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Lemma 4.9 Let \( \Omega \) be an open bounded set in \( {\mathbb{R}}^{d} \). Suppose \( v \) belongs to \( {C}^{1}\left( \bar{\Omega }\right) \) and \( v \) vanishes on \( \partial \Omega \). Then\n\n\[{\int }_{\Omega }{\left| v\left( x\right) \right| }^{2}{dx} \leq {c}_{\Omega }{\int }_{\Omega }{\left| \nabla v\left( ... | Proof. This conclusion could in fact be deduced from the considerations given in Lemma 3.3. We prefer to prove this easy version separately to highlight a simple idea that we shall also use later. It should be noted that the argument yields the estimate \( {c}_{\Omega } \leq d{\left( \Omega \right) }^{2} \), where \( d... | Yes |
Lemma 4.11 Let \( f \) be a continuous function on a compact subset \( \Gamma \) of \( {\mathbb{R}}^{d} \). Then there exists a function \( G \) on \( {\mathbb{R}}^{d} \) that is continuous, and so that \( {\left. G\right| }_{\partial \Gamma } = f \) | Proof. We begin with the observation that if \( {K}_{0} \) and \( {K}_{1} \) are two disjoint compact sets, there exists a continuous function \( 0 \leq g\left( x\right) \leq 1 \) on \( {\mathbb{R}}^{d} \) which takes the value 0 on \( {K}_{0} \) and 1 on \( {K}_{1} \). Indeed, if \( d\left( {x,\Omega }\right) \) denot... | Yes |
Theorem 1.2 If \( {\mu }_{ * } \) is a metric exterior measure on a metric space \( X \) , then the Borel sets in \( X \) are measurable. Hence \( {\mu }_{ * } \) restricted to \( {\mathcal{B}}_{X} \) is a measure. | Proof. By the definition of \( {\mathcal{B}}_{X} \) it suffices to prove that closed sets in \( X \) are Carathéodory measurable. Therefore, let \( F \) denote a closed set and \( A \) a subset of \( X \) with \( {\mu }_{ * }\left( A\right) < \infty \) . For each \( n > 0 \), let\n\n\[ \n{A}_{n} = \left\{ {x \in {F}^{c... | Yes |
Lemma 1.4 If \( {\mu }_{0} \) is a premeasure on an algebra \( \mathcal{A} \), define \( {\mu }_{ * } \) on any subset \( E \) of \( X \) by\n\n\[ \n{\mu }_{ * }\left( E\right) = \inf \left\{ {\mathop{\sum }\limits_{{j = 1}}^{\infty }{\mu }_{0}\left( {E}_{j}\right) : E \subset \mathop{\bigcup }\limits_{{j = 1}}^{\infty... | Proof. Proving that \( {\mu }_{ * } \) is an exterior measure presents no difficulty. To see why the restriction of \( {\mu }_{ * } \) to \( \mathcal{A} \) coincides with \( {\mu }_{0} \), suppose that \( E \in \mathcal{A} \) . Clearly, one always has \( {\mu }_{ * }\left( E\right) \leq {\mu }_{0}\left( E\right) \) sin... | Yes |
Theorem 1.5 Suppose that \( \mathcal{A} \) is an algebra of sets in \( X,{\mu }_{0} \) a premeasure on \( \mathcal{A} \), and \( \mathcal{M} \) the \( \sigma \) -algebra generated by \( \mathcal{A} \) . Then there exists a measure \( \mu \) on \( \mathcal{M} \) that extends \( {\mu }_{0} \) . | Proof. The exterior measure \( {\mu }_{ * } \) induced by \( {\mu }_{0} \) defines a measure \( \mu \) on the \( \sigma \) -algebra of Carathéodory measurable sets. Therefore, by the result in the previous lemma, \( \mu \) is also a measure on \( \mathcal{M} \) that extends \( {\mu }_{0} \) . (We should observe that in... | Yes |
Proposition 3.2 If \( E \) is an arbitrary measurable set in \( X \), then the conclusion of Proposition 3.1 are still valid except that we only assert that \( {E}^{{x}_{2}} \) is \( {\mu }_{1} \) -measurable and \( {\mu }_{1}\left( {E}^{{x}_{2}}\right) \) is defined for almost every \( {x}_{2} \in {X}_{2} \) . | Proof. Consider first the case when \( E \) is a set of measure zero. Then we know by Proposition 1.6 that there is a set \( F \in {\mathcal{A}}_{\sigma \delta } \) such that \( E \subset F \) and \( \left( {{\mu }_{1} \times {\mu }_{2}}\right) \left( F\right) = 0 \) . Since \( {E}^{{x}_{2}} \subset {F}^{{x}_{2}} \) fo... | Yes |
Theorem 3.3 In the setting above, suppose \( f\left( {{x}_{1},{x}_{2}}\right) \) is an integrable function on \( \left( {{X}_{1} \times {X}_{2},{\mu }_{1} \times {\mu }_{2}}\right) \) . (i) For almost every \( {x}_{2} \in {X}_{2} \), the slice \( {f}^{{x}_{2}}\left( {x}_{1}\right) = f\left( {{x}_{1},{x}_{2}}\right) \) ... | Proof. Note that if the desired conclusions hold for finitely many functions, they also hold for their linear combinations. In particular it suffices to assume that \( f \) is non-negative. When \( f = {\chi }_{E} \), where \( E \) is a set of finite measure, what we wish to prove is contained in Proposition 3.2. Hence... | Yes |
Proposition 4.1 The total variation \( \left| \nu \right| \) of a signed measure \( \nu \) is itself a (positive) measure that satisfies \( \nu \leq \left| \nu \right| \) . | Proof. Suppose \( {\left\{ {E}_{j}\right\} }_{j = 1}^{\infty } \) is a countable collection of disjoints sets in \( \mathcal{M} \), and let \( E = \bigcup {E}_{j} \) . It suffices to prove:\n\n(11)\n\n\[ \sum \left| \nu \right| \left( {E}_{j}\right) \leq \left| \nu \right| \left( E\right) \;\text{ and }\;\left| \nu \ri... | Yes |
Proposition 4.2 The assertion (14) implies (12). Conversely, if \( \\left| \\nu \\right| \) is a finite measure, then (12) implies (14). | That (12) is a consequence of (14) is obvious because \( \\mu \\left( E\\right) = 0 \) gives \( \\left| {\\nu \\left( E\\right) }\\right| < \\epsilon \) for every \( \\epsilon > 0 \) . To prove the converse, it suffices to consider the case when \( \\nu \) is positive, upon replacing \( \\nu \) by \( \\left| \\nu \\rig... | Yes |
Lemma 5.2 The following relations hold among the subspaces \( S,{S}_{ * } \) , and \( \overline{{S}_{1}} \) .\n\n(i) \( S = {S}_{ * } \) .\n\n(ii) The orthogonal complement of \( \overline{{S}_{1}} \) is \( S \) . | Proof. First, since \( T \) is an isometry, we have that \( \left( {{Tf},{Tg}}\right) = \left( {f, g}\right) \) for all \( f, g \in \mathcal{H} \), and thus \( {T}^{ * }T = I \) . (See Exercise 22 in Chapter 4.) So if \( {Tf} = f \) then \( {T}^{ * }{Tf} = {T}^{ * }f \), which means that \( f = {T}^{ * }f \) . To prove... | No |
Proposition 6.2 Suppose \( T \) is symmetric. Then \( \parallel T\parallel \leq M \) if and only if \( - {MI} \leq \) \( T \leq {MI} \) . As a result, \( \parallel T\parallel = \max \left( {\left| a\right| ,\left| b\right| }\right) \) . | This is a consequence of (7) in Chapter 4. | No |
Proposition 6.4 If \( {T}_{1} \) and \( {T}_{2} \) are positive operators that commute, then \( {T}_{1}{T}_{2} \) is also positive. | Indeed, if \( S \) is a square root of \( {T}_{1} \) given in the previous proposition, then \( {T}_{1}{T}_{2} = \) \( {SS}{T}_{2} = S{T}_{2}S \), and hence \( \left( {{T}_{1}{T}_{2}f, f}\right) = \left( {S{T}_{2}{Sf}, f}\right) = \left( {{T}_{2}{Sf},{Sf}}\right) \), since \( S \) is symmetric, and thus the last term i... | Yes |
Proposition 6.5 Suppose \( T \) is symmetric and a and \( b \) are given by (33). If \( p\left( t\right) = \) \( \mathop{\sum }\limits_{{k = 0}}^{n}{c}_{k}{t}^{k} \) is a real polynomial which is positive for \( t \in \left\lbrack {a, b}\right\rbrack \), then the operator \( p\left( T\right) = \mathop{\sum }\limits_{{k... | To see this, write \( p\left( t\right) = c\mathop{\prod }\limits_{j}\left( {t - {\rho }_{j}}\right) \mathop{\prod }\limits_{k}\left( {{\rho }_{k}^{\prime } - t}\right) \mathop{\prod }\limits_{\ell }\left( {{\left( t - {\mu }_{\ell }\right) }^{2} + {\nu }_{\ell }}\right) \), where \( c \) is positive and the third facto... | Yes |
Corollary 6.6 If \( p\left( t\right) \) is a real polynomial, then\n\n\[ \parallel p\left( T\right) \parallel \leq \mathop{\sup }\limits_{{t \in \left\lbrack {a, b}\right\rbrack }}\left| {p\left( t\right) }\right| \] | This is an immediate consequence using Proposition 6.2, since \( - M \leq p\left( t\right) \leq M \) , where \( M = \mathop{\sup }\limits_{{t \in \left\lbrack {a, b}\right\rbrack }}\left| {p\left( t\right) }\right| \), and thus \( - {MI} \leq p\left( T\right) \leq {MI} \). | Yes |
Proposition 6.7 Suppose \( \left\{ {T}_{n}\right\} \) is a sequence of positive operators that satisfy \( {T}_{n} \geq {T}_{n + 1} \) for all \( n \) . Then there is a positive operator \( T \), such that \( {T}_{n}f \rightarrow {Tf} \) as \( n \rightarrow \infty \) for every \( f \in \mathcal{H} \) . | Proof. We note that for each fixed \( f \in \mathcal{H} \) the sequence of positive numbers \( \left( {{T}_{n}f, f}\right) \) is decreasing and hence convergent. Now observe that for any positive operator \( S \) with \( \parallel S\parallel \leq M \) we have\n\n(35)\n\n\[ \parallel S\left( f\right) {\parallel }^{2} \l... | Yes |
Proposition 6.8 If \( T \) is symmetric, then \( \sigma \left( T\right) \) is a closed subset of the interval \( \left\lbrack {a, b}\right\rbrack \) given by (33). | Note that if \( z \notin \left\lbrack {a, b}\right\rbrack \), the function \( \Phi \left( t\right) = {\left( t - z\right) }^{-1} \) is continuous on \( \left\lbrack {a, b}\right\rbrack \) and \( \Phi \left( T\right) \left( {T - {zI}}\right) = \left( {T - {zI}}\right) \Phi \left( T\right) = I \), so \( \Phi \left( T\rig... | Yes |
Proposition 6.9 For each \( f \in \mathcal{H} \), the Lebesgue-Stieltjes measure corresponding to \( F\left( \lambda \right) = \left( {E\left( \lambda \right) f, f}\right) \) is supported on \( \sigma \left( T\right) \) . | To put it another way, \( F\left( \lambda \right) \) is constant on each open interval of the complement of \( \sigma \left( T\right) \) . To prove this, let \( J \) be one of the open intervals in the complement of \( \sigma \left( T\right) \) , \( {x}_{0} \in J \), and \( {J}_{0} \) the sub-interval centered at \( {x... | Yes |
Property 1 (Monotonicity) If \( {E}_{1} \subset {E}_{2} \), then \( {m}_{\alpha }^{ * }\left( {E}_{1}\right) \leq {m}_{\alpha }^{ * }\left( {E}_{2}\right) \) . | This is straightforward, since any cover of \( {E}_{2} \) is also a cover of \( {E}_{1} \) . | No |
Property 2 (Sub-additivity) \( {m}_{\alpha }^{ * }\left( {\mathop{\bigcup }\limits_{{j = 1}}^{\infty }{E}_{j}}\right) \leq \mathop{\sum }\limits_{{j = 1}}^{\infty }{m}_{\alpha }^{ * }\left( {E}_{j}\right) \) for any countable family \( \left\{ {E}_{j}\right\} \) of sets in \( {\mathbb{R}}^{d} \) . | For the proof, fix \( \delta \), and choose for each \( j \) a cover \( {\left\{ {F}_{j, k}\right\} }_{k = 1}^{\infty } \) of \( {E}_{j} \) by sets of diameter less than \( \delta \) such that \( \mathop{\sum }\limits_{k}{\left( \operatorname{diam}{F}_{j, k}\right) }^{\alpha } \leq {\mathcal{H}}_{\alpha }^{\delta }\lef... | Yes |
Property 3 If \( d\left( {{E}_{1},{E}_{2}}\right) > 0 \), then \( {m}_{\alpha }^{ * }\left( {{E}_{1} \cup {E}_{2}}\right) = {m}_{\alpha }^{ * }\left( {E}_{1}\right) + {m}_{\alpha }^{ * }\left( {E}_{2}\right) \) . | It suffices to prove that \( {m}_{\alpha }^{ * }\left( {{E}_{1} \cup {E}_{2}}\right) \geq {m}_{\alpha }^{ * }\left( {E}_{1}\right) + {m}_{\alpha }^{ * }\left( {E}_{2}\right) \) since the reverse inequality is guaranteed by sub-additivity. Fix \( \epsilon > 0 \) with \( \epsilon < \) \( d\left( {{E}_{1},{E}_{2}}\right) ... | Yes |
Property 5 Hausdorff measure is invariant under translations\n\n\[ \n{m}_{\alpha }\left( {E + h}\right) = {m}_{\alpha }\left( E\right) \;\text{ for all }h \in {\mathbb{R}}^{d}, \n\] | These conclusions follow once we observe that the diameter of a set \( S \) is invariant under translations and rotations, and satisfies \( \operatorname{diam}\left( {\lambda S}\right) = \) \( \lambda \operatorname{diam}\left( S\right) \) for \( \lambda > 0 \) . | Yes |
Property 6 The quantity \( {m}_{0}\left( E\right) \) counts the number of points in \( E \) , while \( {m}_{1}\left( E\right) = m\left( E\right) \) for all Borel sets \( E \subset \mathbb{R} \) . (Here \( m \) denotes the Lebesgue measure on \( \mathbb{R} \) .) | In fact, note that in one dimension every set of diameter \( \delta \) is contained in an interval of length \( \delta \) (and for an interval its length equals its Lebesgue measure). | No |
Property 7 If \( E \) is a Borel subset of \( {\mathbb{R}}^{d} \), then \( {c}_{d}{m}_{d}\left( E\right) = m\left( E\right) \) for some constant \( {c}_{d} \) that depends only on the dimension \( d \) . | The constant \( {c}_{d} \) equals \( m\left( B\right) /{\left( \operatorname{diam}B\right) }^{d} \), for the unit ball \( B \) ; note that this ratio is the same for all balls \( B \) in \( {\mathbb{R}}^{d} \), and so \( {c}_{d} = {v}_{d}/{2}^{d} \) (where \( {v}_{d} \) denotes the volume of the unit ball). The proof o... | Yes |
Theorem 2.1 The Cantor set \( \mathcal{C} \) has strict Hausdorff dimension \( \alpha = \) \( \log 2/\log 3 \) . | The inequality\n\n\[ \n{m}_{\alpha }\left( \mathcal{C}\right) \leq 1 \n\]\n\nfollows from the construction of \( \mathcal{C} \) and the definitions. Indeed, recall from Chapter 1 that \( \mathcal{C} = \bigcap {C}_{k} \), where each \( {C}_{k} \) is a finite union of \( {2}^{k} \) intervals of length \( {3}^{-k} \) . Gi... | Yes |
Lemma 2.2 Suppose a function \( f \) defined on a compact set \( E \) satisfies a Lipschitz condition with exponent \( \gamma \) . Then\n\n(i) \( {m}_{\beta }\left( {f\left( E\right) }\right) \leq {M}^{\beta }{m}_{\alpha }\left( E\right) \) if \( \beta = \alpha /\gamma \) .\n\n(ii) \( \dim f\left( E\right) \leq \frac{1... | Proof. Suppose \( \left\{ {F}_{k}\right\} \) is a countable family of sets that covers \( E \) . Then \( \left\{ {f\left( {E \cap {F}_{k}}\right) }\right\} \) covers \( f\left( E\right) \) and, moreover, \( f\left( {E \cap {F}_{k}}\right) \) has diameter less than \( M{\left( \operatorname{diam}{F}_{k}\right) }^{\gamma... | Yes |
Lemma 2.3 The Cantor-Lebesgue function \( F \) on \( \mathcal{C} \) satisfies a Lipschitz condition with exponent \( \gamma = \log 2/\log 3 \) . | Proof. The function \( F \) was constructed in Section 3.1 of Chapter 3 as the limit of a sequence \( \left\{ {F}_{n}\right\} \) of piecewise linear functions. The function \( {F}_{n} \) increases by at most \( {2}^{-n} \) on each interval of length \( {3}^{-n} \) . So the slope of \( {F}_{n} \) is always bounded by \(... | Yes |
Theorem 2.5 The Sierpinski triangle \( \mathcal{S} \) has strict Hausdorff dimension \( \alpha = \log 3/\log 2 \) . | The inequality \( {m}_{\alpha }\left( \mathcal{S}\right) \leq 1 \) follows immediately from the construction. Given \( \delta > 0 \), choose \( K \) so that \( {2}^{-K} < \delta \) . Since the set \( {S}_{K} \) covers \( \mathcal{S} \) and consists of \( {3}^{K} \) triangles each of diameter \( {2}^{-K} < \delta \), we... | Yes |
Lemma 2.6 Suppose \( B \) is a ball in the covering \( \mathcal{B} \) that satisfies\n\n\[ \n{2}^{-\ell } \leq \operatorname{diam}B < {2}^{-\ell + 1}\;\text{ for some }\ell \leq k.\n\]\n\nThen \( B \) contains at most \( c{3}^{k - \ell } \) vertices of the \( {k}^{\text{th }} \) generation. | Proof of Lemma 2.6. Let \( {B}^{ * } \) denote the ball with same center as \( B \) but three times its diameter, and let \( {\bigtriangleup }_{k} \) be a triangle of the \( {k}^{\text{th }} \) generation whose vertex \( v \) lies in \( B \) . If \( {\bigtriangleup }_{\ell }^{\prime } \) denotes the triangle of the \( ... | Yes |
Theorem 2.7 The function \( \mathcal{K}\left( t\right) \) satisfies a Lipschitz condition of exponent \( \gamma = \log 3/\log 4 \), that is:\n\n\[ \left| {\mathcal{K}\left( t\right) - \mathcal{K}\left( s\right) }\right| \leq M{\left| t - s\right| }^{\gamma }\;\text{ for all }t, s \in \left\lbrack {0,1}\right\rbrack . \... | We have already observed that \( \left| {{K}_{j + 1}\left( t\right) - {K}_{j}\left( t\right) }\right| \leq {3}^{-j} \) . Since \( {K}_{j} \) travels a distance of \( {3}^{-j} \) in \( {4}^{-j} \) units of time, we see that\n\n\[ \left| {{K}_{j}^{\prime }\left( t\right) }\right| \leq {\left( \frac{4}{3}\right) }^{j}\;\t... | Yes |
Lemma 2.8 Suppose \( \\left\\{ {f}_{j}\\right\\} \) is a sequence of continuous functions on the interval \( \\left\\lbrack {0,1}\\right\\rbrack \) that satisfy\n\n\[ \n\\left| {{f}_{j}\\left( t\\right) - {f}_{j}\\left( s\\right) }\\right| \\leq {A}^{j}\\left| {t - s}\\right| \\;\\text{ for some }A > 1,\n\]\n\nand\n\n\... | Proof. The continuous limit \( f \) is given by the uniformly convergent series\n\n\[ \nf\\left( t\\right) = {f}_{1}\\left( t\\right) + \\mathop{\\sum }\\limits_{{k = 1}}^{\\infty }\\left( {{f}_{k + 1}\\left( t\\right) - {f}_{k}\\left( t\\right) }\\right)\n\]\n\nand therefore\n\n\[ \n\\left| {f\\left( t\\right) - {f}_{... | Yes |
Theorem 2.9 Suppose \( {S}_{1},{S}_{2},\ldots ,{S}_{m} \) are \( m \) similartities, each with the same ratio \( r \) that satisfies \( 0 < r < 1 \) . Then there exists a unique nonempty compact set \( F \) such that\n\n\[ F = {S}_{1}\left( F\right) \cup \cdots \cup {S}_{m}\left( F\right) \] | The proof of this theorem is in the nature of a fixed point argument. We shall begin with some large ball \( B \) and iteratively apply the mappings \( {S}_{1},\ldots ,{S}_{m} \) . The fact that the similarities have ratio \( r < 1 \) will suffice to imply that this process contracts to a unique set \( F \) with the de... | Yes |
Lemma 2.10 There exists a closed ball \( B \) so that \( {S}_{j}\left( B\right) \subset B \) for all \( j = 1,\ldots, m \) . | Proof. Indeed, we note that if \( S \) is a similarity with ratio \( r \), then\n\n\[ \left| {S\left( x\right) }\right| \leq \left| {S\left( x\right) - S\left( 0\right) }\right| + \left| {S\left( 0\right) }\right| \]\n\n\[ \leq r\left| x\right| + \left| {S\left( 0\right) }\right| \text{.} \]\n\nIf we require that \( \l... | Yes |
Lemma 2.11 The distance function dist defined on compact subsets of \( {\mathbb{R}}^{d} \) satisfies\n\n(i) \( \operatorname{dist}\left( {A, B}\right) = 0 \) if and only if \( A = B \) .\n\n(ii) \( \operatorname{dist}\left( {A, B}\right) = \operatorname{dist}\left( {B, A}\right) \) .\n\n(iii) \( \operatorname{dist}\lef... | The proof of the lemma is simple and may be left to the reader. | No |
Theorem 2.12 Suppose \( {S}_{1},{S}_{2},\ldots ,{S}_{m} \) are \( m \) separated similarities with the common ratio \( r \) that satisfies \( 0 < r < 1 \) . Then the set \( F \) has Hausdorff dimension equal to \( \log m/\log \left( {1/r}\right) \) . | We now turn to the proof of Theorem 2.12, which will follow the same approach used in the case of the Sierpinski triangle. If \( \alpha = \log m/\log \left( {1/r}\right) \) , we claim that \( {m}_{\alpha }\left( F\right) < \infty \), hence \( \dim F \leq \alpha \) . Moreover, this inequality holds even without the sepa... | Yes |
Lemma 2.13 Suppose \( B \) is a ball in the covering \( \mathcal{B} \) that satisfies\n\n\[ \n{r}^{\ell } \leq \operatorname{diam}B < {r}^{\ell - 1}\;\text{ for some }\ell \leq k.\n\]\n\nThen \( B \) contains at most \( {\mathrm{{cm}}}^{k - \ell } \) vertices of the \( {k}^{\text{th }} \) generation. | Proof. If \( v \) is a vertex of the \( {k}^{\text{th }} \) generation with \( v \in B \), and \( \mathcal{O}\left( v\right) \) denotes the corresponding open set of the \( {k}^{\text{th }} \) generation, then, for some fixed dilate \( {B}^{ * } \) of \( B \), properties (a) and (b) above guarantee that \( \mathcal{O}\... | Yes |
Proposition 3.3 Chains of quartic intervals satisfy the following properties:\n\n(i) If \( \\left\\{ {I}^{k}\\right\\} \) is a chain of quartic intervals, then there exists a unique \( t \\in \\left\\lbrack {0,1}\\right\\rbrack \) such that \( t \\in \\mathop{\\bigcap }\\limits_{k}{I}^{k} \) . | Proof. Part (i) follows from the fact that \( \\left\\{ {I}^{k}\\right\\} \) is a decreasing sequence of compact sets whose diameters go to 0 . | Yes |
Theorem 3.5 Given a dyadic correspondence \( \Phi \), there exist sets \( {Z}_{1} \subset \left\lbrack {0,1}\right\rbrack \) and \( {Z}_{2} \subset \left\lbrack {0,1}\right\rbrack \times \left\lbrack {0,1}\right\rbrack \), each of measure zero, so that: (i) \( {\Phi }^{ * } \) is a bijection on \( \left\lbrack {0,1}\ri... | Proof. First, let \( {\mathcal{N}}_{1} \) denote the collection of chains of those quartic intervals arising in (iii) of Proposition 3.3, those for which the points in \( I = \left\lbrack {0,1}\right\rbrack \) are not uniquely representable. Similarly, let \( {\mathcal{N}}_{2} \) denote the collection of chains of thos... | Yes |
Lemma 3.6 Let\n\n\\[ \n{E}_{0} = \\left\\{ {x = \\mathop{\\sum }\\limits_{{k = 1}}^{\\infty }{a}_{k}/{4}^{k},\\;}\\right. \\text{where}\\left. {{a}_{k} \\neq {f}_{k}\\text{for all sufficiently large}k}\\right\\} \\text{.} \n\\]\n\nThen \\( m\\left( {E}_{0}\\right) = 0 \\) . | Indeed, if we fix \\( r \\), then \\( m\\left( \\left\\{ {x : {a}_{r} \\neq {f}_{r}}\\right\\} \\right) = 3/4 \\), and\n\n\\[ \nm\\left( \\left\\{ {x : {a}_{r} \\neq {f}_{r}\\text{ and }{a}_{r + 1} \\neq {f}_{r + 1}}\\right\\} \\right) = {\\left( 3/4\\right) }^{2},\\;\\text{ etc. } \n\\]\n\nThus \\( m\\left( \\left\\{ ... | Yes |
Lemma 3.8 If \( \Phi \) is the dyadic correspondence in Lemma 3.7, then \( {\Phi }^{ * }\left( t\right) = \) \( \mathcal{P}\left( t\right) \) for every \( 0 \leq t \leq 1 \) . | Proof. First, we observe that \( {\Phi }^{ * }\left( t\right) \) is unambiguously defined for every \( t \) . Indeed, suppose \( t \in \mathop{\bigcap }\limits_{k}{I}^{k} \) and \( t \in \mathop{\bigcap }\limits_{k}{J}^{k} \) are two chains of quartic intervals; then \( {I}^{k} \) and \( {J}^{k} \) must be adjacent for... | Yes |
Theorem 4.3 There exists a set \( \mathcal{B} \) in \( {\mathbb{R}}^{2} \) that:\n\n(i) is compact,\n\n(ii) has Lebesgue measure zero,\n\n(iii) contains a translate of every unit line segment. | Note that with \( F = \mathcal{B} \) and \( \gamma \in {S}^{1} \) one has \( {m}_{1}\left( {F \cap {\mathcal{P}}_{{t}_{0},\gamma }}\right) \geq 1 \) for some \( {t}_{0} \). If \( {m}_{1}\left( {F \cap {\mathcal{P}}_{t,\gamma }}\right) \) were continuous in \( t \), then this measure would be strictly positive for an in... | No |
Lemma 4.7 If \( f \) is continuous with compact support, then for every \( \gamma \in {S}^{d - 1} \) we have\n\n\[ \widehat{\mathcal{R}}\left( f\right) \left( {\lambda ,\gamma }\right) = \widehat{f}\left( {\lambda \gamma }\right) \] | Proof. For each unit vector \( \gamma \) we use the adapted coordinate system described above: \( x = \left( {{x}_{1},\ldots ,{x}_{d}}\right) \) where \( \gamma \) coincides with the \( {x}_{d} \) direction. We can then write each \( x \in {\mathbb{R}}^{d} \) as \( x = \left( {u, t}\right) \) with \( u \in {\mathbb{R}}... | Yes |
Lemma 4.8 If \( f \) is continuous with compact support, then\n\n\[ \n{\int }_{{S}^{d - 1}}\left( {{\int }_{-\infty }^{\infty }{\left| \widehat{\mathcal{R}}\left( f\right) \left( \lambda ,\gamma \right) \right| }^{2}{\left| \lambda \right| }^{d - 1}{d\lambda }}\right) {d\sigma }\left( \gamma \right) = 2{\int }_{{\mathb... | Proof. The Plancherel formula in Chapter 5 guarantees that\n\n\[ \n2{\int }_{{\mathbb{R}}^{d}}{\left| f\left( x\right) \right| }^{2}{dx} = 2{\int }_{{\mathbb{R}}^{d}}{\left| \widehat{f}\left( \xi \right) \right| }^{2}{d\xi }.\n\]\n\nChanging to polar coordinates \( \xi = {\lambda \gamma } \) where \( \lambda > 0 \) and... | Yes |
\[ \mathop{\sup }\limits_{{t \in \mathbb{R}}}\left| {F\left( t\right) }\right| \leq c\left( {A + B}\right) \] | Proof. The first inequality is obtained by considering separately the two cases \( \left| \lambda \right| \leq 1 \) and \( \left| \lambda \right| > 1 \) . We write\n\n\[ F\left( t\right) = {\int }_{\left| \lambda \right| \leq 1}\widehat{F}\left( \lambda \right) {e}^{2\pi i\lambda t}{d\lambda } + {\int }_{\left| \lambda... | Yes |
Theorem 4.10 If \( f \) is continuous with compact support, then\n\n\[ \n{\int }_{{S}^{1}}{\mathcal{R}}_{\delta }^{ * }\left( f\right) \left( \gamma \right) {d\sigma }\left( \gamma \right) \leq c{\left( \log 1/\delta \right) }^{1/2}\left( {\parallel f{\parallel }_{{L}^{1}\left( {\mathbb{R}}^{2}\right) } + \parallel f{\... | The same argument as in the proof of Theorem 4.5 applies here, except that we need a modified version of Lemma 4.9. More precisely, let us set\n\n\[ \n{F}_{\delta }\left( t\right) = {\int }_{-\infty }^{\infty }\widehat{F}\left( \lambda \right) \left( \frac{{e}^{{2\pi i}\left( {t + \delta }\right) \lambda } - {e}^{{2\pi... | Yes |
Theorem 4.12 The set \( F \) is compact and of two-dimensional measure zero. It contains a translate of any unit line segment whose slope is a number \( s \) that lies outside the intervals \( \left( {-1,2}\right) \) . | Let us see how these two assertions imply the theorem. First, we note that the set \( F \) is closed (and hence compact), because both \( {E}_{0} \) and \( {E}_{1} \) are closed. Next observe that with \( 0 < y < 1 \), the slice \( {F}^{y} \) of the set \( F \) is exactly \( \left( {1 - y}\right) \mathcal{C} + \frac{y}... | Yes |
Proposition 4.13 Suppose \( {\lambda }_{0} \) and \( \ell \) are given, with \( 1 \leq {\lambda }_{0} \leq 4 \) and \( \ell \) a positive integer. Then, there exist a \( \bar{\lambda } \) and a pair \( i,{i}^{\prime } \) with \( i \neq {i}^{\prime } \) such that\n\n(9)\n\n\[ \n{\mathcal{K}}_{i}^{\ell }\left( \bar{\lamb... | This is proved on the basis of the following observation.\n\nLemma 4.14 For every | No |
Lemma 4.14 For every \( {\lambda }_{0} \) there is a pair \( 1 \leq {i}_{1},{i}_{2} \leq 4 \), with \( {i}_{1} \neq {i}_{2} \) such that \( {\mathcal{K}}_{{i}_{1}}\left( {\lambda }_{0}\right) \) and \( {\mathcal{K}}_{{i}_{2}}\left( {\lambda }_{0}\right) \) intersect. | Proof. Indeed, if the \( {\mathcal{K}}_{i} \) are disjoint for \( 1 \leq i \leq 4 \) then for sufficiently small \( \delta \) the \( {\mathcal{K}}_{i}^{\delta } \) are also disjoint. Here we have used the notation that \( {F}^{\delta } \) denotes the set of points of distance less than \( \delta \) from \( F \) . (See ... | Yes |
Theorem 1.4 (Irrationality of the Square Root of 2). For every rational \( \frac{p}{q} \in \mathbb{Q} \), we have \( {\left( \frac{p}{q}\right) }^{2} \neq 2 \) . | Proof. Suppose for contradiction that \( {\left( \frac{p}{q}\right) }^{2} = 2 \) where \( \left( {p, q}\right) = 1 \) . In particular, \( p \) and \( q \) are both not even.\n\nFor the first case, assume \( p \) is odd. But now \( {p}^{2} = 2{q}^{2} \), which is a contradiction. \n\nNow suppose \( p \) is even. Then \(... | Yes |
Theorem 1.5. If \( {\left( \frac{p}{q}\right) }^{n} = k \in \mathbb{N} \) where \( m \in \mathbb{N} \), then \( q = 1 \) . | Proof. Do some blah with prime factorizations. | No |
Theorem 1.13 (Least Upper Bound). If \( \mathcal{S} \) is a collection of cuts in \( \mathbb{Q} \) for which \( \mathcal{S} \) is not empty, and there exists an upper bound of \( \mathcal{S} \), then there is a least upper bound for \( \mathcal{S} \). | Proof. Take\n\n\[ E = \{ r \in \mathbb{Q} : \exists A \mid B \in \mathcal{S} \text{ and } r \in A \} \]\n\nand let \( F = \mathbb{Q} - E \). In other words, \( E = \cup_{A \mid B \in \mathcal{S}} A \). We claim \( E \mid F \) is a least upper bound.\n\nNow \( E \) is not empty because \( \mathcal{S} \) is not empty. Be... | Yes |
Theorem 2.5 (Cauchy Convergence Criterion). For sequences of real numbers, the two definitions are equivalent; that is, a sequence of reals converges if and only if it is Cauchy. | Proof. We prove only the hard direction. It is easy to check that \( \\left\\{ {{x}_{m} : m \\in \\mathbb{N}}\\right\\} \) is bounded (and nonempty), by picking \( \\varepsilon = {1000} \) in the Cauchy condition.\n\nNow let us consider\n\n\[ S\\overset{\\text{ def }}{ = }\\{ x \\in \\mathbb{R} : \\exists \\text{ inf m... | Yes |
Theorem 2.7 (Cauchy-Schwarz). For any \( x \) and \( y \) in \( {\mathbb{R}}^{m} \) , \n\n\[ \langle x, y{\rangle }^{2} \leq \left| x\right| \left| y\right| \] | Proof. Assume \( x, y \neq 0 \), otherwise it’s obvious.\n\nLet \( t \in \mathbb{R} \), and observe\n\n\[ 0 \leq \langle x + {ty}, x + {ty}\rangle = \langle x, x\rangle + {2t}\langle x, y\rangle + {t}^{2}\langle y, y\rangle .\n\]\n\nThe right-hand side is a quadratic in \( t \) which clearly has a nonpositive discrimin... | Yes |
Theorem 3.5 (Cantor's Diagonalisation Argument). \( \mathbb{R} \) is NOT denumerable. | Proof. Insert Cantor's proof here. | No |
Proposition 3.6. \( \mathbb{N} \times \mathbb{N} \) is denumerable. | Proof. A list is \( \left( {1,1}\right) ,\left( {1,2}\right) ,\left( {2,1}\right) ,\left( {1,3}\right) ,\left( {2,2}\right) ,\left( {3,1}\right) \), et cetera is a suitable list. | Yes |
Theorem 3.7. Suppose \( X \) is an infinite set.\n\n(a) If \( \exists f : X \rightarrow \mathbb{N} \) an injection, then \( X \) is denumerable.\n\n(b) If \( \exists g : \mathbb{N} \rightarrow X \) a surjection, then \( X \) is denumerable. | Proof. For part (a), notice that \( f\left( X\right) \) is infinite and nonempty. But \( \mathbb{N} \) has the least criminal property, so we can get a bijection! Namely, let \( \tau \left( n\right) \) denote the \( n \) th smallest element of \( f\left( X\right) \) ; this is defined because \( f\left( X\right) \) is i... | Yes |
Proposition 3.8. Let \( A = { \cup }_{i = 1}^{\infty }{A}_{i} \), where each \( {A}_{i} \) is denumerable. Then \( A \) is denumerable. | Proof. There is a surjection from \( \mathbb{N} \times \mathbb{N} \) to \( \cup {A}_{i} \) by taking \( \left( {i, j}\right) \) to the \( j \) th smallest element of \( {A}_{i} \) . | No |
Theorem 4.6. These two definitions are equivalent. | Proof. First we will show the first definition implies the second. Assume (2) is false, then \( \exists \varepsilon > 0 \) such that \( \forall \delta > 0 \) there exists a \ | No |
Proposition 4.8. Let \( f : M \rightarrow Y \) and \( g : Y \rightarrow P \) be continuous maps between metric spaces. Then \( g \circ f : M \rightarrow P \) is continuous as well. | Proof. Trivial using sequences. | No |
Ellipses are homeomorphic to \( {S}^{1} \) if they inherit a distance metric from \( {\mathbb{R}}^{2} \) | Function: assume they are cocentric, and map \( x \in {S}^{1} \) to \( y \in E \), where \( O, x \), and \( y \) are collinear, and \( O \) lies outside the segment \( {xy} \) | No |
Proposition 5.4. \( \lim S \) is closed, regardless of whether \( S \) itself is closed. That is, \( \lim \lim S = \lim S. \) | Proof. We wish to show that if \( {y}_{m} \rightarrow p \in M \), where \( {y}_{m} \in \lim S \), then \( p \in \lim S \) as well.\n\nFor each \( k,{y}_{k} \) is a limit of some sequence in \( S \), say \( {\left( {x}_{k, n}\right) }_{n \geq 1} \), converging to \( {y}_{k} \) . For each \( k \), we can find \( {x}_{k, ... | Yes |
Corollary 5.5. \( \lim S \) is the smallest closed subset of \( M \) that contains \( S \) . | So, we say that \( \lim S \) is the closure of \( S \) . | No |
Theorem 5.9. Open and closed are dual concepts: if \( S \) is closed then \( M \smallsetminus S \) is open, and vice-versa. | Proof. Let \( {S}^{c} = M \smallsetminus S \) . First we show \( S \) closed implies \( {S}^{c} \) open. Let \( p \in {S}^{c} \) be given. Suppose not, and for each \( r > 0 \) and \( {M}_{r}\left( p\right) \) fails to be in \( {S}^{c} \) . Then for each \( n \), by taking \( r = \frac{1}{n} \) there exists \( {x}_{n} ... | No |
Proposition 5.14. For any metric space \( M \), then\n\n(i) \( \varnothing, M \in \mathcal{T} \).\n\n(ii) Any union of members of \( \mathcal{T} \) belongs to \( \mathcal{T} \), even infinitely many, countable or uncountable.\n\n(iii) Any finite intersection of open sets of \( \mathcal{T} \) is open. | Proof. The first two are trivial. The third fact follows from the fact that the intersection of two open sets is open - if \( p \in U \cap V \), where \( U \) and \( V \) are open, then \( \exists r, s > 0 \) for which \( {M}_{r}\left( p\right) \in U \) and \( {M}_{s}\left( p\right) \in V \), whence \( {M}_{\min \{ r, ... | No |
Let \( S = \{ x \in \mathbb{Q} : x \leq 0\} .S \) is open in \( \mathbb{Q} \) | because if a sequence in \( S \) converges to a limit in the rational numbers converges to something in \( S \) . The fact that there is a sequence converging to irrational numbers is irrelevant because our metric space is \( \mathbb{Q} \) | No |
Theorem 5.17. Every open set \( S \in \mathbb{R} \) is a countable disjoint union of open intervals (including rays). | Proof. Clearly \( S = \varnothing \) \ | No |
Theorem 6.3. \( f : M \rightarrow N \) is continuous if and only if for every open set \( V \subseteq N \), the pre-image of \( V \) is open in \( M \) . | Proof. The easy part is to prove that the open set condition implies continuity. We choose \( V \) to be \( {N}_{\varepsilon }\left( {fp}\right) \) ; the pre-image is some open set \( U \subseteq M \) containing \( p \) . Because \( U \) is open, there is some neighborhood of \( p \) contained in \( U \) ; that gets ma... | Yes |
Corollary 6.5. The open set condition is equivalent to the closed set condition | Proof. Take complements. Use the fact that \( {f}^{\text{pre }}\left( {N \smallsetminus K}\right) = M \smallsetminus {f}^{\text{pre }}\left( K\right) \) . | No |
Proposition 6.7. \( \lim S \) is the smallest closed set containing \( S \), in the sense that if \( S \subseteq K \subseteq M \) and \( K \) is closed, then \( \lim S \subseteq K \) . | Proof. \( S \subseteq K \Rightarrow \lim S \subseteq \lim K = K \) . | Yes |
Proposition 6.8. For any \( S \) , \[ \bar{S} = \mathop{\bigcap }\limits_{{S \subseteq K, K\text{ closed }}}K. \] | Proof. Obvious. | No |
Show that the intersection of infinitely many open sets need not be open. | \[ \mathop{\bigcap }\limits_{{n \geq 1}}\left( {\frac{1}{2} - \frac{1}{{2}^{n}},\frac{1}{2} + \frac{1}{{2}^{n}}}\right) = \left\{ \frac{1}{2}\right\} \] | Yes |
Theorem 7.4 (Inheritance Theorem). Let \( K \subset N \subset M \), where \( M \) and \( N \) are subspaces. Suppose \( K \) is closed in \( N \) if and only if there exists a closed \( L \) in \( M \) such that \( K = L \cap N \) . | Proof. Suppose that \( L \) is closed in \( M \), and consider \( K = L \cap N \) . Consider any sequence \( \left( {x}_{n}\right) \) in \( K \) which converges in \( x \in N \) . Then \( {x}_{n} \rightarrow x \) in \( M \) as well, so \( x \in L \) by closure of \( L \) ; therefore \( x \in L \cap N = K \) .\n\nConver... | Yes |
Corollary 7.5. Open sets in \( N \) are inherited from \( M \) ; that is, a set \( V \) is open in \( N \) if and only if there exists \( U \) open in \( M \) such that \( V = U \cap N \) . | Proof. Take complements. Let \( K = N \smallsetminus V \), and consider closed \( L \) such that \( K = L \cap N \) . Then \( U = M \smallsetminus L \) ; note\n\n\[ M = L \sqcup U\text{.} \]\n\nThen \( N = \left( {L \cap N}\right) \sqcup \left( {U \cap N}\right) \) . Because \( L \cap N \) is closed, we deduce \( U \ca... | Yes |
Proposition 7.7. Let \( K \subset N \subset M \) where \( N \) and \( M \) are metric spaces. Suppose further than \( N \) is closed in \( M \) . Then \( K \) is closed in \( N \) if and only if \( K \) is closed in \( M \). | Proof. Direct corollary of previous theorem. | No |
Proposition 7.11. \( \left( {{x}_{n},{y}_{n}}\right) \rightarrow \left( {x, y}\right) \) in \( M \times N \) equipped with \( {d}_{\max },{d}_{E},{d}_{\text{sum }} \) if and only if \( {x}_{n} \rightarrow x \) and \( {y}_{n} \rightarrow y \) . | Proof. Use \( {d}_{\max } \) and remark that \( \max \left\{ {d\left( {{x}_{n}, x}\right), d\left( {{y}_{n}, y}\right) }\right\} \rightarrow 0 \) if and only if \( d\left( {{x}_{n}, x}\right) \rightarrow 0 \) and \( d\left( {{y}_{n}, y}\right) \rightarrow 0 \) . | Yes |
Proposition 7.12. The map \( + : {\mathbb{R}}^{2} \rightarrow \mathbb{R} \) by \( \left( {x, y}\right) \mapsto x + y \) is continuous. So are -, \( \times \), and \( \div \), although the domain of \( \div \) is \( \mathbb{R} - \{ 0\} \) . | Proof. Many epsilons and deltas appeared. Remark that we can pick any of the three \( d \) ’s; let’s use \( {d}_{\max } \) . + and - are trivial.\n\nHow about \( \times \) ? Suppose \( \left( {{x}_{0},{y}_{0}}\right) \in {\mathbb{R}}^{2} \) and \( \varepsilon > 0 \) we wish to find \( \delta \) such that \( \left| {x -... | Yes |
Lemma 7.13. The product of continuous function is continuous. | Proof. Take \( M \rightarrow {\mathbb{R}}^{2} \rightarrow \mathbb{R} \) . Take \( x \mapsto \left( {f\left( x\right), g\left( x\right) }\right) \mapsto f\left( x\right) g\left( x\right) \) . Each component is continuous, and the composition of continuous functions is also continuous. | Yes |
Corollary 8.5. \( {\mathbb{R}}^{m} \) is complete for each \( m \) . | Proof. Induct. | No |
Proposition 8.7. If \( S \) is closed in \( M \) and \( M \) is complete, then \( S \) is complete. | Proof. If \( \left( {x}_{n}\right) \) is Cauchy in \( S \), it is Cauchy in \( M \), so \( {x}_{n} \rightarrow x \in M \) . Because \( x \) is closed, \( x \in S \) . | Yes |
Proposition 8.10. If \( \left( {x}_{n}\right) \) is Cauchy, then \( \left\{ {{x}_{n} : n \in \mathbb{N}}\right\} \) is bounded. | Proof. Totally obvious! Take \( \varepsilon = {1000} \), and then note that there are finitely many points not within a distance of 2013 of a giving sufficiently large point. Here it is written out. IF \( d\left( {{x}_{n},{x}_{m}}\right) < {1000} \) for all \( m, n \geq N \) - and such an \( N \) exists by the Cauchy c... | No |
Theorem 8.14. Every compact is closed and bounded. | Proof. Let \( \left( {x}_{n}\right) \) be a sequence in \( A \) that converges ta \( p \in M \) . Now \( \left( {x}_{n}\right) \) has a subsequence such that \( {x}_{{n}_{k}} \rightarrow x \in A \) . But we must have \( p = x \) . Thus \( p \in A \) . Consequently \( \left( {x}_{n}\right) \) is closed.\n\nNow suppose f... | Yes |
Theorem 8.17. \( \left\lbrack {a, b}\right\rbrack \subset \mathbb{R} \) is compact. | Proof. Let \( {\left( {x}_{n}\right) }_{n \in \mathbb{N}} \) be a sequence. Consider\n\n\[ S = \left\{ {c \in \left\lbrack {a, b}\right\rbrack : {x}_{n} \geq c\text{ for infinitely many }n}\right\} .\n\]\n\nObviously \( a \in S \) bounded by \( b \) . So let \( x = \sup S \) .\n\nBy our selection of \( x \), we find \(... | Yes |
Proposition 8.18. Let \( A \subset M \) and \( B \subset N \) . If \( A \) and \( B \) are compact, then so is \( A \times B \) as a subset of \( M \times N \) . | Proof. Let \( \left( {{x}_{n},{y}_{n}}\right) \) be a sequence in \( A \times B \) . There is a subsequence \( {x}_{{n}_{k}} \rightarrow x \in A \) . Now \( \left( {y}_{{n}_{k}}\right) \) is a sequence in \( B \), so there is a subsequence \( {y}_{{n}_{{k}_{\ell }}} \rightarrow y \in B \) . | No |
Theorem 8.19. Let \( S \subset A \) be a closed subset of \( A \subset M \) . If \( A \) is compact then \( S \) is compact. | Proof. Repeat the proof of Cauchy-ness. | No |
Theorem 9.2 (Bolzano-Weierstrass Theorem). Every bounded sequence \( \left( {p}_{n}\right) \) in \( {\mathbb{R}}^{m} \) has a convergent subsequence. | Proof. There exists a box which contains \( \left( {p}_{n}\right) \) . The box is compact. | No |
Theorem 9.4. If \( \left( {A}_{n}\right) \) is nested decreasing, and each \( {A}_{n} \) is compact, then \( A = \bigcap {A}_{n} \) is compact. | Proof. This is a closed subset of the compact \( {A}_{1} \) . | No |
Proposition 9.5. If each \( {A}_{n} \) is nonempty, then \( A = \bigcap {A}_{n} \) is nonempty. | Proof. Choose an arbitrary \( {a}_{n} \in {A}_{n} \) for each \( n \) . Then \( \left( {a}_{n}\right) \) is a sequence in \( {A}_{1} \), so \( \left( {a}_{n}\right) \) has a convergent subsequence converging to some point in \( {A}_{1} \) .\n\nConsider a sequence \( {\left( {a}_{{n}_{k}}\right) }_{k} \) converging to s... | No |
Theorem 9.6. If \( {A}_{n} \) is a nested decreasing nonempty compact and the diameter 4 approaches zero, then \( \bigcap {A}_{n} \) is a singleton. | Proof. By the previous proposition, then \( A \) is nonempty. Finally remark that any set of more than two points has positive diameter. Because the diameters decrease, we find \( \operatorname{diam}\left( A\right) \leq \operatorname{diam}\left( {A}_{n}\right) \) which forces \( \operatorname{diam}{A}_{n} \rightarrow 0... | No |
Proposition 9.7. Let \( f : M \rightarrow N \) be continuous and \( A \subset M \) compact. Then \( f\left( A\right) \) is compact. | Proof. Let \( \left( {y}_{n}\right) \) be any sequence in \( B \) . For each \( n \) consider any point \( {x}_{n} \in A \) such that \( f\left( {x}_{n}\right) \in A \) ; because \( B = f\left( A\right) \) at least one such point exists. Then \( {x}_{{n}_{k}} \rightarrow p \in A \) for some \( {n}_{k} \) . By continuit... | No |
Proposition 9.8. Let \( f : M \rightarrow N \) be a continuous bijection. If \( M \) is compact, then \( {f}^{-1} \) is continuous, and \( f \) is a homeomorphism. | Proof. Suppose on the contrary that fore some \( \left( {y}_{n}\right) \in N,{y}_{n} \rightarrow y \in N \) but \( {f}^{-1}\left( {y}_{n}\right) \) does not converge to \( {f}^{-1}\left( y\right) \) . Let \( {x}_{n} = {f}^{-1}\left( {y}_{n}\right) \) for each \( n \) and \( x = {f}^{-1}\left( y\right) \) . Evidently th... | Yes |
Now we prove our theorem from last time. Again we use sequences. | Proof. Suppose not. Then there exists \( f : M \rightarrow N \) continuous, with \( M \) compact, but \( f \) is not uniformly continuous. That means we can find an \( \varepsilon > 0 \) such that for each \( \delta > 0 \) , we can find \( x, y \in M \) such that \( {d}_{M}\left( {x, y}\right) < \delta \) but \( {d}_{N... | Yes |
Proposition 10.6. If \( f : M \rightarrow N \) is continuous, and \( M \) is connected, then \( f\left( M\right) \) is connected as well. | Proof. We show that if \( f\left( M\right) \) is disconnected as \( A \sqcup B \), then \( M \) is disconnected. Note that \( A \) and \( B \) are clopen in \( f\left( M\right) \) . Now \( {f}^{\text{pre }}\left( A\right) \) is clopen, as is \( {f}^{\text{pre }}\left( B\right) \) . Yet they are disjoint. Hence \( M \) ... | Yes |
Consider\n\n\[ S = \left\{ {x, y \in {\mathbb{R}}^{2} \mid 0 < x \leq 1, y = \sin {x}^{-1}}\right\} . \]\n\nThis is a connected set. What is its closure? | It turns out that \( \bar{S} - S \) is \( \{ 0\} \times \left\lbrack {-1,1}\right\rbrack \) , and so counterintuitively, the set\n\n\[ T = S \cup \{ 0\} \times \left\lbrack {-1,1}\right\rbrack \]\n\nis connected! | Yes |
Proposition 10.17. Let \( S = \mathop{\bigcup }\limits_{\alpha }{S}_{\alpha } \), where each \( {S}_{\alpha } \subseteq M \) and \( \exists p \in S \) such that \( p \in {S}_{\alpha } \) for each \( \alpha \) . Then \( S \) is connected. | Proof. If not, suppose \( S = A \sqcup B \) and assume without loss of generality that \( p \in A \) . Observe\n\n\[ \left( {A \cap {S}_{\alpha }}\right) \sqcup \left( {B \cap {S}_{\alpha }}\right) = {S}_{\alpha } \]\n\nIt is not hard to check that these are both clopen in \( {S}_{\alpha } \) because \( A \) and \( B \... | Yes |
Proposition 10.19. Let \( U \) be open in \( {\mathbb{R}}^{m} \) . If \( U \) is connected, then \( U \) is path-connected. | Proof. Pick \( p \in U \) . Consider\n\n\[ \nV = \{ q \in U : \exists \text{ path in }U\text{ from }p\text{ to }q\} . \n\]\n\nWe claim \( V \) is clopen in \( U \) . The fact that \( V \) is open follows from the fact that \( U \) is open. For closedness, assume \( q \) is a limit of \( V \) . Again by openness, we can... | Yes |
Consider \( B = \left\{ {\frac{1}{n} \mid n = 1,2,\ldots }\right\} \) living in \( \mathbb{R} \), and let \( A = B \cup \{ 0\} \) . Show that \( A \) is open covering compact. | Proof. Some open set \( \left( {u, v}\right) \in \mathcal{U} \) covers 0 . Only finitely many elements of \( A \) now exist outside \( \left( {-u, v}\right) \), and we can cover the elements of \( A \) exceeding \( v \) easily.\n\nOn the other hand, one can construct open intervals around each element of \( B \) which ... | No |
Theorem 11.4. A is open covering compact if and only if it is sequentially compact. | Proof that covering compact implies sequentially compact. Assume \( \left( {a}_{n}\right) \) is a sequence in \( A \), yet no subsequence converges. That implies that for each \( p \in A \), some neighborhood of \( p \) contains only finitely many points of \( \left( {a}_{n}\right) \) ; that is,\n\n\[ \forall p \in A\e... | No |
Lemma 11.8 (Lebesgue Number Lemma). If \( A \) is sequentially compact set, then all open coverings of \( A \) have a Lebesgue number. | Proof. Suppose on the contrary that for every \( \lambda > 0 \), there exists a \( p \in A \) such that \( {M}_{\lambda }\left( p\right) \) is not contained in any scrap. Then by taking \( \lambda = \frac{1}{n} \) for \( n = 1,2,\ldots \), we can find a point \( {a}_{n} \in A \) such that \( {M}_{\frac{1}{n}} \) is not... | Yes |
Theorem 11.10. Suppose that \( M \) is complete. Then \( A \) is closed and totally bounded if and only if \( A \) is compact. | Proof. First suppose \( A \) is compact. Then \( A \) is closed, and for any \( r > 0 \), the open covering \( \mathcal{U} = \left\{ {{M}_{r}\left( p\right) \mid p \in A}\right\} \) has a finite subcover as desired. | No |
Theorem 12.3. Let \( M \) be a perfect, nonempty complete metric space. Then \( M \) is uncountable. | Proof. Assume not. Evidently \( M \) is denumerable, so let \( M = \left\{ {{x}_{1},{x}_{2},{x}_{3},\ldots }\right\} \) . Define \[ {\widehat{M}}_{r}\left( p\right) = \{ x \in M \mid d\left( {x, p}\right) \leq r\} \] which we colloquially call a closed neighborhood. Choose any point \( {y}_{1} \in M - \left\{ {x}_{1}\r... | Yes |
Theorem 12.6. There exists a subset \( C \subset \mathbb{R} \) which is nonempty, compact, perfect, and totally disconnected. | Proof. We check this.\n\n(a) \( 0 \in C \), so \( C \) is nonempty.\n\n(b) \( C \) is closed (as it is the intersection of closed sets)\n\n(c) We wish to show that for any \( p \in C, r > 0 \), we have \( \left( {p - r, p + r}\right) \cap C \) is infinite. Just select \( n \) such that \( {3}^{-n} < \varepsilon \) . Si... | Yes |
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