nand there exists a subsequence of \( \\left\\{ {a}_{{n}_{k}}\\right\\} \) of \( \\left\\{ {a}_{n}\\right\\} \) such that\n\n\[ \n\\mathop{\\lim }\\limits_{{k \\rightarrow \\infty }}{a}_{{n}_{k}} = \\ell \n\]
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Proof: Suppose \( \\mathop{\\limsup }\\limits_{{n \\rightarrow \\infty }}{a}_{n} = \\ell \) . Then \( \\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}{s}_{n} = \\ell \), where \( {s}_{n} \) is defined as in (2.8). For any \( \\varepsilon > 0 \), there exists \( N \\in \\mathbb{N} \) such that\n\n\[ \n\\ell - \\varepsilon < {s}_{n} < \\ell + \\varepsilon \\text{ for all }n \\geq N.\n\]\n\nThis implies \( {s}_{N} = \\sup \\left\\{ {{a}_{n} : n \\geq N}\\right\\} < \\ell + \\varepsilon \) . Thus,\n\n\[ \n{a}_{n} < \\ell + \\varepsilon \\text{ for all }n \\geq N.\n\]\n\nMoreover, for \( \\varepsilon = 1 \), there exists \( {N}_{1} \\in \\mathbb{N} \) such that\n\n\[ \n\\ell - 1 < {s}_{{N}_{1}} = \\sup \\left\\{ {{a}_{n} : n \\geq {N}_{1}}\\right\\} < \\ell + 1.\n\]\n\nThus, there exists \( {n}_{1} \\in \\mathbb{N} \) such that\n\n\[ \n\\ell - 1 < {a}_{{n}_{1}} < \\ell + 1 \n\]\n\nFor \( \\varepsilon = \\frac{1}{2} \), there exists \( {N}_{2} \\in \\mathbb{N} \) and \( {N}_{2} > {n}_{1} \) such that\n\n\[ \n\\ell - \\frac{1}{2} < {s}_{{N}_{2}} = \\sup \\left\\{ {{a}_{n} : n \\geq {N}_{2}}\\right\\} < \\ell + \\frac{1}{2}.\n\]\n\nThus, there exists \( {n}_{2} > {n}_{1} \) such that\n\n\[ \n\\ell - \\frac{1}{2} < {a}_{{n}_{2}} < \\ell + \\frac{1}{2} \n\]\n\nIn this way, we can construct a strictly increasing sequence \( \\left\\{ {n}_{k}\\right\\} \) of positive integers such that\n\n\[ \n\\ell - \\frac{1}{k} < {a}_{{n}_{k}} < \\ell + \\frac{1}{k} \n\]\n\nTherefore, \( \\mathop{\\lim }\\limits_{{k \\rightarrow \\infty }}{a}_{{n}_{k}} = \\ell \) .\n\nWe now prove the converse. Given any \( \\varepsilon > 0 \), there exists \( N \\in \\mathbb{N} \) such that\n\n\[ \n{a}_{n} < \\ell + \\varepsilon \\text{ and }\\ell - \\varepsilon < {a}_{{n}_{k}} < \\ell + \\varepsilon \n\]\n\nfor all \( n \\geq N \) and \( k \\geq N \) . Let any \( m \\geq N \), we have\n\n\[ \n{s}_{m} = \\sup \\left\\{ {{a}_{k} : k \\geq m}\\right\\} \\leq \\ell + \\varepsilon .\n\]\n\nBy Lemma 2.1.8, \( {n}_{m} \\geq m \), so we also have\n\n\[ \n{s}_{m} = \\sup \\left\\{ {{a}_{k} : k \\geq m}\\right\\} \\geq {a}_{{n}_{m}} > \\ell - \\varepsilon .\n\]\n\nTherefore, \( \\mathop{\\lim }\\limits_{{m \\rightarrow \\infty }}{s}_{m} = \\mathop{\\limsup }\\limits_{{m \\rightarrow \\infty }}{a}_{n} = \\ell \) . \( ▱ \)
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Yes
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Theorem 2.5.5 Let \( \\left\\{ {a}_{n}\\right\\} \) be a sequence and \( \\ell \\in \\mathbb{R} \) . The following are equivalent:\n\n(a) \( \\mathop{\\liminf }\\limits_{{n \\rightarrow \\infty }}{a}_{n} = \\ell \) .\n\n(b) For any \( \\varepsilon > 0 \), there exists \( N \\in \\mathbb{N} \) such that\n\n\[ \n{a}_{n} > \\ell - \\varepsilon \\text{ for all }n \\geq N,\n\]\n\nand there exists a subsequence of \( \\left\\{ {a}_{{n}_{k}}\\right\\} \) of \( \\left\\{ {a}_{n}\\right\\} \) such that\n\n\[ \n\\mathop{\\lim }\\limits_{{k \\rightarrow \\infty }}{a}_{{n}_{k}} = \\ell \n\]
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The following corollary follows directly from Theorems 2.5.4 and 2.5.5.
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No
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Corollary 2.5.7 Let \( \\left\\{ {a}_{n}\\right\\} \) be a sequence.\n\n(a) Suppose \( \\mathop{\\limsup }\\limits_{{n \\rightarrow \\infty }}{a}_{n} = \\ell \) and \( \\left\\{ {a}_{{n}_{k}}\\right\\} \) is a subsequence of \( \\left\\{ {a}_{n}\\right\\} \) with\n\n\[ \n\\mathop{\\lim }\\limits_{{k \\rightarrow \\infty }}{a}_{{n}_{k}} = {\\ell }^{\\prime }\n\]\n\nThen \( {\\ell }^{\\prime } \\leq \\ell \) .
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Proof: We prove only (a) because the proof of (b) is similar. By Theorem 2.5.4 and the definition of limits, for any \( \\varepsilon > 0 \), there exists \( N \\in \\mathbb{N} \) such that\n\n\[ \n{a}_{n} < \\ell + \\varepsilon \\text{ and }{\\ell }^{\\prime } - \\varepsilon < {a}_{{n}_{k}} < {\\ell }^{\\prime } + \\varepsilon\n\]\n\nfor all \( n \\geq N \) and \( k \\geq N \) . Since \( {n}_{N} \\geq N \), this implies\n\n\[ \n{\\ell }^{\\prime } - \\varepsilon < {a}_{{n}_{N}} < \\ell + \\varepsilon\n\]\n\nThus, \( {\\ell }^{\\prime } < \\ell + {2\\varepsilon } \) and, hence, \( {\\ell }^{\\prime } \\leq \\ell \) because \( \\varepsilon \) is arbitrary. \( ▱ \)
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Yes
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Theorem 2.5.9 Suppose \( \left\{ {a}_{n}\right\} \) is a sequence such that \( {a}_{n} > 0 \) for every \( n \in \mathbb{N} \) and\n\n\[ \mathop{\limsup }\limits_{{n \rightarrow \infty }}\frac{{a}_{n + 1}}{{a}_{n}} = \ell < 1 \]\n\nThen \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = 0 \) .
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Proof: Choose \( \varepsilon > 0 \) such that \( \ell + \varepsilon < 1 \) . Then there exists \( N \in \mathbb{N} \) such that\n\n\[ \frac{{a}_{n + 1}}{{a}_{n}} < \ell + \varepsilon \text{ for all }n \geq N. \]\n\nLet \( q = \ell + \varepsilon \) . Then \( 0 < q < 1 \) . By induction,\n\n\[ 0 < {a}_{n} \leq {q}^{n - N}{a}_{N}\text{ for all }n \geq N. \]\n\nSince \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{q}^{n - N}{a}_{N} = 0 \), one has \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = 0 \) . \( ▱ \)
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Yes
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Theorem 2.6.1 The following hold:\n\n(a) The subsets \( \varnothing \) and \( \mathbb{R} \) are open.\n\n(b) The union of any collection of open subsets of \( \mathbb{R} \) is open.\n\n(c) The intersection of a finite number of open subsets of \( \mathbb{R} \) is open.
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Proof: The proof of (a) is straightforward.\n\n(b) Suppose \( \left\{ {{G}_{\alpha } : \alpha \in I}\right\} \) is an arbitrary collection of open subsets of \( \mathbb{R} \) . That means \( {G}_{\alpha } \) is open for every \( \alpha \in I \) . Let us show that the set\n\n\[ G = \mathop{\bigcup }\limits_{{\alpha \in I}}{G}_{\alpha } \]\n\nis open. Take any \( a \in G \) . Then there exists \( {\alpha }_{0} \in I \) such that\n\n\[ a \in {G}_{{\alpha }_{0}} \]\n\nSince \( {G}_{{\alpha }_{0}} \) is open, there exists \( \delta > 0 \) such that\n\n\[ B\left( {a;\delta }\right) \subset {G}_{{\alpha }_{0}}. \]\n\nThis implies\n\n\[ B\left( {a;\delta }\right) \subset G \]\n\nbecause \( {G}_{{\alpha }_{0}} \subset G \) . Thus, \( G \) is open.\n\n(c) Suppose \( {G}_{i}, i = 1,\ldots, n \), are open subsets of \( \mathbb{R} \) . Let us show that the set\n\n\[ G = \mathop{\bigcap }\limits_{{i = 1}}^{n}{G}_{i} \]\n\nis also open. Take any \( a \in G \) . Then \( a \in {G}_{i} \) for \( i = 1,\ldots, n \) . Since each \( {G}_{i} \) is open, there exists \( {\delta }_{i} > 0 \) such that\n\n\[ B\left( {a;{\delta }_{i}}\right) \subset {G}_{i} \]\n\nLet \( \delta = \min \left\{ {{\delta }_{i} : i = 1,\ldots, n}\right\} \) . Then \( \delta > 0 \) and\n\n\[ B\left( {a;\delta }\right) \subset G\text{.} \]\n\nThus, \( G \) is open. \( ▱ \)
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No
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The following hold:\n\n(a) The sets \( \varnothing \) and \( \mathbb{R} \) are closed.\n\n(b) The intersection of any collection of closed subsets of \( \mathbb{R} \) is closed.\n\n(c) The union of a finite number of closed subsets of \( \mathbb{R} \) is closed.
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Proof: The proofs for these are simple using the De Morgan's law. Let us prove, for instance, (b). Let \( \left\{ {{S}_{\alpha } : \alpha \in I}\right\} \) be a collection of closed sets. We will prove that the set\n\n\[ S = \mathop{\bigcap }\limits_{{\alpha \in I}}{S}_{\alpha } \]\n\nis also closed. We have\n\n\[ {S}^{c} = {\left( \mathop{\bigcap }\limits_{{\alpha \in I}}{S}_{\alpha }\right) }^{c} = \mathop{\bigcup }\limits_{{\alpha \in I}}{S}_{\alpha }^{c} \]\n\nThus, \( {S}^{c} \) is open because it is a union of opens sets in \( \mathbb{R} \) (Theorem 2.6.1(b)). Therefore, \( S \) is closed. \( ▱ \)
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No
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Theorem 2.6.3 A subset \( A \) of \( \mathbb{R} \) is closed if and only if for any sequence \( \left\{ {a}_{n}\right\} \) in \( A \) that converges to a point \( a \in \mathbb{R} \), it follows that \( a \in A \) .
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Proof: Suppose \( A \) is a closed subset of \( \mathbb{R} \) and \( \left\{ {a}_{n}\right\} \) is a sequence in \( A \) that converges to \( a \) . Suppose by contradiction that \( a \notin A \) . Since \( A \) is closed, there exists \( \varepsilon > 0 \) such that \( B\left( {a;\varepsilon }\right) = \left( {a - \varepsilon, a + \varepsilon }\right) \subset {A}^{c} \) . Since \( \left\{ {a}_{n}\right\} \) converges to \( a \), there exists \( N \in \mathbb{N} \) such that\n\n\[ a - \varepsilon < {a}_{N} < a + \varepsilon . \]\n\nThis implies \( {a}_{N} \in {A}^{c} \), a contradiction.\n\nLet us now prove the converse. Suppose by contradiction that \( A \) is not closed. Then \( {A}^{c} \) is not open. Since \( {A}^{c} \) is not open, there exists \( a \in {A}^{c} \) such that for any \( \varepsilon > 0 \), one has \( B\left( {a;\varepsilon }\right) \cap A \neq \varnothing \) . In particular, for such an \( a \) and for each \( n \in \mathbb{N} \), there exists \( {a}_{n} \in B\left( {a;\frac{1}{n}}\right) \cap A \) . It is clear that the sequence \( \left\{ {a}_{n}\right\} \) is in \( A \) and it is convergent to \( a \) (because \( \left| {{a}_{n} - a}\right| < \frac{1}{n} \), for all \( n \in \mathbb{N} \) ). This is a contradiction since \( a \notin A \) . Therefore, \( A \) is closed. \( ▱ \)
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Yes
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Theorem 2.6.4 If \( A \) is a nonempty subset of \( \mathbb{R} \) that is closed and bounded above, then \( \max A \) exists. Similarly, if \( A \) is a nonempty subset of \( \mathbb{R} \) that is closed and bounded below, then \( \min A \) exists
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Proof: Let \( A \) be a nonempty closed set that is bounded above. Then sup \( A \) exists. Let \( m = \sup A \) . To complete the proof, we will show that \( m \in A \) . Assume by contradiction that \( m \notin A \) . Then \( m \in {A}^{c} \) , which is an open set. So there exists \( \delta > 0 \) such that\n\n\[ \left( {m - \delta, m + \delta }\right) \subset {A}^{c}.\n\]\n\nThis means there exists no \( a \in A \) with\n\n\[ m - \delta < a \leq m.\text{.}\n\]\n\nThis contradicts the fact that \( m \) is the least upper bound of \( A \) (see Proposition 1.5.1). Therefore, \( \max A \) exists. \( ▱ \)
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Yes
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Theorem 2.6.5 A subset \( A \) of \( \mathbb{R} \) is compact if and only if it is closed and bounded.
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Proof: Suppose \( A \) is a compact subset of \( \mathbb{R} \) . Let us first show that \( A \) is bounded. Suppose, by contradiction, that \( A \) is not bounded. Then for every \( n \in \mathbb{N} \), there exists \( {a}_{n} \in A \) such that\n\n\[ \left| {a}_{n}\right| \geq n \]\n\nSince \( A \) is compact, there exists a subsequence \( \left\{ {a}_{{n}_{k}}\right\} \) that converges to some \( a \in A \) . Then\n\n\[ \left| {a}_{{n}_{k}}\right| \geq {n}_{k} \geq k\;\text{ for all }k \]\n\nTherefore, \( \mathop{\lim }\limits_{{k \rightarrow \infty }}\left| {a}_{{n}_{k}}\right| = \infty \) . This is a contradiction because \( \left\{ \left| {a}_{{n}_{k}}\right| \right\} \) converges to \( \left| a\right| \) . Thus \( A \) is bounded.\n\nLet us now show that \( A \) is closed. Let \( \left\{ {a}_{n}\right\} \) be a sequence in \( A \) that converges to a point \( a \in \mathbb{R} \) . By the definition of compactness, \( \left\{ {a}_{n}\right\} \) has a subsequence \( \left\{ {a}_{{n}_{k}}\right\} \) that converges to \( b \in A \) . Then \( a = b \in A \) and, hence, \( A \) is closed by Theorem 2.6.3.\n\nFor the converse, suppose \( A \) is closed and bounded and let \( \left\{ {a}_{n}\right\} \) be a sequence in \( A \) . Since \( A \) is bounded, the sequence is bounded and, by the Bolzano-Weierstrass theorem (Theorem 2.4.1), it\n\n\( {}^{1} \) This definition of compactness is more commonly referred to as sequential compactness. has a convergent subsequence, \( \left\{ {a}_{{n}_{k}}\right\} \) . Say, \( \mathop{\lim }\limits_{{k \rightarrow \infty }}{a}_{{n}_{k}} = a \) . It now follows from Theorem 2.6.3 that \( a \in A \) . This shows that \( A \) is compact as desired. \( ▱ \)
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Yes
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Theorem 2.6.6 Any infinite bounded subset of \( \mathbb{R} \) has at least one limit point.
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Proof: Let \( A \) be an infinite subset of \( \mathbb{R} \) and let \( \left\{ {a}_{n}\right\} \) be a sequence of \( A \) such that\n\n\[ \n{a}_{m} \neq {a}_{n}\text{ for }m \neq n \]\n\n(see Theorem 1.2.7). Since \( \left\{ {a}_{n}\right\} \) is bounded, by the Bolzano-Weierstrass theorem (Theorem 2.4.1), it has a convergent subsequence \( \left\{ {a}_{{n}_{k}}\right\} \) . Set \( b = \mathop{\lim }\limits_{{k \rightarrow \infty }}{a}_{{n}_{k}} \) . Given \( \delta > 0 \), there exists \( K \in \mathbb{N} \) such that \( {a}_{{n}_{k}} \in B\left( {b;\delta }\right) \) for \( k \geq K \) . Since the set \( \left\{ {{a}_{{n}_{k}} : k \geq K}\right\} \) is infinite, it follows that \( b \) is a limit point of A. \( ▱ \)
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Yes
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Theorem 2.6.7 Let \( D \) be a subset of \( \mathbb{R} \) . A subset \( V \) of \( D \) is open in \( D \) if and only if there exists an open subset \( G \) of \( \mathbb{R} \) such that\n\n\[ V = D \cap G. \]
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Proof: Suppose \( V \) is open in \( D \) . By definition, for every \( a \in V \), there exists \( {\delta }_{a} > 0 \) such that\n\n\[ B\left( {a;{\delta }_{a}}\right) \cap D \subset V. \]\n\nDefine\n\n\[ G = { \cup }_{a \in V}B\left( {a;{\delta }_{a}}\right) \]\n\nThen \( G \) is a union of open subsets of \( \mathbb{R} \), so \( G \) is open. Moreover,\n\n\[ V \subset G \cap D = { \cup }_{a \in V}\left\lbrack {B\left( {a;{\delta }_{a}}\right) \cap D}\right\rbrack \subset V. \]\n\nTherefore, \( V = G \cap D \) .\n\nLet us now prove the converse. Suppose \( V = G \cap D \), where \( G \) is an open set. For any \( a \in V \), we have \( a \in G \), so there exists \( \delta > 0 \) such that\n\n\[ B\left( {a;\delta }\right) \subset G. \]\n\nIt follows that\n\n\[ B\left( {a;\delta }\right) \cap D \subset G \cap D = V. \]\n\nThe proof is now complete. \( ▱ \)
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Yes
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Theorem 2.6.9 Let \( D \) be a subset of \( \mathbb{R} \) . A subset \( K \) of \( D \) is closed in \( D \) if and only if there exists a closed subset \( F \) of \( \mathbb{R} \) such that\n\n\[ K = D \cap F. \]
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Proof: Suppose \( K \) is a closed set in \( D \) . Then \( D \smallsetminus K \) is open in \( D \) . By Theorem 2.6.7, there exists an open set \( G \) such that\n\n\[ D \smallsetminus K = D \cap G. \]\n\nIt follows that\n\n\[ K = D \smallsetminus \left( {D \smallsetminus K}\right) = D \smallsetminus \left( {D \cap G}\right) = D \smallsetminus G = D \cap {G}^{c}. \]\n\nLet \( F = {G}^{c} \) . Then \( F \) is a closed subset of \( \mathbb{R} \) and \( K = D \cap F \) .\n\nConversely, suppose that there exists a closed subset \( F \) of \( \mathbb{R} \) such that \( K = D \cap F \) . Then\n\n\[ D \smallsetminus K = D \smallsetminus \left( {D \cap F}\right) = D \smallsetminus F = D \cap {F}^{c}. \]\n\nSince \( {F}^{c} \) is an open subset of \( \mathbb{R} \), applying Theorem 2.6.7 again, one has that \( D \smallsetminus K \) is open in \( D \) . Therefore, \( K \) is closed in \( D \) by definition. \( ▱ \)
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Yes
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Corollary 2.6.10 Let \( D \) be a subset of \( \mathbb{R} \) . A subset \( K \) of \( D \) is closed in \( D \) if and only if for every sequence \( \left\{ {x}_{k}\right\} \) in \( K \) that converges to a point \( \bar{x} \in D \) it follows that \( \bar{x} \in K \) .
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Proof: Let \( D \) be a subset of \( \mathbb{R} \) . Suppose \( K \) is closed in \( D \) . By Theorem 2.6.9, there exists a closed subset \( F \) of \( \mathbb{R} \) such that\n\n\[ K = D \cap F\text{.}\]\n\nLet \( \left\{ {x}_{k}\right\} \) be a sequence in \( K \) that converges to a point \( \bar{x} \in D \) . Since \( \left\{ {x}_{k}\right\} \) is also a sequence in \( F \) and \( F \) is a closed subset of \( \mathbb{R},\bar{x} \in F \) . Thus, \( \bar{x} \in D \cap F = K \) .\n\nLet us prove the converse. Suppose by contradiction that \( K \) is not closed in \( D \) or \( D \smallsetminus K \) is not open in \( D \) . Then there exists \( \bar{x} \in D \smallsetminus K \) such that for every \( \delta > 0 \), one has\n\n\[ B\left( {\bar{x};\delta }\right) \cap D \nsubseteq D \smallsetminus K.\]\n\nIn particular, for every \( k \in \mathbb{N} \),\n\n\[ B\left( {\bar{x};\frac{1}{k}}\right) \cap D \nsubseteq D \smallsetminus K.\]\n\nFor each \( k \in \mathbb{N} \), choose \( {x}_{k} \in B\left( {\bar{x};\frac{1}{k}}\right) \cap D \) such that \( {x}_{k} \notin D \smallsetminus K \) . Then \( \left\{ {x}_{k}\right\} \) is a sequence in \( K \) and, moreover, \( \left\{ {x}_{k}\right\} \) converges to \( \bar{x} \in \widetilde{D} \) . Then \( \bar{x} \in K \) . This is a contradiction. We conclude that \( K \) is closed in \( D \) . \( ▱ \)
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Yes
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Theorem 3.1.2 — Sequential Characterization of Limits. Let \( f:D \rightarrow \mathbb{R} \) and let \( \bar{x} \) be a limit point of \( D \) . Then\n\n\[ \mathop{\lim }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) = \ell \]\n\n(3.1)\n\nif and only if\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}f\left( {x}_{n}\right) = \ell \]\n\n(3.2)\n\nfor every sequence \( \left\{ {x}_{n}\right\} \) in \( D \) such that \( {x}_{n} \neq \bar{x} \) for every \( n \) and \( \left\{ {x}_{n}\right\} \) converges to \( \bar{x} \) .
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Proof: Suppose (3.1) holds. Let \( \left\{ {x}_{n}\right\} \) be a sequence in \( D \) with \( {x}_{n} \neq \bar{x} \) for every \( n \) and such that \( \left\{ {x}_{n}\right\} \) converges to \( \bar{x} \) . Given any \( \varepsilon > 0 \), there exists \( \delta > 0 \) such that \( \left| {f\left( x\right) - \ell }\right| < \varepsilon \) whenever \( x \in D \) and \( 0 < \left| {x - \bar{x}}\right| < \delta \) . Then there exists \( N \in \mathbb{N} \) with \( 0 < \left| {{x}_{n} - \bar{x}}\right| < \delta \) for all \( n \geq N \) . For such \( n \), we have\n\n\[ \left| {f\left( {x}_{n}\right) - \ell }\right| < \varepsilon \]\n\nThis implies (3.2).\n\nConversely, suppose (3.1) is false. Then there exists \( {\varepsilon }_{0} > 0 \) such that for every \( \delta > 0 \), there exists \( x \in D \) with \( 0 < \left| {x - \bar{x}}\right| < \delta \) and \( \left| {f\left( x\right) - \ell }\right| \geq {\varepsilon }_{0} \) . Thus, for every \( n \in \mathbb{N} \), there exists \( {x}_{n} \in D \) with \( 0 < \left| {{x}_{n} - \bar{x}}\right| < \frac{1}{n} \) and \( \left| {f\left( {x}_{n}\right) - \ell }\right| \geq {\varepsilon }_{0} \) . By the squeeze theorem (Theorem 2.1.6), the sequence \( \left\{ {x}_{n}\right\} \) converges to \( \bar{x} \) . Moreover, \( {x}_{n} \neq \bar{x} \) for every \( n \) . This shows that (3.2) is false. It follows that (3.2) implies (3.1) and the proof is complete.
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Yes
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Corollary 3.1.3 Let \( f : D \rightarrow \mathbb{R} \) and let \( \bar{x} \) be a limit point of \( D \) . If \( f \) has a limit at \( \bar{x} \), then this limit is unique.
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Proof: Suppose by contradiction that \( f \) has two different limits \( {\ell }_{1} \) and \( {\ell }_{2} \) . Let \( \left\{ {x}_{n}\right\} \) be a sequence in \( D \smallsetminus \{ \bar{x}\} \) that converges to \( \bar{x} \) . By Theorem 3.1.2, the sequence \( \left\{ {f\left( {x}_{n}\right) }\right\} \) converges to two different limits \( {\ell }_{1} \) and \( {\ell }_{2} \) . This is a contradiction to Theorem 2.1.3. \( ▱ \)
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Yes
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Corollary 3.1.4 Let \( f : D \rightarrow \mathbb{R} \) and let \( \bar{x} \) be a limit point of \( D \) . Then \( f \) does not have a limit at \( \bar{x} \) if and only if there exists a sequence \( \left\{ {x}_{n}\right\} \) in \( D \) such that \( {x}_{n} \neq \bar{x} \) for every \( n,\left\{ {x}_{n}\right\} \) converges to \( \bar{x} \), and \( \left\{ {f\left( {x}_{n}\right) }\right\} \) does not converge.
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Example 3.1.5 Consider the Dirichlet function \( f : \mathbb{R} \rightarrow \mathbb{R} \) given by \[ f\left( x\right) = \left\{ \begin{array}{ll} 1, & \text{ if }x \in \mathbb{Q}; \\ 0, & \text{ if }x \in {\mathbb{Q}}^{c}. \end{array}\right. \] Then \( \mathop{\lim }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) \) does not exist for any \( \bar{x} \in \mathbb{R} \) . Indeed, fix \( \bar{x} \in \mathbb{R} \) and choose two sequences \( \left\{ {r}_{n}\right\} \) , \( \left\{ {s}_{n}\right\} \) converging to \( \bar{x} \) such that \( {r}_{n} \in \mathbb{Q} \) and \( {s}_{n} \notin \mathbb{Q} \) for all \( n \in \mathbb{N} \) . Define a new sequence \( \left\{ {x}_{n}\right\} \) by \[ {x}_{n} = \left\{ \begin{array}{ll} {r}_{k}, & \text{ if }n = {2k} \\ {s}_{k}, & \text{ if }n = {2k} - 1 \end{array}\right. \] It is clear that \( \left\{ {x}_{n}\right\} \) converges to \( \bar{x} \) . Moreover, since \( \left\{ {f\left( {r}_{n}\right) }\right\} \) converges to 1 and \( \left\{ {f\left( {s}_{n}\right) }\right\} \) converges to 0, Theorem 2.1.9 implies that the sequence \( \left\{ {f\left( {x}_{n}\right) }\right\} \) does not converge. It follows from the sequential characterization of limits that \( \mathop{\lim }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) \) does not exist.
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Yes
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Theorem 3.1.5 Let \( f, g : D \rightarrow \mathbb{R} \) and let \( \bar{x} \) be a limit point of \( D \) . Suppose that\n\n\[ \mathop{\lim }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) = {\ell }_{1},\mathop{\lim }\limits_{{x \rightarrow \bar{x}}}g\left( x\right) = {\ell }_{2}, \]\n\nand that there exists \( \delta > 0 \) such that\n\n\[ f\left( x\right) \leq g\left( x\right) \text{ for all }x \in B\left( {\bar{x};\delta }\right) \cap D, x \neq \bar{x}. \]\n\nThen \( {\ell }_{1} \leq {\ell }_{2} \) .
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Proof: Let \( \left\{ {x}_{n}\right\} \) be a sequence in \( B\left( {\bar{x};\delta }\right) \cap D = \left( {\bar{x} - \delta ,\bar{x} + \delta }\right) \cap D \) that converges to \( \bar{x} \) and \( {x}_{n} \neq \bar{x} \) for all \( n \) . By Theorem 3.1.2,\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}f\left( {x}_{n}\right) = {\ell }_{1}\text{ and }\mathop{\lim }\limits_{{n \rightarrow \infty }}g\left( {x}_{n}\right) = {\ell }_{2}. \]\n\nSince \( f\left( {x}_{n}\right) \leq g\left( {x}_{n}\right) \) for all \( n \in \mathbb{N} \), applying Theorem 2.1.5, we obtain \( {\ell }_{1} \leq {\ell }_{2} \) .
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Yes
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Theorem 3.1.6 Let \( f, g : D \rightarrow \mathbb{R} \) and let \( \bar{x} \) be a limit point of \( D \) . Suppose\n\n\[ \mathop{\lim }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) = {\ell }_{1},\mathop{\lim }\limits_{{x \rightarrow \bar{x}}}g\left( x\right) = {\ell }_{2} \]\n\nand \( {\ell }_{1} < {\ell }_{2} \) . Then there exists \( \delta > 0 \) such that\n\n\[ f\left( x\right) < g\left( x\right) \text{ for all }x \in B\left( {\bar{x};\delta }\right) \cap D, x \neq \bar{x}. \]
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Proof: Choose \( \varepsilon > 0 \) such that \( {\ell }_{1} + \varepsilon < {\ell }_{2} - \varepsilon \) (equivalently, such that \( \varepsilon < \frac{{\ell }_{2} - {\ell }_{1}}{2} \) ). Then there exists \( \delta > 0 \) such that\n\n\[ {\ell }_{1} - \varepsilon < f\left( x\right) < {\ell }_{1} + \varepsilon \text{ and }{\ell }_{2} - \varepsilon < g\left( x\right) < {\ell }_{2} + \varepsilon \]\n\nfor all \( x \in B\left( {\bar{x};\delta }\right) \cap D, x \neq \bar{x} \) . Thus,\n\n\[ f\left( x\right) < {\ell }_{1} + \varepsilon < {\ell }_{2} - \varepsilon < g\left( x\right) \text{ for all }x \in B\left( {\bar{x};\delta }\right) \cap D, x \neq \bar{x}. \]\n\nThe proof is now complete.
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Yes
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Theorem 3.1.7 Let \( f, g, h : D \rightarrow \mathbb{R} \) and let \( \bar{x} \) be a limit point of \( D \) . Suppose there exists \( \delta > 0 \) such that \( f\left( x\right) \leq g\left( x\right) \leq h\left( x\right) \) for all \( x \in B\left( {\bar{x};\delta }\right) \cap D, x \neq \bar{x} \) . If \( \mathop{\lim }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) = \mathop{\lim }\limits_{{x \rightarrow \bar{x}}}h\left( x\right) = \ell \), then \( \mathop{\lim }\limits_{{x \rightarrow \bar{x}}}g\left( x\right) = \ell \) .
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Proof: The proof is straightforward using Theorem 2.1.6 and Theorem 3.1.2. \( ▱ \)
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No
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Theorem 3.2.1 Let \( f, g : D \rightarrow \mathbb{R} \) and let \( c \in \mathbb{R} \) . Suppose \( \bar{x} \) is a limit point of \( D \) and\n\n\[ \mathop{\lim }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) = \ell ,\mathop{\lim }\limits_{{x \rightarrow \bar{x}}}g\left( x\right) = m.\]\n\nThen\n\n(a) \( \mathop{\lim }\limits_{{x \rightarrow \bar{x}}}\left( {f + g}\right) \left( x\right) = \ell + m \) ,\n\n(b) \( \mathop{\lim }\limits_{{x \rightarrow \bar{x}}}\left( {fg}\right) \left( x\right) = \ell m \) ,\n\n(c) \( \mathop{\lim }\limits_{{x \rightarrow \bar{x}}}\left( {cf}\right) \left( x\right) = c\ell \) ,\n\n(d) \( \mathop{\lim }\limits_{{x \rightarrow \bar{x}}}\left( \frac{f}{g}\right) \left( x\right) = \frac{\ell }{m} \) provided that \( m \neq 0 \) .
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Proof: Let us first prove (a). Let \( \left\{ {x}_{n}\right\} \) be a sequence in \( D \) that converges to \( \bar{x} \) and \( {x}_{n} \neq \bar{x} \) for every \( n \) . By Theorem 3.1.2,\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}f\left( {x}_{n}\right) = \ell \text{ and }\mathop{\lim }\limits_{{n \rightarrow \infty }}g\left( {x}_{n}\right) = m.\]\n\nIt follows from Theorem 2.2.1 that\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {f\left( {x}_{n}\right) + g\left( {x}_{n}\right) }\right) = \ell + m.\]\n\nApplying Theorem 3.1.2 again, we get \( \mathop{\lim }\limits_{{x \rightarrow \bar{x}}}\left( {f + g}\right) \left( x\right) = \ell + m \) . The proofs of (b) and (c) are similar.\n\nLet us now show that if \( m \neq 0 \), then \( \bar{x} \) is a limit point of \( \widetilde{D} \) . Since \( \bar{x} \) is a limit point of \( D \), there is a sequence \( \left\{ {u}_{k}\right\} \) in \( D \) converging to \( \bar{x} \) such that \( {u}_{k} \neq \bar{x} \) for every \( k \) . Since \( m \neq 0 \), it follows from an easy application of Theorem 3.1.6 that there exists \( \delta > 0 \) with\n\n\[ g\left( x\right) \neq 0\text{whenever}0 < \left| {x - \bar{x}}\right| < \delta, x \in D.\]\n\nThis implies\n\n\[ x \in \widetilde{D}\text{whenever}0 < \left| {x - \bar{x}}\right| < \delta, x \in D.\]\n\nThen \( {u}_{k} \in \widetilde{D} \) for all \( k \) sufficiently large, and hence \( \bar{x} \) is a limit point of \( \widetilde{D} \) . The rest of the proof of (d) can be completed easily following the proof of (a). \( ▱ \)
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Yes
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Theorem 3.2.2 (Cauchy’s criterion) Let \( f : D \rightarrow \mathbb{R} \) and let \( \bar{x} \) be a limit point of \( D \) . Then \( f \) has a limit at \( \bar{x} \) if and only if for any \( \varepsilon > 0 \), there exists \( \delta > 0 \) such that\n\n\[ \left| {f\left( r\right) - f\left( s\right) }\right| < \varepsilon \text{whenever}r, s \in D\text{and}0 < \left| {r - \bar{x}}\right| < \delta ,0 < \left| {s - \bar{x}}\right| < \delta \text{.} \]
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Proof: Suppose \( \mathop{\lim }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) = \ell \) . Given \( \varepsilon > 0 \), there exists \( \delta > 0 \) such that\n\n\[ \left| {f\left( x\right) - \ell }\right| < \frac{\varepsilon }{2}\text{ whenever }x \in D\text{ and }0 < \left| {x - \bar{x}}\right| < \delta . \]\n\nThus, for \( r, s \in D \) with \( 0 < \left| {r - \bar{x}}\right| < \delta \) and \( 0 < \left| {s - \bar{x}}\right| < \delta \), we have\n\n\[ \left| {f\left( r\right) - f\left( s\right) }\right| \leq \left| {f\left( r\right) - \ell }\right| + \left| {\ell - f\left( s\right) }\right| < \varepsilon . \]\n\nLet us prove the converse. Fix a sequence \( \left\{ {u}_{n}\right\} \) in \( D \) such with \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{u}_{n} = \bar{x} \) and \( {u}_{n} \neq \bar{x} \) for every \( n \) . Given \( \varepsilon > 0 \), there exists \( \delta > 0 \) such that\n\n\[ \left| {f\left( r\right) - f\left( s\right) }\right| < \varepsilon \text{whenever}r, s \in D\text{and}0 < \left| {r - \bar{x}}\right| < \delta ,0 < \left| {s - \bar{x}}\right| < \delta \text{.} \]\n\nThen there exists \( N \in \mathbb{N} \) satisfying\n\n\[ 0 < \left| {{u}_{n} - \bar{x}}\right| < \delta \text{ for all }n \geq N. \]\n\nThis implies\n\n\[ \left| {f\left( {u}_{n}\right) - f\left( {u}_{m}\right) }\right| < \varepsilon \text{ for all }m, n \geq N. \]\n\nThus, \( \left\{ {f\left( {u}_{n}\right) }\right\} \) is a Cauchy sequence, and hence there exists \( \ell \in \mathbb{R} \) such that\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}f\left( {u}_{n}\right) = \ell \]\n\nWe now prove that \( f \) has limit \( \ell \) at \( \bar{x} \) using Theorem 3.1.2. Let \( \left\{ {x}_{n}\right\} \) be a sequence in \( D \) such that \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n} = \bar{x} \) and \( {x}_{n} \neq \bar{x} \) for every \( n \) . By the previous argument, there exists \( {\ell }^{\prime } \in \mathbb{R} \) such that\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}f\left( {x}_{n}\right) = {\ell }^{\prime } \]\n\nFix any \( \varepsilon > 0 \) and let \( \delta > 0 \) satisfy (3.3). There exists \( K \in \mathbb{N} \) such that\n\n\[ \left| {{u}_{n} - \bar{x}}\right| < \delta \text{ and }\left| {{x}_{n} - \bar{x}}\right| < \delta \]\n\nfor all \( n \geq K \) . Then \( \left| {f\left( {u}_{n}\right) - f\left( {x}_{n}\right) }\right| < \varepsilon \) for such \( n \) . Letting \( n \rightarrow \infty \), we have \( \left| {\ell - {\ell }^{\prime }}\right| \leq \varepsilon \) . Thus, \( \ell = {\ell }^{\prime } \]\nsince \( \varepsilon \) is arbitrary. It now follows from Theorem 3.1.2 that \( \mathop{\lim }\limits_{{x \rightarrow \bar{x}}}f\left( x\right) = \ell \) . \( ▱ \)
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Yes
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Theorem 3.2.4 Suppose \( f : \left( {a, b}\right) \rightarrow \mathbb{R} \) is increasing on \( \left( {a, b}\right) \) and \( \bar{x} \in \left( {a, b}\right) \) . Then \( \mathop{\lim }\limits_{{x \rightarrow {\bar{x}}^{ - }}}f\left( x\right) \) and \( \mathop{\lim }\limits_{{x \rightarrow {\bar{x}}^{ + }}}f\left( x\right) \) exist. Moreover,\n\n\[ \mathop{\sup }\limits_{{a < x < \bar{x}}}f\left( x\right) = \mathop{\lim }\limits_{{x \rightarrow {\bar{x}}^{ - }}}f\left( x\right) \leq f\left( \bar{x}\right) \leq \mathop{\lim }\limits_{{x \rightarrow {\bar{x}}^{ + }}}f\left( x\right) = \mathop{\inf }\limits_{{\bar{x} < x < b}}f\left( x\right) . \]
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Proof: Since \( f\left( x\right) \leq f\left( \bar{x}\right) \) for all \( x \in \left( {a,\bar{x}}\right) \), the set\n\n\[ \{ f\left( x\right) : x \in \left( {a,\bar{x}}\right) \} \]\n\nis nonempty and bounded above. Thus,\n\n\[ \ell = \mathop{\sup }\limits_{{a < x < \bar{x}}}f\left( x\right) \]\n\nis a real number. We will show that \( \mathop{\lim }\limits_{{x \rightarrow {\bar{x}}^{ - }}}f\left( x\right) = \ell \) . For any \( \varepsilon > 0 \), by the definition of the least upper bound, there exists \( a < {x}_{1} < \bar{x} \) such that\n\n\[ \ell - \varepsilon < f\left( {x}_{1}\right) \]\n\nLet \( \delta = \bar{x} - {x}_{1} > 0 \) . Using the increasing monotonicity, we get\n\n\[ \ell - \varepsilon < f\left( {x}_{1}\right) \leq f\left( x\right) \leq \ell < \ell + \varepsilon \text{ for all }x \in \left( {{x}_{1},\bar{x}}\right) = {B}_{ - }\left( {\bar{x};\delta }\right) . \]\n\nTherefore, \( \mathop{\lim }\limits_{{x \rightarrow {\bar{x}}^{ - }}}f\left( x\right) = \ell \) . The rest of the proof of the theorem is similar. \( ▱ \)
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No
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Theorem 3.3.2 Let \( f : D \rightarrow \mathbb{R} \) and let \( {x}_{0} \in D \) be a limit point of \( D \) . Then \( f \) is continuous at \( {x}_{0} \) if and only if\n\n\[ \mathop{\lim }\limits_{{x \rightarrow {x}_{0}}}f\left( x\right) = f\left( {x}_{0}\right) \]
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- Example 3.3.2 Let \( f : \mathbb{R} \rightarrow \mathbb{R} \) be given by \( f\left( x\right) = 3{x}^{2} - {2x} + 1 \) . Fix \( {x}_{0} \in \mathbb{R} \) . Since, from the results of the previous theorem, we have\n\n\[ \mathop{\lim }\limits_{{x \rightarrow {x}_{0}}}f\left( x\right) = \mathop{\lim }\limits_{{x \rightarrow {x}_{0}}}\left( {3{x}^{2} - {2x} + 1}\right) = 3{x}_{0}^{2} - 2{x}_{0} + 1 = f\left( {x}_{0}\right) ,\]\n\n it follows that \( f \) is continuous at \( {x}_{0} \) .
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Yes
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Theorem 3.3.4 Let \( f, g : D \rightarrow \mathbb{R} \) and let \( {x}_{0} \in D \) . Suppose \( f \) and \( g \) are continuous at \( {x}_{0} \) . Then\n\n(a) \( f + g \) and \( {fg} \) are continuous at \( {x}_{0} \) .\n\n(b) \( {cf} \) is continuous at \( {x}_{0} \) for any constant \( c \) .\n\n(c) If \( g\left( {x}_{0}\right) \neq 0 \), then \( \frac{f}{g} \) (defined on \( \widetilde{D} = \{ x \in D : g\left( x\right) \neq 0\} \) ) is continuous at \( {x}_{0} \) .
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Proof: We prove (a) and leave the other parts as an exercise. We will use Theorem 3.3.3. Let \( \left\{ {x}_{k}\right\} \) be a sequence in \( D \) that converges to \( {x}_{0} \) . Since \( f \) and \( g \) are continuous at \( {x}_{0} \), by Theorem 3.3.3 we obtain that \( \left\{ {f\left( {x}_{k}\right) }\right\} \) converges to \( f\left( {x}_{0}\right) \) and \( \left\{ {g\left( {x}_{k}\right) }\right\} \) converges to \( g\left( {x}_{0}\right) \) . By Theorem 2.2.1 (a), we get that \( \left\{ {f\left( {x}_{k}\right) + g\left( {x}_{k}\right) }\right\} \) converges to \( f\left( {x}_{0}\right) + g\left( {x}_{0}\right) \) . Therefore,\n\n\[ \mathop{\lim }\limits_{{k \rightarrow \infty }}\left( {f + g}\right) \left( {x}_{k}\right) = \mathop{\lim }\limits_{{k \rightarrow \infty }}f\left( {x}_{k}\right) + g\left( {x}_{k}\right) = f\left( {x}_{0}\right) + g\left( {x}_{0}\right) = \left( {f + g}\right) \left( {x}_{0}\right) . \]\n\nSince \( \left\{ {x}_{k}\right\} \) was arbitrary, using Theorem 3.3.3 again we conclude \( f + g \) is continuous at \( {x}_{0} \) . \( ▱ \)
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No
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Theorem 3.4.1 Let \( D \) be a nonempty compact subset of \( \mathbb{R} \) and let \( f : D \rightarrow \mathbb{R} \) be a continuous function. Then \( f\left( D\right) \) is a compact subset of \( \mathbb{R} \) . In particular, \( f\left( D\right) \) is closed and bounded.
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Proof: Take any sequence \( \left\{ {y}_{n}\right\} \) in \( f\left( D\right) \) . Then for each \( n \), there exists \( {a}_{n} \in D \) such that \( {y}_{n} = f\left( {a}_{n}\right) \) . Since \( D \) is compact, there exists a subsequence \( \left\{ {a}_{{n}_{k}}\right\} \) of \( \left\{ {a}_{n}\right\} \) and a point \( a \in D \) such that\n\n\[ \mathop{\lim }\limits_{{k \rightarrow \infty }}{a}_{{n}_{k}} = a \in D \]\n\nIt now follows from Theorem 3.3.3 that\n\n\[ \mathop{\lim }\limits_{{k \rightarrow \infty }}{y}_{{n}_{k}} = \mathop{\lim }\limits_{{k \rightarrow \infty }}f\left( {a}_{{n}_{k}}\right) = f\left( a\right) \in f\left( D\right) . \]\n\nTherefore, \( f\left( D\right) \) is compact.\n\nThe final conclusion follows from Theorem 2.6.5 \( ▱ \)
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Yes
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Theorem 3.4.2 — Extreme Value Theorem. Suppose \( f : D \rightarrow \mathbb{R} \) is continuous and \( D \) is a compact set. Then \( f \) has an absolute minimum and an absolute maximum on \( D \) .
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Proof: Since \( D \) is compact, \( A = f\left( D\right) \) is closed and bounded (see Theorem 2.6.5). Let\n\n\[ m = \inf A = \mathop{\inf }\limits_{{x \in D}}f\left( x\right) . \]\n\nIn particular, \( m \in \mathbb{R} \) . For every \( n \in \mathbb{N} \), there exists \( {a}_{n} \in A \) such that\n\n\[ m \leq {a}_{n} < m + 1/n \]\n\nFor each \( n \), since \( {a}_{n} \in A = f\left( D\right) \), there exists \( {x}_{n} \in D \) such that \( {a}_{n} = f\left( {x}_{n}\right) \) and, hence,\n\n\[ m \leq f\left( {x}_{n}\right) < m + 1/n \]\n\nBy the compactness of \( D \), there exists an element \( \bar{x} \in D \) and a subsequence \( \left\{ {x}_{{n}_{k}}\right\} \) that converges to\n\n\( \bar{x} \in D \) as \( k \rightarrow \infty \) . Because\n\n\[ m \leq f\left( {x}_{{n}_{k}}\right) < m + \frac{1}{{n}_{k}}\text{ for every }k, \]\n\nby the squeeze theorem (Theorem 2.1.6) we conclude \( \mathop{\lim }\limits_{{k \rightarrow \infty }}f\left( {x}_{{n}_{k}}\right) = m \) . On the other hand, by continuity we have \( \mathop{\lim }\limits_{{k \rightarrow \infty }}f\left( {x}_{{n}_{k}}\right) = f\left( \bar{x}\right) \) . We conclude that \( f\left( \bar{x}\right) = m \leq f\left( x\right) \) for every \( x \in D \) . Thus, \( f \) has an absolute minimum at \( \bar{x} \) . The proof is similar for the case of absolute maximum.
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Yes
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