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Proposition 4.2.7. Let \( I \) be an interval and let \( f : I \rightarrow \mathbb{R} \) be a differentiable function.\n\n(i) \( f \) is increasing if and only if \( {f}^{\prime }\left( x\right) \geq 0 \) for all \( x \in I \) . | Proof. Let us prove the first item. Suppose \( f \) is increasing, then for all \( x, c \in I \) with \( x \neq c \) we have\n\n\[ \frac{f\left( x\right) - f\left( c\right) }{x - c} \geq 0.\]\n\nTaking a limit as \( x \) goes to \( c \) we see that \( {f}^{\prime }\left( c\right) \geq 0 \) .\n\nFor the other direction,... | Yes |
Proposition 4.2.8. Let \( I \) be an interval and let \( f : I \rightarrow \mathbb{R} \) be a differentiable function.\n\n(i) If \( {f}^{\prime }\left( x\right) > 0 \) for all \( x \in I \), then \( f \) is strictly increasing.\n\n(ii) If \( {f}^{\prime }\left( x\right) < 0 \) for all \( x \in I \), then \( f \) is str... | The proof of (i) is left as an exercise. Then (ii) follows from (i) by considering \( - f \) instead. | No |
Proposition 4.2.9. Let \( f : \\left( {a, b}\\right) \\rightarrow \\mathbb{R} \) be continuous. Let \( c \\in \\left( {a, b}\\right) \) and suppose \( f \) is differentiable on \( \\left( {a, c}\\right) \) and \( \\left( {c, b}\\right) \) .\n\n(i) If \( {f}^{\\prime }\\left( x\\right) \\leq 0 \) for \( x \\in \\left( {... | Proof. We prove the first item leaving the second to the reader. Take \( x \\in \\left( {a, c}\\right) \) and \( \\left\{ {y}_{n}\\right\} \) a sequence such that \( x < {y}_{n} < c \) and \( \\lim {y}_{n} = c \) . By the preceding proposition, \( f \) is decreasing on \( \\left( {a, c}\\right) \) so \( f\\left( x\\rig... | No |
Theorem 4.2.11 (Darboux). Let \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be differentiable. Suppose \( y \in \mathbb{R} \) is such that \( {f}^{\prime }\left( a\right) < y < {f}^{\prime }\left( b\right) \) or \( {f}^{\prime }\left( a\right) > y > {f}^{\prime }\left( b\right) \) . Then there exist... | Proof. Suppose \( {f}^{\prime }\left( a\right) < y < {f}^{\prime }\left( b\right) \) . Define\n\n\[ g\left( x\right) \mathrel{\text{:=}} {yx} - f\left( x\right) .\n\]\n\nThe function \( g \) is continuous on \( \left\lbrack {a, b}\right\rbrack \), and so \( g \) attains a maximum at some \( c \in \left\lbrack {a, b}\ri... | Yes |
Theorem 4.3.2 (Taylor). Suppose \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) is a function with \( n \) continuous derivatives on \( \left\lbrack {a, b}\right\rbrack \) and such that \( {f}^{\left( n + 1\right) } \) exists on \( \left( {a, b}\right) \) . Given distinct points \( {x}_{0} \) and \( x... | Proof. Find a number \( {M}_{x,{x}_{0}} \) (depending on \( x \) and \( {x}_{0} \) ) solving the equation\n\n\[ f\left( x\right) = {P}_{n}^{{x}_{0}}\left( x\right) + {M}_{x,{x}_{0}}{\left( x - {x}_{0}\right) }^{n + 1}. \]\n\nDefine a function \( g\left( s\right) \) by\n\n\[ g\left( s\right) \mathrel{\text{:=}} f\left( ... | Yes |
Proposition 4.3.3 (Second derivative test). Suppose \( f : \left( {a, b}\right) \rightarrow \mathbb{R} \) is twice continuously differentiable, \( {x}_{0} \in \left( {a, b}\right) ,{f}^{\prime }\left( {x}_{0}\right) = 0 \) and \( {f}^{\prime \prime }\left( {x}_{0}\right) > 0 \) . Then \( f \) has a strict relative mini... | Proof. As \( {f}^{\prime \prime } \) is continuous, there exists a \( \delta > 0 \) such that \( {f}^{\prime \prime }\left( c\right) > 0 \) for all \( c \in \left( {{x}_{0} - \delta ,{x}_{0} + \delta }\right) \), see Exercise 3.2.11. Take \( x \in \left( {{x}_{0} - \delta ,{x}_{0} + \delta }\right), x \neq {x}_{0} \) .... | No |
Lemma 4.4.1. Let \( I, J \subset \mathbb{R} \) be intervals. If \( f : I \rightarrow J \) is strictly monotone (hence one-to-one), onto \( \left( {f\left( I\right) = J}\right) \), differentiable at \( {x}_{0} \in I \), and \( {f}^{\prime }\left( {x}_{0}\right) \neq 0 \), then the inverse \( {f}^{-1} \) is differentiabl... | Proof. By Proposition 3.6.6, \( f \) has a continuous inverse. For convenience call the inverse \( g : J \rightarrow I \) . Let \( {x}_{0},{y}_{0} \) be as in the statement. For any \( x \in I \) write \( y \mathrel{\text{:=}} f\left( x\right) \) . If \( x \neq {x}_{0} \) and so \( y \neq {y}_{0} \), we find \[ \frac{g... | Yes |
Theorem 4.4.2 (Inverse function theorem). Let \( f : \left( {a, b}\right) \rightarrow \mathbb{R} \) be a continuously differentiable function, \( {x}_{0} \in \left( {a, b}\right) \) a point where \( {f}^{\prime }\left( {x}_{0}\right) \neq 0 \) . Then there exists an open interval \( I \subset \left( {a, b}\right) \) wi... | Proof. Without loss of generality, suppose \( {f}^{\prime }\left( {x}_{0}\right) > 0 \) . As \( {f}^{\prime } \) is continuous, there must exist an open interval \( I = \left( {{x}_{0} - \delta ,{x}_{0} + \delta }\right) \) such that \( {f}^{\prime }\left( x\right) > 0 \) for all \( x \in I \) . See Exercise 3.2.11.\n\... | No |
Corollary 4.4.3. Given any \( n \in \mathbb{N} \) and any \( x \geq 0 \) there exists a unique number \( y \geq 0 \) (denoted \( {x}^{1/n} \mathrel{\text{:=}} y) \), such that \( {y}^{n} = x \) . Furthermore, the function \( g : \left( {0,\infty }\right) \rightarrow \left( {0,\infty }\right) \) defined by \( g\left( x\... | Proof. For \( x = 0 \) the existence of a unique root is trivial.\n\nLet \( f : \left( {0,\infty }\right) \rightarrow \left( {0,\infty }\right) \) be defined by \( f\left( y\right) \mathrel{\text{:=}} {y}^{n} \) . The function \( f \) is continuously differentiable and \( {f}^{\prime }\left( y\right) = n{y}^{n - 1} \),... | No |
Proposition 5.1.2. Let \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be a bounded function. Let \( m, M \in \mathbb{R} \) be such that for all \( x \) we have \( m \leq f\left( x\right) \leq M \) . For any partition \( P \) of \( \left\lbrack {a, b}\right\rbrack \) we have\n\n\[ m\left( {b - a}\righ... | Proof. Let \( P \) be a partition. Then note that \( m \leq {m}_{i} \) for all \( i \) and \( {M}_{i} \leq M \) for all \( i \) . Also \( {m}_{i} \leq {M}_{i} \) for all \( i \) . Finally \( \mathop{\sum }\limits_{{i = 1}}^{n}\Delta {x}_{i} = \left( {b - a}\right) \) . Therefore,\n\n\[ m\left( {b - a}\right) = m\left( ... | Yes |
Proposition 5.1.7. Let \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be a bounded function, and let \( P \) be a partition of \( \left\lbrack {a, b}\right\rbrack \) . Let \( \widetilde{P} \) be a refinement of \( P \) . Then\n\n\[ L\left( {P, f}\right) \leq L\left( {\widetilde{P}, f}\right) \;\text{... | Proof. The tricky part of this proof is to get the notation correct. Let \( \widetilde{P} \mathrel{\text{:=}} \left\{ {{\widetilde{x}}_{0},{\widetilde{x}}_{1},\ldots ,{\widetilde{x}}_{m}}\right\} \) be a refinement of \( P \mathrel{\text{:=}} \left\{ {{x}_{0},{x}_{1},\ldots ,{x}_{n}}\right\} \) . Then \( {x}_{0} = {\wi... | No |
Proposition 5.1.8. Let \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be a bounded function. Let \( m, M \in \mathbb{R} \) be such that for all \( x \in \left\lbrack {a, b}\right\rbrack \) we have \( m \leq f\left( x\right) \leq M \) . Then\n\n\[ m\left( {b - a}\right) \leq {\int }_{a}^{b}f \leq \ove... | Proof. By Proposition 5.1.2 we have for any partition \( P \)\n\n\[ m\left( {b - a}\right) \leq L\left( {P, f}\right) \leq U\left( {P, f}\right) \leq M\left( {b - a}\right) . \]\n\nThe inequality \( m\left( {b - a}\right) \leq L\left( {P, f}\right) \) implies \( m\left( {b - a}\right) \leq {\int }_{a}^{b}f \) . Also \(... | Yes |
Proposition 5.1.13. Let \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be a bounded function. Then \( f \) is Riemann integrable if for every \( \varepsilon > 0 \), there exists a partition \( P \) such that\n\n\[ U\left( {P, f}\right) - L\left( {P, f}\right) < \varepsilon . \]\n | Proof. If for every \( \varepsilon > 0 \) such a \( P \) exists, then we have:\n\n\[ 0 \leq \overline{{\int }_{a}^{b}}f - \underline{{\int }_{a}^{b}}f \leq U\left( {P, f}\right) - L\left( {P, f}\right) < \varepsilon . \]\n\nTherefore, \( \overline{{\int }_{a}^{b}}f = \underline{{\int }_{a}^{b}}f \), and \( f \) is inte... | Yes |
Proposition 5.2.2. Let \( a < b < c \) . A function \( f : \left\lbrack {a, c}\right\rbrack \rightarrow \mathbb{R} \) is Riemann integrable if and only if \( f \) is Riemann integrable on \( \left\lbrack {a, b}\right\rbrack \) and \( \left\lbrack {b, c}\right\rbrack \) . If \( f \) is Riemann integrable, then\n\n\[{\in... | Proof. Suppose \( f \in \mathcal{R}\left\lbrack {a, c}\right\rbrack \), then \( \overline{{\int }_{a}^{c}}f = \underline{{\int }_{a}^{c}}f = {\int }_{a}^{c}f \) . We apply the lemma to get\n\n\[{\int }_{a}^{c}f = \underline{{\int }_{a}^{c}}f = \underline{{\int }_{a}^{b}}f + \underline{{\int }_{b}^{c}}f \leq \overline{{... | Yes |
Proposition 5.2.4 (Linearity). Let \( f \) and \( g \) be in \( \mathcal{R}\left\lbrack {a, b}\right\rbrack \) and \( \alpha \in \mathbb{R} \). (i) \( \alpha \) fis in \( \mathcal{R}\left\lbrack {a, b}\right\rbrack \) and \[ {\int }_{a}^{b}{\alpha f}\left( x\right) {dx} = \alpha {\int }_{a}^{b}f\left( x\right) {dx}. \] | Proof. Let us prove the first item for \( \alpha \geq 0 \) . Let \( P \) be a partition of \( \left\lbrack {a, b}\right\rbrack \) . Let \( {m}_{i} \mathrel{\text{:=}} \inf \{ f\left( x\right) : x \in \) \( \left. \left\lbrack {{x}_{i - 1},{x}_{i}}\right\rbrack \right\} \) as usual. Since \( \alpha \) is nonnegative, we... | Yes |
Proposition 5.2.5. Let \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) and \( g : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be bounded functions. Then\n\n\[ \overline{{\int }_{a}^{b}}\left( {f + g}\right) \leq \overline{{\int }_{a}^{b}}f + \overline{{\int }_{a}^{b}}g,\;\text{ and }\;\... | The proof of the proposition above is Exercise 5.2.16. It follows as supremum of a sum is less than or equal to the sum of suprema and similarly for infima, see Exercise 1.3.7. | No |
Proposition 5.2.6 (Monotonicity). Let \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) and \( g : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be bounded, and \( f\left( x\right) \leq g\left( x\right) \) for all \( x \in \left\lbrack {a, b}\right\rbrack \) . Then\n\n\[ \underline{{\int }_... | Proof. Let \( P = \left\{ {{x}_{0},{x}_{1},\ldots ,{x}_{n}}\right\} \) be a partition of \( \left\lbrack {a, b}\right\rbrack \) . Then let\n\n\[ {m}_{i} \mathrel{\text{:=}} \inf \left\{ {f\left( x\right) : x \in \left\lbrack {{x}_{i - 1},{x}_{i}}\right\rbrack }\right\} \;\text{and}\;{\widetilde{m}}_{i} \mathrel{\text{:... | Yes |
Lemma 5.2.7. If \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) is a continuous function, then \( f \in \mathcal{R}\left\lbrack {a, b}\right\rbrack \) . | Proof. As \( f \) is continuous on a closed bounded interval, it is uniformly continuous. Let \( \varepsilon > 0 \) be given. Find a \( \delta > 0 \) such that \( \left| {x - y}\right| < \delta \) implies \( \left| {f\left( x\right) - f\left( y\right) }\right| < \frac{\varepsilon }{b - a} \) . Let \( P = \left\{ {{x}_{... | Yes |
Lemma 5.2.8. Let \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be a bounded function, \( \left\{ {a}_{n}\right\} \) and \( \left\{ {b}_{n}\right\} \) be sequences such that \( a < {a}_{n} < {b}_{n} < b \) for all \( n \), with \( \lim {a}_{n} = a \) and \( \lim {b}_{n} = b \) . Suppose \( f \in \mat... | Proof. Let \( M > 0 \) be a real number such that \( \left| {f\left( x\right) }\right| \leq M \) . As \( \left( {b - a}\right) \geq \left( {{b}_{n} - {a}_{n}}\right) \) , \[ - M\left( {b - a}\right) \leq - M\left( {{b}_{n} - {a}_{n}}\right) \leq {\int }_{{a}_{n}}^{{b}_{n}}f \leq M\left( {{b}_{n} - {a}_{n}}\right) \leq ... | Yes |
Theorem 5.2.9. Let \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be a bounded function with finitely many discontinuities. Then \( f \in \mathcal{R}\left\lbrack {a, b}\right\rbrack \) . | Proof. We divide the interval into finitely many intervals \( \left\lbrack {{a}_{i},{b}_{i}}\right\rbrack \) so that \( f \) is continuous on the interior \( \left( {{a}_{i},{b}_{i}}\right) \) . If \( f \) is continuous on \( \left( {{a}_{i},{b}_{i}}\right) \), then it is continuous and hence integrable on \( \left\lbr... | Yes |
Proposition 5.2.10. Let \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be Riemann integrable. Let \( g : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be a function such that \( f\left( x\right) = g\left( x\right) \) for all \( x \in \left\lbrack {a, b}\right\rbrack \smallsetminus S \), ... | Sketch of proof. Using additivity of the integral, we split up the interval \( \left\lbrack {a, b}\right\rbrack \) into smaller intervals such that \( f\left( x\right) = g\left( x\right) \) holds for all \( x \) except at the endpoints (details are left to the reader). Therefore, without loss of generality suppose \( f... | No |
Theorem 5.3.1. Let \( F : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be a continuous function, differentiable on \( \left( {a, b}\right) \) . Let \( f \in \mathcal{R}\left\lbrack {a, b}\right\rbrack \) be such that \( f\left( x\right) = {F}^{\prime }\left( x\right) \) for \( x \in \left( {a, b}\right) \... | Proof. Let \( P = \left\{ {{x}_{0},{x}_{1},\ldots ,{x}_{n}}\right\} \) be a partition of \( \left\lbrack {a, b}\right\rbrack \) . For each interval \( \left\lbrack {{x}_{i - 1},{x}_{i}}\right\rbrack \), use the mean value theorem to find a \( {c}_{i} \in \left( {{x}_{i - 1},{x}_{i}}\right) \) such that\n\n\[f\left( {c}... | Yes |
Theorem 5.3.3. Let \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be a Riemann integrable function. Define\n\n\[ F\left( x\right) \mathrel{\text{:=}} {\int }_{a}^{x}f \]\n\nFirst, \( F \) is continuous on \( \left\lbrack {a, b}\right\rbrack \) . Second, if \( f \) is continuous at \( c \in \left\lbra... | Proof. As \( f \) is bounded, there is an \( M > 0 \) such that \( \left| {f\left( x\right) }\right| \leq M \) for all \( x \in \left\lbrack {a, b}\right\rbrack \) . Suppose \( x, y \in \left\lbrack {a, b}\right\rbrack \) with \( x > y \) . Then\n\n\[ \left| {F\left( x\right) - F\left( y\right) }\right| = \left| {{\int... | Yes |
Theorem 5.3.5 (Change of variables). Let \( g : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be a continuously differentiable function, let \( f : \left\lbrack {c, d}\right\rbrack \rightarrow \mathbb{R} \) be continuous, and suppose \( g\left( \left\lbrack {a, b}\right\rbrack \right) \subset \left\lbrack ... | Proof. As \( g,{g}^{\prime } \), and \( f \) are continuous, we know \( f\left( {g\left( x\right) }\right) {g}^{\prime }\left( x\right) \) is a continuous function on \( \left\lbrack {a, b}\right\rbrack \) , therefore it is Riemann integrable. Similarly, \( f \) is integrable on any subinterval of \( \left\lbrack {c, d... | Yes |
Proposition 5.5.2 ( \( p \) -test for integrals). The improper integral\n\n\[ \n{\int }_{1}^{\infty }\frac{1}{{x}^{p}}{dx} \]\n\nconverges to \( \frac{1}{p - 1} \) if \( p > 1 \) and diverges if \( 0 < p \leq 1 \) .\n\nThe improper integral\n\n\[ \n{\int }_{0}^{1}\frac{1}{{x}^{p}}{dx} \]\n\nconverges to \( \frac{1}{1 -... | Proof. The proof follows by application of the fundamental theorem of calculus. Let us do the proof for \( p > 1 \) for the infinite right endpoint and leave the rest to the reader. Hint: You should handle \( p = 1 \) separately.\n\nSuppose \( p > 1 \) . Then\n\n\[ \n{\int }_{1}^{b}\frac{1}{{x}^{p}}{dx} = {\int }_{1}^{... | No |
Proposition 5.5.3. Let \( f : \lbrack a,\infty ) \rightarrow \mathbb{R} \) be a function that is Riemann integrable on \( \left\lbrack {a, b}\right\rbrack \) for all \( b > a \) . Given any \( b > a,{\int }_{b}^{\infty }f \) converges if and only if \( {\int }_{a}^{\infty }f \) converges, in which case\n\n\[{\int }_{a}... | Proof. Let \( c > b \) . Then\n\n\[{\int }_{a}^{c}f = {\int }_{a}^{b}f + {\int }_{b}^{c}f.\]\n\nTaking the limit \( c \rightarrow \infty \) finishes the proof. | No |
Proposition 5.5.4. Suppose \( f : \lbrack a,\infty ) \rightarrow \mathbb{R} \) is nonnegative \( \left( {f\left( x\right) \geq 0\text{for all}x}\right) \) and such that \( f \) is Riemann integrable on \( \left\lbrack {a, b}\right\rbrack \) for all \( b > a \) .\n\n(i)\n\n\[{\int }_{a}^{\infty }f = \sup \left\{ {{\int ... | Proof. We start with the first item. As \( f \) is nonnegative, \( {\int }_{a}^{x}f \) is increasing as a function of \( x \) . If the supremum is infinite, then for every \( M \in \mathbb{R} \) we find \( N \) such that \( {\int }_{a}^{N}f \geq M \) . As \( {\int }_{a}^{x}f \) is increasing, \( {\int }_{a}^{x}f \geq M... | Yes |
Proposition 5.5.5 (Comparison test for improper integrals). Let \( f : \lbrack a,\infty ) \rightarrow \mathbb{R} \) and \( g : \lbrack a,\infty ) \rightarrow \mathbb{R} \) be functions that are Riemann integrable on \( \left\lbrack {a, b}\right\rbrack \) for all \( b > a \). Suppose that for all \( x \geq a \) we have\... | Proof. Let us start with the first item. For any \( b \) and \( c \), such that \( a \leq b \leq c \), we have \( - g\left( x\right) \leq \) \( f\left( x\right) \leq g\left( x\right) \), and so\n\n\[ {\int }_{b}^{c} - g \leq {\int }_{b}^{c}f \leq {\int }_{b}^{c}g \]\n\nIn other words, \( \left| {{\int }_{b}^{c}f}\right... | Yes |
Proposition 5.5.10. If \( f : \mathbb{R} \rightarrow \mathbb{R} \) is a function integrable on every bounded interval \( \left\lbrack {a, b}\right\rbrack \) . Then\n\n\[ \mathop{\lim }\limits_{{a \rightarrow - \infty }}\mathop{\lim }\limits_{{b \rightarrow \infty }}{\int }_{a}^{b}f\;\text{ converges if and only if }\;\... | Proof. Without loss of generality assume \( a < 0 \) and \( b > 0 \) . Suppose the first expression converges. Then\n\n\[ \mathop{\lim }\limits_{{a \rightarrow - \infty }}\mathop{\lim }\limits_{{b \rightarrow \infty }}{\int }_{a}^{b}f = \mathop{\lim }\limits_{{a \rightarrow - \infty }}\mathop{\lim }\limits_{{b \rightar... | Yes |
Proposition 5.5.13. Suppose \( f : \lbrack k,\infty ) \rightarrow \mathbb{R} \) is a decreasing nonnegative function where \( k \in \mathbb{Z} \) . Then\n\n\[ \mathop{\sum }\limits_{{n = k}}^{\infty }f\left( n\right) \;\text{ converges if and only if }\;{\int }_{k}^{\infty }f\;\text{ converges. }\n\]\n\nIn this case\n\... | Proof. Let \( \varepsilon > 0 \) be given. And suppose \( {\int }_{k}^{\infty }f \) converges. Let \( \ell, m \in \mathbb{Z} \) be such that \( m > \ell \geq k \) . Because \( f \) is decreasing we have \( {\int }_{n}^{n + 1}f \leq f\left( n\right) \leq {\int }_{n - 1}^{n}f \) . Therefore\n\n\[ {\int }_{\ell }^{m}f = \... | Yes |
Proposition 6.1.10. A sequence of bounded functions \( {f}_{n} : S \rightarrow \mathbb{R} \) converges uniformly to \( f : S \rightarrow \mathbb{R} \) , if and only if\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{\begin{Vmatrix}{f}_{n} - f\end{Vmatrix}}_{u} = 0 \] | Proof. First suppose \( \lim {\begin{Vmatrix}{f}_{n} - f\end{Vmatrix}}_{u} = 0 \) . Let \( \varepsilon > 0 \) be given. Then there exists an \( N \) such that for \( n \geq N \) we have \( {\begin{Vmatrix}{f}_{n} - f\end{Vmatrix}}_{u} < \varepsilon \) . As \( {\begin{Vmatrix}{f}_{n} - f\end{Vmatrix}}_{u} \) is the supr... | Yes |
Proposition 6.1.13. Let \( {f}_{n} : S \rightarrow \mathbb{R} \) be bounded functions. Then \( \left\{ {f}_{n}\right\} \) is Cauchy in the uniform norm if and only if there exists an \( f : S \rightarrow \mathbb{R} \) and \( \left\{ {f}_{n}\right\} \) converges uniformly to \( f \) . | Proof. Let us first suppose \( \left\{ {f}_{n}\right\} \) is Cauchy in the uniform norm. Let us define \( f \) . Fix \( x \), then the sequence \( \left\{ {{f}_{n}\left( x\right) }\right\} \) is Cauchy because\n\n\[ \left| {{f}_{m}\left( x\right) - {f}_{k}\left( x\right) }\right| \leq {\begin{Vmatrix}{f}_{m} - {f}_{k}\... | Yes |
Theorem 6.2.2. Let \( \\left\\{ {f}_{n}\\right\\} \) be a sequence of continuous functions \( {f}_{n} : S \\rightarrow \\mathbb{R} \) converging uniformly to \( f : S \\rightarrow \\mathbb{R} \). Then \( f \) is continuous. | Proof. Let \( x \\in S \) be fixed. Let \( \\left\\{ {x}_{n}\\right\\} \) be a sequence in \( S \) converging to \( x \).\n\nLet \( \\varepsilon > 0 \) be given. As \( \\left\\{ {f}_{k}\\right\\} \) converges uniformly to \( f \), we find a \( k \\in \\mathbb{N} \) such that\n\n\[ \n\\left| {{f}_{k}\\left( y\\right) - ... | Yes |
Theorem 6.2.4. Let \( \left\{ {f}_{n}\right\} \) be a sequence of Riemann integrable functions \( {f}_{n} : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) converging uniformly to \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) . Then \( f \) is Riemann integrable and\n\n\[{\int }_{a}^{b}f ... | Proof. Let \( \varepsilon > 0 \) be given. As \( {f}_{n} \) goes to \( f \) uniformly, we find an \( M \in \mathbb{N} \) such that for all \( n \geq M \) we have \( \left| {{f}_{n}\left( x\right) - f\left( x\right) }\right| < \frac{\varepsilon }{2\left( {b - a}\right) } \) for all \( x \in \left\lbrack {a, b}\right\rbr... | Yes |
Theorem 6.2.10. Let \( I \) be a bounded interval and let \( {f}_{n} : I \rightarrow \mathbb{R} \) be continuously differentiable functions. Suppose \( \left\{ {f}_{n}^{\prime }\right\} \) converges uniformly to \( g : I \rightarrow \mathbb{R} \), and suppose \( {\left\{ {f}_{n}\left( c\right) \right\} }_{n = 1}^{\inft... | Proof. Define \( f\left( c\right) \mathrel{\text{:=}} \mathop{\lim }\limits_{{n \rightarrow \infty }}{f}_{n}\left( c\right) \) . As \( {f}_{n}^{\prime } \) are continuous and hence Riemann integrable, then via the fundamental theorem of calculus, we find that for \( x \in I \) ,\n\n\[ \n{f}_{n}\left( x\right) = {f}_{n}... | Yes |
Proposition 6.2.11. Let \( \mathop{\sum }\limits_{{n = 0}}^{\infty }{c}_{n}{\left( x - a\right) }^{n} \) be a convergent power series with a radius of convergence \( 0 < \rho \leq \infty \) . Then the series converges uniformly in \( \left\lbrack {a - r, a + r}\right\rbrack \) for any \( 0 < r < \rho \) . | Proof. Let \( I \mathrel{\text{:=}} \left( {a - \rho, a + \rho }\right) \) if \( \rho < \infty \), or let \( I \mathrel{\text{:=}} \mathbb{R} \) if \( \rho = \infty \) . Take \( 0 < r < \rho \) . The series converges absolutely for any \( x \in I \), in particular if \( x = a + r \) . Therefore \( \mathop{\sum }\limits... | Yes |
Corollary 6.2.12. Let \( \mathop{\sum }\limits_{{n = 0}}^{\infty }{c}_{n}{\left( x - a\right) }^{n} \) be a convergent power series with a radius of convergence \( 0 < \rho \leq \infty \) . Let \( I \mathrel{\text{:=}} \left( {a - \rho, a + \rho }\right) \) if \( \rho < \infty \) or \( I \mathrel{\text{:=}} \mathbb{R} ... | Proof. Take \( 0 < r < \rho \) . The partial sums \( \mathop{\sum }\limits_{{n = 0}}^{k}{c}_{n}{\left( x - a\right) }^{n} \) converge uniformly on \( \left\lbrack {a - r, a + r}\right\rbrack \) . For any fixed \( x \in \left\lbrack {a - r, a + r}\right\rbrack \), the convergence is also uniform on \( \left\lbrack {a, x... | Yes |
Corollary 6.2.13. Let \( \mathop{\sum }\limits_{{n = 0}}^{\infty }{c}_{n}{\left( x - a\right) }^{n} \) be a convergent power series with a radius of convergence \( 0 < \rho \leq \infty \) . Let \( I \mathrel{\text{:=}} \left( {a - \rho, a + \rho }\right) \) if \( \rho < \infty \) or \( I \mathrel{\text{:=}} \mathbb{R} ... | Proof. Take \( 0 < r < \rho \) . We have uniform convergence of the series on \( \left\lbrack {a - r, a + r}\right\rbrack \), but we need uniform convergence of the derivative. Let\n\n\[ \nR \mathrel{\text{:=}} \mathop{\limsup }\limits_{{n \rightarrow \infty }}{\left| {c}_{n}\right| }^{1/n}.\n\]\n\nAs the series is con... | Yes |
Theorem 6.3.2 (Picard’s theorem on existence and uniqueness). Let \( I, J \subset \mathbb{R} \) be closed bounded intervals, let \( {I}^{ \circ } \) and \( {J}^{ \circ } \) be their interiors \( {}^{ \dagger } \), and let \( \left( {{x}_{0},{y}_{0}}\right) \in {I}^{ \circ } \times {J}^{ \circ } \) . Suppose \( F : I \t... | Proof. Suppose we could find a solution \( f \) . Using the fundamental theorem of calculus we integrate the equation \( {f}^{\prime }\left( x\right) = F\left( {x, f\left( x\right) }\right), f\left( {x}_{0}\right) = {y}_{0} \), and write (6.1) as the integral equation\n\n\[ f\left( x\right) = {y}_{0} + {\int }_{{x}_{0}... | No |
Lemma 7.1.4 (Cauchy-Schwarz inequality*). If \( x = \left( {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right) \in {\mathbb{R}}^{n}, y = \left( {{y}_{1},{y}_{2},\ldots ,{y}_{n}}\right) \in {\mathbb{R}}^{n} \) , then\n\n\[ \n{\left( \mathop{\sum }\limits_{{j = 1}}^{n}{x}_{j}{y}_{j}\right) }^{2} \leq \left( {\mathop{\sum }\limits_{... | Proof. Any square of a real number is nonnegative. Hence any sum of squares is nonnegative:\n\n\[ \n0 \leq \mathop{\sum }\limits_{{j = 1}}^{n}\mathop{\sum }\limits_{{k = 1}}^{n}{\left( {x}_{j}{y}_{k} - {x}_{k}{y}_{j}\right) }^{2} \n\]\n\n\[ \n= \mathop{\sum }\limits_{{j = 1}}^{n}\mathop{\sum }\limits_{{k = 1}}^{n}\left... | Yes |
Proposition 7.2.6. Let \( \left( {X, d}\right) \) be a metric space.\n\n(i) ∅ and \( X \) are open.\n\n(ii) If \( {V}_{1},{V}_{2},\ldots ,{V}_{k} \) are open, then \n\n\n\nis also open. That is, a finite intersection... | Proof. The sets \( X \) and \( \varnothing \) are obviously open in \( X \) .\n\nLet us prove (ii). If \( x \in \mathop{\bigcap }\limits_{{j = 1}}^{k}{V}_{j} \), then \( x \in {V}_{j} \) for all \( j \) . As \( {V}_{j} \) are all open, for every \( j \) there exists a \( {\delta }_{j} > 0 \) such that \( B\left( {x,{\d... | Yes |
Proposition 7.2.9. Let \( \left( {X, d}\right) \) be a metric space, \( x \in X \), and \( \delta > 0 \) . Then \( B\left( {x,\delta }\right) \) is open and \( C\left( {x,\delta }\right) \) is closed. | Proof. Let \( y \in B\left( {x,\delta }\right) \) . Let \( \alpha \mathrel{\text{:=}} \delta - d\left( {x, y}\right) \) . As \( \alpha > 0 \), consider \( z \in B\left( {y,\alpha }\right) \) . Then\n\n\[ d\left( {x, z}\right) \leq d\left( {x, y}\right) + d\left( {y, z}\right) < d\left( {x, y}\right) + \alpha = d\left( ... | Yes |
Proposition 7.2.11. Suppose \( \left( {X, d}\right) \) is a metric space, and \( Y \subset X \) . Then \( U \subset Y \) is open in \( Y \) (i.e., in the subspace topology), if and only if there exists an open set \( V \subset X \) (so open in \( X \) ), such that \( V \cap Y = U. | Proof. Suppose \( V \subset X \) is open and \( x \in V \cap Y \) . Let \( U \mathrel{\text{:=}} V \cap Y \) . As \( V \) is open, there exists a \( \delta > 0 \) such that \( {B}_{X}\left( {x,\delta }\right) \subset V \) . Then\n\n\[ \n{B}_{Y}\left( {x,\delta }\right) = {B}_{X}\left( {x,\delta }\right) \cap Y \subset ... | Yes |
Proposition 7.2.12. Suppose \( \left( {X, d}\right) \) is a metric space, \( V \subset X \) is open and \( E \subset X \) is closed.\n\n(i) \( U \subset V \) is open in the subspace topology if and only if \( U \) is open in \( X \) . | Proof. Let us prove (i) and leave (ii) to an exercise.\n\nIf \( U \subset V \) is open in the subspace topology, by Proposition 7.2.11, there exists a set \( W \subset X \) open in \( X \), such that \( U = W \cap V \) . Intersection of two open sets is open so \( U \) is open in \( X \) .\n\nNow suppose \( U \) is ope... | No |
Proposition 7.2.14. Let \( \left( {X, d}\right) \) be a metric space. A nonempty set \( S \subset X \) is disconnected if and only if there exist open sets \( {U}_{1} \) and \( {U}_{2} \) in \( X \), such that \( {U}_{1} \cap {U}_{2} \cap S = \varnothing ,{U}_{1} \cap S \neq \varnothing ,{U}_{2} \cap S \neq \varnothing... | Proof. The proof follows by Proposition 7.2.11. If \( {U}_{1} \) and \( {U}_{2} \) as in the statement are open in \( X \) , then \( {U}_{1} \cap S \) and \( {U}_{2} \cap S \) are open in \( S \) . From the discussion above it follows that \( S \) is disconnected.\n\nFor the other direction start with nonempty disjoint... | Yes |
Proposition 7.2.16. A nonempty set \( S \subset \mathbb{R} \) is connected if and only if it is an interval or a single point. | Proof. Suppose \( S \) is connected. If \( S \) is a single point, then we are done. So suppose \( x < y \) and \( x, y \in S \) . If \( z \in \mathbb{R} \) is such that \( x < z < y \), then \( \left( {-\infty, z}\right) \cap S \) is nonempty and \( \left( {z,\infty }\right) \cap S \) is nonempty. The two sets are dis... | Yes |
Proposition 7.2.19. Let \( \left( {X, d}\right) \) be a metric space and \( A \subset X \) . The closure \( \bar{A} \) is closed. Furthermore, if \( A \) is closed, then \( \bar{A} = A \) . | Proof. The closure is an intersection of closed sets, so \( \bar{A} \) is closed. If \( A \) is closed, then \( A \) is a closed set that contains \( A \), and so \( \bar{A} \subset A \) . And since \( A \subset \bar{A} \), we must have \( A = \bar{A} \) . | Yes |
Proposition 7.2.22. Let \( \left( {X, d}\right) \) be a metric space and \( A \subset X \) . Then \( x \in \bar{A} \) if and only if for every \( \delta > 0 \) , \( B\left( {x,\delta }\right) \cap A \neq \varnothing \) . | Proof. Let us prove the two contrapositives. Let us show that \( x \notin \bar{A} \) if and only if there exists a \( \delta > 0 \) such that \( B\left( {x,\delta }\right) \cap A = \varnothing \) .\n\nFirst suppose \( x \notin \bar{A} \) . We know \( \bar{A} \) is closed. Thus there is a \( \delta > 0 \) such that \( B... | Yes |
Proposition 7.2.26. Let \( \left( {X, d}\right) \) be a metric space and \( A \subset X \) . Then \( {A}^{ \circ } \) is open and \( \partial A \) is closed. | Proof. Given \( x \in {A}^{ \circ } \), there is a \( \delta > 0 \) such that \( B\left( {x,\delta }\right) \subset A \) . If \( z \in B\left( {x,\delta }\right) \), then as open balls are open, there is an \( \varepsilon > 0 \) such that \( B\left( {z,\varepsilon }\right) \subset B\left( {x,\delta }\right) \subset A \... | Yes |
Proposition 7.2.27. Let \( \\left( {X, d}\\right) \) be a metric space and \( A \\subset X \) . Then \( x \\in \\partial A \) if and only if for every \( \\delta > 0, B\\left( {x,\\delta }\\right) \\cap A \) and \( B\\left( {x,\\delta }\\right) \\cap {A}^{c} \) are both nonempty. | Proof. Suppose \( x \\in \\partial A = \\bar{A} \\smallsetminus {A}^{ \\circ } \) and let \( \\delta > 0 \) be arbitrary. By Proposition 7.2.22, \( B\\left( {x,\\delta }\\right) \) contains a point of \( A \) . If \( B\\left( {x,\\delta }\\right) \) contained no points of \( {A}^{c} \), then \( x \) would be in \( {A}^... | Yes |
Proposition 7.3.3. A convergent sequence in a metric space has a unique limit. | Proof. Suppose the sequence \( \left\{ {x}_{n}\right\} \) has limits \( x \) and \( y \) . Take an arbitrary \( \varepsilon > 0 \) . From the definition find an \( {M}_{1} \) such that for all \( n \geq {M}_{1}, d\left( {{x}_{n}, x}\right) < \varepsilon /2 \) . Similarly find an \( {M}_{2} \) such that for all \( n \ge... | Yes |
Proposition 7.3.9. Let \( {\left\{ {x}_{j}\right\} }_{j = 1}^{\infty } \) be a sequence in \( {\mathbb{R}}^{n} \), where we write \( {x}_{j} = \left( {{x}_{j,1},{x}_{j,2},\ldots ,{x}_{j, n}}\right) \in {\mathbb{R}}^{n} \). Then \( {\left\{ {x}_{j}\right\} }_{j = 1}^{\infty } \) converges if and only if \( {\left\{ {x}_... | Proof. Suppose the sequence \( {\left\{ {x}_{j}\right\} }_{j = 1}^{\infty } \) converges to \( y = \left( {{y}_{1},{y}_{2},\ldots ,{y}_{n}}\right) \in {\mathbb{R}}^{n} \). Given \( \varepsilon > 0 \), there exists an \( M \), such that for all \( j \geq M \) we have \[ d\left( {y,{x}_{j}}\right) < \varepsilon \] Fix so... | Yes |
Proposition 7.3.11. Let \( \left( {X, d}\right) \) be a metric space and \( \left\{ {x}_{n}\right\} \) a sequence in \( X \) . Then \( \left\{ {x}_{n}\right\} \) converges to \( x \in X \) if and only if for every open neighborhood \( U \) of \( x \), there exists an \( M \in \mathbb{N} \) such that for all \( n \geq M... | Proof. First suppose \( \left\{ {x}_{n}\right\} \) converges. Let \( U \) be an open neighborhood of \( x \), then there exists an \( \varepsilon > 0 \) such that \( B\left( {x,\varepsilon }\right) \subset U \) . As the sequence converges, find an \( M \in \mathbb{N} \) such that for all \( n \geq M \) we have \( d\lef... | Yes |
Proposition 7.3.13. Let \( \left( {X, d}\right) \) be a metric space and \( A \subset X \) . Then \( x \in \bar{A} \) if and only if there exists a sequence \( \left\{ {x}_{n}\right\} \) of elements in \( A \) such that \( \lim {x}_{n} = x \) . | Proof. Let \( x \in \bar{A} \) . For every \( n \in \mathbb{N} \), by Proposition 7.2.22 there exists a point \( {x}_{n} \in B\left( {x,1/n}\right) \cap A \) . As \( d\left( {x,{x}_{n}}\right) < 1/n \), we have \( \lim {x}_{n} = x \) . | No |
Proposition 7.4.2. A convergent sequence in a metric space is Cauchy. | Proof. Suppose \( \left\{ {x}_{n}\right\} \) converges to \( x \) . Given \( \varepsilon > 0 \) there is an \( M \) such that for \( n \geq M \) we have \( d\left( {x,{x}_{n}}\right) < \varepsilon /2 \) . Hence for all \( n, k \geq M \) we have \( d\left( {{x}_{n},{x}_{k}}\right) \leq d\left( {{x}_{n}, x}\right) + d\le... | Yes |
Proposition 7.4.4. The space \( {\mathbb{R}}^{n} \) with the standard metric is a complete metric space. | Proof. Let \( {\left\{ {x}_{j}\right\} }_{j = 1}^{\infty } \) be a Cauchy sequence in \( {\mathbb{R}}^{n} \), where we write \( {x}_{j} = \left( {{x}_{j,1},{x}_{j,2},\ldots ,{x}_{j, n}}\right) \in {\mathbb{R}}^{n} \). As the sequence is Cauchy, given \( \varepsilon > 0 \), there exists an \( M \) such that for all \( i... | Yes |
Proposition 7.4.9. Let \( \left( {X, d}\right) \) be a metric space. A compact set \( K \subset X \) is closed and bounded. | Proof. First, we prove that a compact set is bounded. Fix \( p \in X \) . We have the open cover\n\n\[ K \subset \mathop{\bigcup }\limits_{{n = 1}}^{\infty }B\left( {p, n}\right) = X \]\n\nIf \( K \) is compact, then there exists some set of indices \( {n}_{1} < {n}_{2} < \ldots < {n}_{k} \) such that\n\n\[ K \subset \... | Yes |
Lemma 7.4.10 (Lebesgue covering lemma*). Let \( \\left( {X, d}\\right) \) be a metric space and \( K \\subset X \) . Suppose every sequence in \( K \) has a subsequence convergent in \( K \) . Given an open cover \( {\\left\\{ {U}_{\\lambda }\\right\\} }_{\\lambda \\in I} \) of \( K \), there exists a \( \\delta > 0 \)... | Proof. We prove the lemma by contrapositive. If the conclusion is not true, then there is an open cover \( {\\left\\{ {U}_{\\lambda }\\right\\} }_{\\lambda \\in I} \) of \( K \) with the following property. For every \( n \\in \\mathbb{N} \) there exists an \( {x}_{n} \\in K \) such that \( B\\left( {{x}_{n},1/n}\\righ... | Yes |
Proposition 7.4.13. Let \( \\left( {X, d}\\right) \) be a metric space and let \( K \\subset X \) be compact. If \( E \\subset K \) is a closed set, then \( E \) is compact. | Proof. Let \( \\left\\{ {x}_{n}\\right\\} \) be a sequence in \( E \) . It is also a sequence in \( K \) . Therefore, it has a convergent subsequence \( \\left\\{ {x}_{{n}_{j}}\\right\\} \) that converges to some \( x \\in K \) . As \( E \) is closed the limit of a sequence in \( E \) is also in \( E \) and so \( x \\i... | Yes |
Proposition 7.5.2. Let \( \left( {X,{d}_{X}}\right) \) and \( \left( {Y,{d}_{Y}}\right) \) be metric spaces. Then \( f : X \rightarrow Y \) is continuous at \( c \in X \) if and only if for every sequence \( \left\{ {x}_{n}\right\} \) in \( X \) converging to \( c \), the sequence \( \left\{ {f\left( {x}_{n}\right) }\r... | Proof. Suppose \( f \) is continuous at \( c \) . Let \( \left\{ {x}_{n}\right\} \) be a sequence in \( X \) converging to \( c \) . Given \( \varepsilon > 0 \) , there is a \( \delta > 0 \) such that \( {d}_{X}\left( {x, c}\right) < \delta \) implies \( {d}_{Y}\left( {f\left( x\right), f\left( c\right) }\right) < \var... | Yes |
Lemma 7.5.5. Let \( \\left( {X,{d}_{X}}\\right) \) and \( \\left( {Y,{d}_{Y}}\\right) \) be metric spaces and \( f : X \\rightarrow Y \) a continuous function. If \( K \\subset X \) is a compact set, then \( f\\left( K\\right) \) is a compact set. | Proof. A sequence in \( f\\left( K\\right) \) can be written as \( {\\left\{ f\\left( {x}_{n}\\right) \\right\} }_{n = 1}^{\\infty } \), where \( {\\left\{ {x}_{n}\\right\} }_{n = 1}^{\\infty } \) is a sequence in \( K \) . The set \( K \) is compact and therefore there is a subsequence \( {\\left\{ {x}_{{n}_{j}}\\righ... | Yes |
Theorem 7.5.6. Let \( \left( {X, d}\right) \) be a compact metric space and \( f : X \rightarrow \mathbb{R} \) a continuous function. Then \( f \) is bounded and in fact \( f \) achieves an absolute minimum and an absolute maximum on \( X \) . | Proof. As \( X \) is compact and \( f \) is continuous, then \( f\left( X\right) \subset \mathbb{R} \) is compact. Hence \( f\left( X\right) \) is closed and bounded. In particular, \( \sup f\left( X\right) \in f\left( X\right) \) and \( \inf f\left( X\right) \in f\left( X\right) \), because both the sup and the inf ca... | Yes |
Lemma 7.5.7. Let \( \\left( {X,{d}_{X}}\\right) \) and \( \\left( {Y,{d}_{Y}}\\right) \) be metric spaces. A function \( f : X \\rightarrow Y \) is continuous at \( c \\in X \) if and only if for every open neighborhood \( U \) of \( f\\left( c\\right) \) in \( Y \), the set \( {f}^{-1}\\left( U\\right) \) contains an ... | Proof. First suppose that \( f \) is continuous at \( c \) . Let \( U \) be an open neighborhood of \( f\\left( c\\right) \) in \( Y \), then \( {B}_{Y}\\left( {f\\left( c\\right) ,\\varepsilon }\\right) \\subset U \) for some \( \\varepsilon > 0 \) . By continuity of \( f \), there exists a \( \\delta > 0 \) such that... | Yes |
Theorem 7.5.8. Let \( \\left( {X,{d}_{X}}\\right) \) and \( \\left( {Y,{d}_{Y}}\\right) \) be metric spaces. A function \( f : X \\rightarrow Y \) is continuous if and only if for every open \( U \\subset Y,{f}^{-1}\\left( U\\right) \) is open in \( X \) . | The proof follows from Lemma 7.5.7 and is left as an exercise. | No |
Theorem 7.5.11. Let \( \left( {X,{d}_{X}}\right) \) and \( \left( {Y,{d}_{Y}}\right) \) be metric spaces. Suppose \( f : X \rightarrow Y \) is continuous and \( X \) is compact. Then \( f \) is uniformly continuous. | Proof. Let \( \varepsilon > 0 \) be given. For each \( c \in X \), pick \( {\delta }_{c} > 0 \) such that \( {d}_{Y}\left( {f\left( x\right), f\left( c\right) }\right) < \varepsilon /2 \) whenever \( x \in B\left( {c,{\delta }_{c}}\right) \) . The balls \( B\left( {c,{\delta }_{c}}\right) \) cover \( X \), and the spac... | Yes |
Proposition 7.5.12. If \( f : \left\lbrack {a, b}\right\rbrack \times \left\lbrack {c, d}\right\rbrack \rightarrow \mathbb{R} \) is a continuous function, then \( g : \left\lbrack {c, d}\right\rbrack \rightarrow \mathbb{R} \) defined by\n\n\[ g\left( y\right) \mathrel{\text{:=}} {\int }_{a}^{b}f\left( {x, y}\right) {dx... | Proof. Fix \( y \in \left\lbrack {c, d}\right\rbrack \), and let \( \left\{ {y}_{n}\right\} \) be a sequence in \( \left\lbrack {c, d}\right\rbrack \) converging to \( y \) . Let \( \varepsilon > 0 \) be given. As \( f \) is continuous on \( \left\lbrack {a, b}\right\rbrack \times \left\lbrack {c, d}\right\rbrack \), w... | Yes |
Lemma 7.5.17. Let \( \left( {X,{d}_{X}}\right) \) and \( \left( {Y,{d}_{Y}}\right) \) be metric spaces, \( S \subset X, p \in X \) a cluster point of \( S \), and let \( f : S \rightarrow Y \) be a function.\n\nThen \( f\left( x\right) \) converges to \( L \in Y \) as \( x \) goes to \( p \) if and only if for every se... | By applying Proposition 7.5.2 or the definition directly we find (exercise) as in chapter 3, that for cluster points \( p \) of \( S \subset X \), the function \( f : S \rightarrow Y \) is continuous at \( p \) if and only if\n\n\[ \mathop{\lim }\limits_{{x \rightarrow p}}f\left( x\right) = f\left( p\right) \] | No |
Theorem 7.6.2 (Contraction mapping principle or Banach fixed point theorem*). Let \( \left( {X, d}\right) \) be a nonempty complete metric space and \( f : X \rightarrow X \) a contraction. Then \( f \) has a unique fixed point. | Proof. Pick any \( {x}_{0} \in X \) . Define a sequence \( \left\{ {x}_{n}\right\} \) by \( {x}_{n + 1} \mathrel{\text{:=}} f\left( {x}_{n}\right) \) . \n\n\[ \nd\left( {{x}_{n + 1},{x}_{n}}\right) = d\left( {f\left( {x}_{n}\right), f\left( {x}_{n - 1}\right) }\right) \leq {kd}\left( {{x}_{n},{x}_{n - 1}}\right) \leq \... | Yes |
Theorem 3.2.1 There are infinitely many primes. | Proof. Suppose this were not the case. That is, suppose there are only finitely many primes. Then there must be a last, largest prime, call it \( p \) . Consider the number\n\n\[ N = p! + 1 = \left( {p \cdot \left( {p - 1}\right) \cdots \cdot 3 \cdot 2 \cdot 1}\right) + 1. \]\n\nNow \( N \) is certainly larger than \( ... | Yes |
Lemma 4.1.5 Handshake Lemma. In any graph, the sum of the degrees of vertices in the graph is always twice the number of edges. | The handshake lemma \( {}^{2} \) is sometimes called the degree sum formula, and can be written symbolically as\n\n\[ \mathop{\sum }\limits_{{v \in V}}d\left( v\right) = {2e} \]\n\nHere we are using the notation \( d\left( v\right) \) for the degree of the vertex \( v \) . | No |
Proposition 4.1.8 In any graph, the number of vertices with odd degree must be even. | Proof. Suppose there were a graph with an odd number of vertices with odd degree. Then the sum of the degrees in the graph would be odd, which is impossible, by the handshake lemma. QED | Yes |
A graph \( T \) is a tree if and only if between every pair of distinct vertices of \( T \) there is a unique path. | Proof. This is an \ | No |
Proposition 4.2.3 Any tree with at least two vertices has at least two vertices of degree one. | Proof. We give a proof by contradiction. Let \( T \) be a tree with at least two vertices, and suppose, contrary to stipulation, that there are not two vertices of degree one.\n\nLet \( P \) be a path in \( T \) of longest possible length. Let \( u \) and \( v \) be the endpoints of the path. Since \( T \) does not hav... | Yes |
Proposition 4.2.4 Let \( T \) be a tree with \( v \) vertices and \( e \) edges. Then \( e = v - 1 \). | Proof. We will give a proof by induction on the number of vertices in the tree. That is, we will prove that every tree with \( v \) vertices has exactly \( v - 1 \) edges, and then use induction to show this is true for all \( v \geq 1 \). For the base case, consider all trees with \( v = 1 \) vertices. There is only o... | Yes |
Theorem 4.3.1 \( {K}_{5} \) is not planar. | Proof. The proof is by contradiction. So assume that \( {K}_{5} \) is planar. Then the graph must satisfy Euler’s formula for planar graphs. \( {K}_{5} \) has 5 vertices and 10 edges, so we get\n\n\[ 5 - {10} + f = 2 \]\n\nwhich says that if the graph is drawn without any edges crossing, there would be \( f = 7 \) face... | Yes |
Theorem 4.3.2 \( {K}_{3,3} \) is not planar. | Proof. Again, we proceed by contradiction. Suppose \( {K}_{3,3} \) were planar. Then by Euler’s formula there will be 5 faces, since \( v = 6, e = 9 \), and \( 6 - 9 + f = 2 \) . How many boundaries surround these 5 faces? Let \( B \) be this number. Since each edge is used as a boundary twice, we have \( B = {2e} \). ... | Yes |
Theorem 4.3.4 There are exactly five regular polyhedra. | Proof. Recall that all the faces of a regular polyhedron are identical regular polygons, and that each vertex has the same degree. Consider four cases, depending on the type of regular polygon.\n\nCase 1: Each face is a triangle. Let \( f \) be the number of faces. There are then \( {3f}/2 \) edges. Using Euler’s formu... | Yes |
Theorem 4.6.1 Hall’s Marriage Theorem. Let \( G \) be a bipartite graph with sets \( A \) and \( B \) . Then \( G \) has a matching of \( A \) if and only if\n\n\[ \left| {N\left( S\right) }\right| \geq \left| S\right| \]\n\nfor all \( S \subseteq A \) . | There are quite a few different proofs of this theorem - a quick internet search will get you started. | No |
Proposition 1.2. Let \( A, B \), and \( C \) be sets. Then\n\n1. \( A \cup A = A, A \cap A = A \), and \( A \smallsetminus A = \varnothing \) ; | Proof. We will prove (1) and (3) and leave the remaining results to be proven in the exercises.\n\n(1) Observe that\n\n\[ A \cup A = \{ x : x \in A\\text{ or }x \in A\} \]\n\n\[ = \{ x : x \in A\} \]\n\n\[ = A \]\n\nand\n\n\[ A \cap A = \{ x : x \in A\\text{ and }x \in A\} \]\n\n\[ = \{ x : x \in A\} \]\n\n\[ = A\\text... | No |
Theorem 1.3 De Morgan's Laws. Let \( A \) and \( B \) be sets. Then\n\n1. \( {\left( A \cup B\right) }^{\prime } = {A}^{\prime } \cap {B}^{\prime } \) ;\n\n2. \( {\left( A \cap B\right) }^{\prime } = {A}^{\prime } \cup {B}^{\prime } \) . | Proof. (1) If \( A \cup B = \varnothing \), then the theorem follows immediately since both \( A \) and \( B \) are the empty set. Otherwise, we must show that \( {\left( A \cup B\right) }^{\prime } \subset {A}^{\prime } \cap {B}^{\prime } \) and \( {\left( A \cup B\right) }^{\prime } \supset {A}^{\prime } \cap {B}^{\p... | Yes |
\[ \left( {A \smallsetminus B}\right) \cap \left( {B \smallsetminus A}\right) = \varnothing . \] | To see that this is true, observe that\n\n\[ \left( {A \smallsetminus B}\right) \cap \left( {B \smallsetminus A}\right) = \left( {A \cap {B}^{\prime }}\right) \cap \left( {B \cap {A}^{\prime }}\right) \]\n\n\[ = A \cap {A}^{\prime } \cap B \cap {B}^{\prime } \]\n\n\[ = \varnothing \text{.} \] | Yes |
If \( A = \{ x, y\}, B = \{ 1,2,3\} \), and \( C = \varnothing \), then \( A \times B \) is the set | \[ \{ \left( {x,1}\right) ,\left( {x,2}\right) ,\left( {x,3}\right) ,\left( {y,1}\right) ,\left( {y,2}\right) ,\left( {y,3}\right) \} \] and \[ A \times C = \varnothing \text{.} \] | Yes |
Let \( f\left( x\right) = {x}^{2} \) and \( g\left( x\right) = {2x} + 5 \) . Then | \[
\left( {f \circ g}\right) \left( x\right) = f\left( {g\left( x\right) }\right) = {\left( 2x + 5\right) }^{2} = 4{x}^{2} + {20x} + {25}
\]
and
\[
\left( {g \circ f}\right) \left( x\right) = g\left( {f\left( x\right) }\right) = 2{x}^{2} + 5.
\]
In general, order makes a difference; that is, in most cases \( f \circ... | Yes |
Sometimes it is the case that \( f \circ g = g \circ f \) . Let \( f\left( x\right) = {x}^{3} \) and \( g\left( x\right) = \sqrt[3]{x} \) . | Then\n\n\[ \left( {f \circ g}\right) \left( x\right) = f\left( {g\left( x\right) }\right) = f\left( \sqrt[3]{x}\right) = {\left( \sqrt[3]{x}\right) }^{3} = x \]\n\nand\n\n\[ \left( {g \circ f}\right) \left( x\right) = g\left( {f\left( x\right) }\right) = g\left( {x}^{3}\right) = \sqrt[3]{{x}^{3}} = x. \] | Yes |
Given a \( 2 \times 2 \) matrix\n\n\[ A = \left( \begin{array}{ll} a & b \\ c & d \end{array}\right) \]\n\nwe can define a map \( {T}_{A} : {\mathbb{R}}^{2} \rightarrow {\mathbb{R}}^{2} \) by\n\n\[ {T}_{A}\left( {x, y}\right) = \left( {{ax} + {by},{cx} + {dy}}\right) \]\n\nfor \( \left( {x, y}\right) \) in \( {\mathbb{... | This is actually matrix multiplication; that is,\n\n\[ \left( \begin{array}{ll} a & b \\ c & d \end{array}\right) \left( \begin{array}{l} x \\ y \end{array}\right) = \left( \begin{array}{l} {ax} + {by} \\ {cx} + {dy} \end{array}\right) . \] | Yes |
Theorem 1.15. Let \( f : A \rightarrow B, g : B \rightarrow C \), and \( h : C \rightarrow D \) . Then\n\n1. The composition of mappings is associative; that is, \( \left( {h \circ g}\right) \circ f = h \circ \left( {g \circ f}\right) \) ; | Proof. We will prove (1) and (3). Part (2) is left as an exercise. Part (4) follows directly from (2) and (3).\n\n(1) We must show that\n\n\[ h \circ \left( {g \circ f}\right) = \left( {h \circ g}\right) \circ f. \]\n\nFor \( a \in A \) we have\n\n\[ \left( {h \circ \left( {g \circ f}\right) }\right) \left( a\right) = ... | No |
The natural logarithm and the exponential functions, \( f\left( x\right) = \ln x \) and \( {f}^{-1}\left( x\right) = {e}^{x} \), are inverses of each other provided that we are careful about choosing domains. | Observe that\n\n\[ f\left( {{f}^{-1}\left( x\right) }\right) = f\left( {e}^{x}\right) = \ln {e}^{x} = x \]\n\nand\n\n\[ {f}^{-1}\left( {f\left( x\right) }\right) = {f}^{-1}\left( {\ln x}\right) = {e}^{\ln x} = x \]\n\nwhenever composition makes sense. | Yes |
Suppose that\n\n\[ A = \\left( \\begin{array}{ll} 3 & 1 \\\\ 5 & 2 \\end{array}\\right) \]\n\nThen \( A \) defines a map from \( {\\mathbb{R}}^{2} \) to \( {\\mathbb{R}}^{2} \) by\n\n\[ {T}_{A}\\left( {x, y}\\right) = \\left( {{3x} + y,{5x} + {2y}}\\right) . \]\n\nWe can find an inverse map of \( {T}_{A} \) by simply i... | In this example,\n\n\[ {A}^{-1} = \\left( \\begin{matrix} 2 & - 1 \\\\ - 5 & 3 \\end{matrix}\\right) \]\n\nhence, the inverse map is given by\n\n\[ {T}_{A}^{-1}\\left( {x, y}\\right) = \\left( {{2x} - y, - {5x} + {3y}}\\right) . \]\n\nIt is easy to check that\n\n\[ {T}_{A}^{-1} \\circ {T}_{A}\\left( {x, y}\\right) = {T... | Yes |
Given the permutation\n\n\[ \pi = \left( \begin{array}{lll} 1 & 2 & 3 \\ 2 & 3 & 1 \end{array}\right) \]\n\non \( S = \{ 1,2,3\} \), it is easy to see that the permutation defined by\n\n\[ {\pi }^{-1} = \left( \begin{array}{lll} 1 & 2 & 3 \\ 3 & 1 & 2 \end{array}\right) \]\n\nis the inverse of \( \pi \) . | In fact, any bijective mapping possesses an inverse, as we will see in the next theorem. | No |
Theorem 1.20. A mapping is invertible if and only if it is both one-to-one and onto. | Proof. Suppose first that \( f : A \rightarrow B \) is invertible with inverse \( g : B \rightarrow A \) . Then \( g \circ f = i{d}_{A} \) is the identity map; that is, \( g\left( {f\left( a\right) }\right) = a \) . If \( {a}_{1},{a}_{2} \in A \) with \( f\left( {a}_{1}\right) = f\left( {a}_{2}\right) \), then \( {a}_{... | Yes |
Let \( p, q, r \), and \( s \) be integers, where \( q \) and \( s \) are nonzero. Define \( p/q \sim r/s \) if \( {ps} = {qr} \) . Clearly \( \sim \) is reflexive and symmetric. To show that it is also transitive, suppose that \( p/q \sim r/s \) and \( r/s \sim t/u \), with \( q, s \), and \( u \) all nonzero. | Then \( {ps} = {qr} \) and \( {ru} = {st} \) . Therefore, \[ {psu} = {qru} = {qst}. \] Since \( s \neq 0,{pu} = {qt} \) . Consequently, \( p/q \sim t/u \) . | Yes |
Suppose that \( f \) and \( g \) are differentiable functions on \( \mathbb{R} \) . We can define an equivalence relation on such functions by letting \( f\left( x\right) \sim g\left( x\right) \) if \( {f}^{\prime }\left( x\right) = {g}^{\prime }\left( x\right) \) . It is clear that \( \sim \) is both reflexive and sym... | From calculus we know that \( f\left( x\right) - g\left( x\right) = {c}_{1} \) and \( g\left( x\right) - h\left( x\right) = {c}_{2} \), where \( {c}_{1} \) and \( {c}_{2} \) are both constants. Hence,\n\n\[ f\left( x\right) - h\left( x\right) = \left( {f\left( x\right) - g\left( x\right) }\right) + \left( {g\left( x\ri... | Yes |
Let \( A \) and \( B \) be \( 2 \times 2 \) matrices with entries in the real numbers. We can define an equivalence relation on the set of \( 2 \times 2 \) matrices, by saying \( A \sim B \) if there exists an invertible matrix \( P \) such that \( {PA}{P}^{-1} = B \). | Let \( I \) be the \( 2 \times 2 \) identity matrix; that is, \[ I = \left( \begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right) \] Then \( {IA}{I}^{-1} = {IAI} = A \) ; therefore, the relation is reflexive. To show symmetry, suppose that \( A \sim B \) . Then there exists an invertible matrix \( P \) such that \( {PA}{... | Yes |
Theorem 1.25. Given an equivalence relation \( \sim \) on a set \( X \), the equivalence classes of \( X \) form a partition of \( X \) . Conversely, if \( \mathcal{P} = \left\{ {X}_{i}\right\} \) is a partition of a set \( X \), then there is an equivalence relation on \( X \) with equivalence classes \( {X}_{i} \) . | Proof. Suppose there exists an equivalence relation \( \sim \) on the set \( X \) . For any \( x \in X \) , the reflexive property shows that \( x \in \left\lbrack x\right\rbrack \) and so \( \left\lbrack x\right\rbrack \) is nonempty. Clearly \( X = \mathop{\bigcup }\limits_{{x \in X}}\left\lbrack x\right\rbrack \) . ... | Yes |
Example 1.30. Let \( r \) and \( s \) be two integers and suppose that \( n \in \mathbb{N} \) . We say that \( r \) is congruent to \( s \) modulo \( n \), or \( r \) is congruent to \( s{\;\operatorname{mod}\;n} \), if \( r - s \) is evenly divisible by \( n \) ; that is, \( r - s = {nk} \) for some \( k \in \mathbb{Z... | Certainly any integer \( r \) is equivalent to itself since \( r - r = 0 \) is divisible by \( n \) . We will now show that the relation is symmetric. If \( r \equiv s \left( {\;\operatorname{mod}\;n}\right) \), then \( r - s = - \left( {s - r}\right) \) is divisible by \( n \) . So \( s - r \) is divisible by \( n \) ... | Yes |
For all integers \( n \geq 3,{2}^{n} > n + 4 \) | Since\n\n\[\n8 = {2}^{3} > 3 + 4 = 7\n\]\n\nthe statement is true for \( {n}_{0} = 3 \) . Assume that \( {2}^{k} > k + 4 \) for \( k \geq 3 \) . Then \( {2}^{k + 1} = 2 \cdot {2}^{k} > \n\n\( 2\left( {k + 4}\right) \) . But\n\n\[\n2\left( {k + 4}\right) = {2k} + 8 > k + 5 = \left( {k + 1}\right) + 4\n\]\n\nsince \( k \... | Yes |
Every integer \( {10}^{n + 1} + 3 \cdot {10}^{n} + 5 \) is divisible by 9 for \( n \in \mathbb{N} \) . | For \( n = 1 \) , \[ {10}^{1 + 1} + 3 \cdot {10} + 5 = {135} = 9 \cdot {15} \] is divisible by 9 . Suppose that \( {10}^{k + 1} + 3 \cdot {10}^{k} + 5 \) is divisible by 9 for \( k \geq 1 \) . Then \[ {10}^{\left( {k + 1}\right) + 1} + 3 \cdot {10}^{k + 1} + 5 = {10}^{k + 2} + 3 \cdot {10}^{k + 1} + {50} - {45} \] \[ =... | Yes |
We will prove the binomial theorem using mathematical induction; that is,\n\n\[ \n{\left( a + b\right) }^{n} = \mathop{\sum }\limits_{{k = 0}}^{n}\left( \begin{array}{l} n \\ k \end{array}\right) {a}^{k}{b}^{n - k} \n\]\n\nwhere \( a \) and \( b \) are real numbers, \( n \in \mathbb{N} \), and\n\n\[ \n\left( \begin{arr... | We first show that\n\n\[ \n\left( \begin{matrix} n + 1 \\ k \end{matrix}\right) = \left( \begin{array}{l} n \\ k \end{array}\right) + \left( \begin{matrix} n \\ k - 1 \end{matrix}\right) \n\]\n\nThis result follows from\n\n\[ \n\left( \begin{array}{l} n \\ k \end{array}\right) + \left( \begin{matrix} n \\ k - 1 \end{ma... | Yes |
Lemma 2.7. The Principle of Mathematical Induction implies that 1 is the least positive natural number. | Proof. Let \( S = \{ n \in \mathbb{N} : n \geq 1\} \) . Then \( 1 \in S \) . Assume that \( n \in S \) . Since \( 0 < 1 \), it must be the case that \( n = n + 0 < n + 1 \) . Therefore, \( 1 \leq n < n + 1 \) . Consequently, if \( n \in S \), then \( n + 1 \) must also be in \( S \), and by the Principle of Mathematica... | Yes |
Theorem 2.8. The Principle of Mathematical Induction implies the Principle of Well-Ordering. That is, every nonempty subset of \( \mathbb{N} \) contains a least element. | Proof. We must show that if \( S \) is a nonempty subset of the natural numbers, then \( S \) contains a least element. If \( S \) contains 1, then the theorem is true by Lemma 2.7. Assume that if \( S \) contains an integer \( k \) such that \( 1 \leq k \leq n \), then \( S \) contains a least element. We will show th... | No |
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