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Example 1.10.6 Let \( {T}_{0},{T}_{1},{T}_{2},\ldots \) be well-founded trees on \( \mathbb{N} \) . Then\n\n\[ \mathop{\bigvee }\limits_{n}{T}_{n} = \{ e\} \bigcup \left\{ {i \hat{} s : s \in {T}_{i}, i \in \mathbb{N}}\right\} \]\n\n is a well-founded tree. (See Figure 1.5.) | Null | No |
Proposition 1.10.7 (König’s infinity lemma, [57]) Let \( T \) be a finitely splitting, infinite tree on \( A \) . Then \( T \) is ill-founded. | Proof. Let \( T \) be a finitely splitting, infinite tree on \( A \) . Let \( \left( {a}_{0}\right) \) be a node of \( T \) with infinitely many extensions in \( T \) . Since \( T \) is finitely splitting (and \( e \in T),\{ a \in A : \left( a\right) \in T\} \) is finite. Further, \( T \) is infinite. So, \( \left( {a}... | Yes |
Proposition 1.10.8 Let \( T \) be a tree on a finite set \( A \) . Then\n\n\[ \left\lbrack T\right\rbrack \neq \varnothing \Leftrightarrow \left( {\forall k \in \mathbb{N}}\right) \left( {\exists u \in T}\right) \left( {\left| u\right| = k}\right) . \] | Proof. \ | No |
Proposition 1.10.10 A tree \( T \) on a well-ordered set \( A \) is well-founded if and only if \( { \leq }_{KB} \) is a well-order on \( T \) . | Proof. Let \( T \) be ill-founded. Take any \( \alpha \) in \( \left\lbrack T\right\rbrack \) . Then \( \left( {\alpha \mid k}\right) \) is a descending sequence in \( \left( {T,{ \leq }_{KB}}\right) \) . This proves the \ | No |
Proposition 1.11.1 (Proof by induction on well-founded trees) Let \( T \) be a well-founded tree and for \( u \in T \), let \( {P}_{u} \) be a mathematical proposition. Then\n\n\[ \left( {\forall u \in T}\right) \left( {\left( {\left( {\forall v \in {T}_{u}\smallsetminus \{ e\} }\right) {P}_{u \land v}}\right) \Rightar... | Proof. Suppose there is a node \( u \) of \( T \) such that \( {P}_{u} \) does not hold. Take an extension \( w \) of \( u \) in \( T \) such that \( {P}_{w} \) does not hold and if \( v \succ w \) and \( v \in T \) then \( {P}_{v} \) holds. Since \( T \) is well-founded, such a \( w \) exists. Thus for every extension... | Yes |
Proposition 1.11.2 (Definition by induction on well-founded trees) Let \( T \) be a well-founded tree on a set \( A, X \) a set, and \( \mathcal{F} \) the set of all maps with domain \( {T}_{u} \smallsetminus \{ e\} \) and range contained in \( X \), where \( u \) varies over \( T \) . Given any map \( G : \mathcal{F} ... | Proof. (Existence) For \( u \in T \), let \( {P}_{u} \) be the proposition \ | No |
Example 1.11.4 \( \rho \left( T\right) = \left| s\right| \) if \( T = \{ s\left| {i : i < }\right| s \mid \}, s \in {\mathbb{N}}^{ < \mathbb{N}} \) . | Null | No |
Proposition 1.12.1 Let \( \\left\\{ {{A}_{s} : s \\in {\\mathbb{N}}^{ < \\mathbb{N}}}\\right\\} \) be a system of subsets of a set \( X \) . Put\n\n\[ \nB = \\mathop{\\bigcap }\\limits_{k}\\mathop{\\bigcup }\\limits_{{\\left| s\\right| = k}}\\left\\lbrack {{A}_{s} \\times \\sum \\left( s\\right) }\\right\\rbrack \n\] \... | The proof of the above proposition is routine and is left as an exercise. | No |
Proposition 1.12.2 For every family \( \mathcal{F} \) of subsets of \( X \) , | Proof. (i) \( \mathcal{F} \subseteq \mathcal{A}\left( \mathcal{F}\right) \) . Let \( A \in \mathcal{F} \) . Take\n\n\[{A}_{s} = A, s \in {\mathbb{N}}^{ < \mathbb{N}}.\]\n\nClearly, \( A = \mathcal{A}\left( \left\{ {A}_{s}\right\} \right) \in \mathcal{A}\left( \mathcal{F}\right) \) .\n\n(ii) \( {\mathcal{F}}_{\sigma } \... | Yes |
Lemma 1.12.3 Let \( \\left\\{ {{A}_{s} : s \\in {A}^{ < \\mathbb{N}}}\\right\\} \) be a system of sets such that \( {A}_{s} \\cap {A}_{t} = \\varnothing \) whenever \( s \\bot t \) . Then | Proof. Let \( x \\in {\\mathcal{A}}_{A}\\left( \\left\\{ {A}_{s}\\right\\} \\right) \) . By the definition of \( {\\mathcal{A}}_{A} \), there is an \( \\alpha \\in {A}^{\\mathbb{N}} \) such that \( x \\in {A}_{\\alpha \\mid n} \) for all \( n \) . So \( x \\in \\mathop{\\bigcap }\\limits_{n}\\mathop{\\bigcup }\\limits_... | Yes |
Proposition 1.12.4 If \( A \) is a finite set and \( \left\{ {{A}_{s} : s \in {A}^{ < \mathbb{N}}}\right\} \) regular, then | \[ {\mathcal{A}}_{A}\left( \left\{ {A}_{s}\right\} \right) = \mathop{\bigcap }\limits_{n}\mathop{\bigcup }\limits_{{\left| s\right| = n}}{A}_{s} \] Proof. We have seen in the proof of 1.12.3 that \[ {\mathcal{A}}_{A}\left( \left\{ {A}_{s}\right\} \right) \subseteq \mathop{\bigcap }\limits_{n}\mathop{\bigcup }\limits_{{... | Yes |
Corollary 1.12.5 Let \( {\left( \mathcal{F}\right) }_{s} = {\left( \mathcal{F}\right) }_{d} = \mathcal{F} \) ; i.e., \( \mathcal{F} \) is closed under finite intersections and finite unions. Then \( {\mathcal{A}}_{2}\left( \mathcal{F}\right) = {\mathcal{F}}_{\delta } \) . In particular, \( {\mathcal{A}}_{2} \) does not... | Null | No |
Corollary 1.13.2 For any family \( \mathcal{F} \) of sets, \( \mathcal{A}\left( \mathcal{F}\right) \) is closed under countable intersections and countable unions. | Null | No |
Example 2.1.1 Let \( X = {\mathbb{R}}^{n}, n \) a positive integer. For \( x = \left( {{x}_{1},\cdots ,{x}_{n}}\right) \) and \( y = \left( {{y}_{1},\cdots ,{y}_{n}}\right) \) in \( {\mathbb{R}}^{n} \), let\n\n\[ \n{d}_{1}\left( {x, y}\right) = \left| {x - y}\right| = \sqrt{\mathop{\sum }\limits_{{i = 1}}^{n}{\left( {x... | Null | No |
Example 2.1.2 Let \( X = {\mathbb{R}}^{\mathbb{N}}, x = \left( {{x}_{0},{x}_{1},\ldots }\right) \) and \( y = \left( {{y}_{0},{y}_{1},\ldots }\right) \) . Define\n\n\[ d\left( {x, y}\right) = \mathop{\sum }\limits_{n}\frac{1}{{2}^{n + 1}}\min \left\{ {\left| {{x}_{n} - {y}_{n}}\right| ,1}\right\} .\n\]\n\nThen \( d \) ... | Null | No |
If \( X \) is any set and\n\n\[ d\left( {x, y}\right) = \left\{ \begin{array}{ll} 0 & \text{ if }x = y \\ 1 & \text{ otherwise } \end{array}\right. \]\n\nthen \( d \) defines a metric on \( X \), called the discrete metric. | Null | No |
Example 2.1.4 Let \( \\left( {{X}_{0},{d}_{0}}\\right) ,\\left( {{X}_{1},{d}_{1}}\\right) ,\\left( {{X}_{2},{d}_{2}}\\right) ,\\ldots \) be metric spaces and \( X = \\mathop{\\prod }\\limits_{n}{X}_{n} \) . Fix \( x = \\left( {{x}_{0},{x}_{1},\\ldots }\\right) \) and \( y = \\left( {{y}_{0},{y}_{1},\\ldots }\\right) \)... | Null | No |
Proposition 2.1.9 A metrizable space is separable if and only if it is second countable. | Null | No |
Proposition 2.1.12 Let \( X \) be a separable metric space and \( \alpha \) an ordinal. Then every nondecreasing family \( \left\{ {{U}_{\beta } : \beta < \alpha }\right\} \) of nonempty open sets is countable. | Proof. Fix a countable base \( \left\{ {V}_{n}\right\} \) for \( X \) . Let \( \beta < \alpha \) be such that \( {U}_{\beta + 1} \smallsetminus \) \( {U}_{\beta } \neq \varnothing \) . Let \( n\left( \beta \right) \) be the first integer \( m \) such that\n\n\[ \n{V}_{m}\bigcap {U}_{\beta }^{c} \neq \varnothing \& {V}_... | No |
Proposition 2.1.18 (Urysohn’s lemma) Suppose \( {A}_{0},\;{A}_{1}\; \) are two nonempty, disjoint closed subsets of a metrizable space \( X \) . Then there is a continuous function \( u : X \rightarrow \left\lbrack {0,1}\right\rbrack \) such that\n\n\[ u\left( x\right) = \left\{ \begin{array}{ll} 0 & \text{ if }x \in {... | Proof. Let \( d \) be a compatible metric on \( X \) . Take\n\n\[ u\left( x\right) = \frac{d\left( {x,{A}_{0}}\right) }{d\left( {x,{A}_{0}}\right) + d\left( {x,{A}_{1}}\right) }.\] | Yes |
Proposition 2.1.19 For every nonempty closed subset \( A \) of a metrizable space \( X \) there is a continuous function \( f : X \rightarrow \left\lbrack {0,1}\right\rbrack \) such that \( A = \) \( {f}^{-1}\left( 0\right) \) . | Proof. Write \( A = \mathop{\bigcap }\limits_{{n = 0}}^{\infty }{U}_{n} \), where the \( {U}_{n} \) ’s are open (2.1.16). By 2.1.18, for each \( n \in \mathbb{N} \), there is a continuous \( {f}_{n} : X \rightarrow \left\lbrack {0,1}\right\rbrack \) such that\n\n\[ \n{f}_{n}\left( x\right) = \left\{ \begin{array}{ll} 0... | Yes |
Proposition 2.1.25 Suppose \( X \) is a metric space and \( f : X \rightarrow \mathbb{R} \) an upper-semicontinuous map such that there is a continuous map \( g : X \rightarrow \mathbb{R} \) such that \( f \leq g \) ; i.e., \( f\left( x\right) \leq g\left( x\right) \) for all \( x \) . Then there is a sequence of conti... | Proof. Let \( r \) be any rational number. Set\n\n\[ \n{U}_{r} = \{ x \in X : f\left( x\right) < r < g\left( x\right) \} .\n\]\n\nSince \( f \) is upper-semicontinuous and \( g \) continuous, \( {U}_{r} \) is open. Let \( \left( {F}_{n}^{r}\right) \) be a sequence of closed sets such that \( {U}_{r} = \mathop{\bigcup }... | Yes |
Proposition 2.1.27 The product of countably many second countable (equivalently separable) metric spaces is second countable. | Proof. Let \( {X}_{0},{X}_{1},\ldots \) be second countable. Let \( X = \mathop{\prod }\limits_{i}{X}_{i} \) . We show that \( X \) has a countable subbase. The result then follows from 2.1.8. Let \( \left\{ {{U}_{in} : n \in \mathbb{N}}\right\} \) be a base for \( {X}_{i} \) . Then, by the definition of the product to... | Yes |
Proposition 2.1.29 (Cantor intersection theorem) A metric space \( \\left( {X, d}\\right) \) is complete if and only if for every decreasing sequence \( {F}_{0} \\supseteq {F}_{1} \\supseteq {F}_{2} \\subseteq \\cdots \) of nonempty closed subsets of \( X \) with diameter \( \\left( {F}_{n}\\right) \\rightarrow 0 \), t... | Proof. Assume that \( \\left( {X, d}\\right) \) is complete. Let \( \\left( {F}_{n}\\right) \) be a decreasing sequence of nonempty closed sets with diameter converging to 0 . Choose \( {x}_{n} \\in {F}_{n} \) . Since \( \\operatorname{diameter}\\left( {F}_{n}\\right) \\rightarrow 0,\\left( {x}_{n}\\right) \) is Cauchy... | Yes |
Proposition 2.1.31 Let \( \left( {{X}_{0},{d}_{0}}\right) ,\left( {{X}_{1},{d}_{1}}\right) ,\left( {{X}_{2},{d}_{2}}\right) ,\ldots \) be complete metric spaces, \( X = \mathop{\prod }\limits_{n}{X}_{n} \), and \( d \) the product metric on \( X \) . Then \( \left( {X, d}\right) \) is complete. | Proof. Let \( {\alpha }_{0},{\alpha }_{1},{\alpha }_{2},\ldots \) be a Cauchy sequence in \( X \) . Then for each \( k \) , \( {\alpha }_{0}\left( k\right) ,{\alpha }_{1}\left( k\right) ,{\alpha }_{2}\left( k\right) ,\ldots \) is a Cauchy sequence in \( {X}_{k} \) . As \( {X}_{k} \) is complete, we get an \( \alpha \le... | Yes |
Theorem 2.1.32 Any second countable metrizable space \( X \) can be embedded in the Hilbert cube \( \mathbb{H} \) . | Proof. Let \( \left( {U}_{n}\right) \) be a countable base for \( X \) . For each pair of integers \( n \) , \( m \) with \( \operatorname{cl}\left( {U}_{n}\right) \subseteq {U}_{m} \), choose a continuous \( {f}_{nm} : X \rightarrow \left\lbrack {0,1}\right\rbrack \) such that\n\n\[ \n{f}_{nm}\left( x\right) = \left\{... | Yes |
Proposition 2.1.34 Every nonempty open set \( U \) in \( \mathbb{R} \) is a countable union of pairwise disjoint nonempty open intervals. | Proof. Let \( x \in U \) and let \( {I}_{x} \) be the union of all open intervals containing \( x \) and contained in \( U \) . Clearly, for any \( x, y \), either \( {I}_{x} = {I}_{y} \) or \( {I}_{x} \cap {I}_{y} = \varnothing \) . Since \( \mathbb{R} \) is separable, \( \left\{ {{I}_{x} : x \in U}\right\} \) is coun... | Yes |
Proposition 2.1.35 (Sierpiński) The open unit interval \( \\left( {0,1}\\right) \) cannot be expressed as a countable disjoint union of nonempty closed subsets of \( \\mathbb{R} \) . | Proof. Let \( {A}_{0},{A}_{1},{A}_{2},\\ldots \) be a sequence of pairwise disjoint nonempty closed sets in \( \\mathbb{R} \), each contained in \( \\left( {0,1}\\right) \). We show that \( \\bigcup {A}_{i} \\neq \\left( {0,1}\\right) \). Suppose \( \\bigcup {A}_{i} = \\left( {0,1}\\right) \). Then for \( k \\in \\math... | Yes |
Theorem 2.2.1 (Alexandrov) Every \( {G}_{\delta } \) subset \( G \) of a completely metrizable (Polish) space \( X \) is completely metrizable (Polish). | Proof. Fix a complete metric \( d \) on \( X \) compatible with its topology. We first prove the result when \( G \) is open. Consider the function \( f : G \rightarrow X \times \mathbb{R} \) defined by\n\n\[ f\left( x\right) = \left( {x,\frac{1}{d\left( {x, X \smallsetminus G}\right) }}\right) ,\;x \in G. \]\n\nNote t... | Yes |
Proposition 2.2.3 Let \( f : A \rightarrow Z \) be a continuous map from a subset \( A \) of a metrizable space \( W \) to a completely metrizable space \( Z \) . Then \( f \) can be extended continuously to a \( {G}_{\delta } \) set containing \( A \) . | Proof. Take a bounded complete metric \( \rho \) on \( Z \) compatible with its topology. For any \( x \in \operatorname{cl}\left( A\right) \), let\n\n\[ \n{O}_{f}\left( x\right) = \inf \{ \operatorname{diameter}\left( {f\left( {A \cap V}\right) }\right) : V\text{ open,}x \in V\} .\n\]\n\nWe call \( {O}_{f}\left( x\rig... | Yes |
Theorem 2.2.6 (Lavrentiev) Let \( X, Y \) be completely metrizable spaces, \( A \subseteq X, B \subseteq Y \), and \( f : A \rightarrow B \) a homeomorphism onto \( B \) . Then \( f \) can be extended to a homeomorphism between two \( {G}_{\delta } \) sets containing \( A \) and \( B \) . | Proof. Let \( g = {f}^{-1} \) . By 2.2.3, choose a \( {G}_{\delta } \) set \( {A}^{\prime } \supseteq A \) and a contiunuous extension \( {f}^{\prime } : {A}^{\prime } \rightarrow Y \) of \( f \) . Similarly, choose a \( {G}_{\delta } \) set \( {B}^{\prime } \supseteq B \) and a continuous extension \( {g}^{\prime } : ... | Yes |
Theorem 2.2.7 Let \( X \) be a completely metrizable space and \( Y \) a completely metrizable subspace. Then \( Y \) is a \( {G}_{\delta } \) set in \( X \) . | Proof. The result follows from 2.2.6 by taking \( A = B = Y \) and \( f : A \rightarrow \) \( B \) the identity map. | Yes |
Lemma 2.2.9 Every second countable, zero-dimensional metrizable space \( X \) can be embedded in \( \mathcal{C} \) . | Proof. Fix a countable base \( \left\{ {{U}_{n} : n \in \mathbb{N}}\right\} \) for \( X \) such that each \( {U}_{n} \) is clopen. Define \( f : X \rightarrow \mathcal{C} \) by\n\n\[ f\left( x\right) = \left( {{\chi }_{{U}_{0}}\left( x\right) ,{\chi }_{{U}_{1}}\left( x\right) ,{\chi }_{{U}_{2}}\left( x\right) ,\ldots }... | Yes |
Proposition 2.2.11 Every zero-dimensional Polish space is homeomorphic to a \( {G}_{\delta } \) subset of \( \mathcal{C} \) . | The Cantor space is clearly embedded in \( {\mathbb{N}}^{\mathbb{N}} \) . Hence every zero-dimensional Polish space is homeomorphic to a \( {G}_{\delta } \) subset of \( {\mathbb{N}}^{\mathbb{N}} \) . | No |
Proposition 2.2.13 Let \( A \) be any set with the discrete topology. Suppose \( {A}^{\mathbb{N}} \) is equipped with the product toplogy and \( C \) is any subset of \( {A}^{\mathbb{N}} \) . Then \( C \) is closed if and only if it is the body of a tree \( T \) on \( A \) . | Proof. Let \( T \) be a tree on \( A \) . We show that \( {A}^{\mathbb{N}} \smallsetminus \left\lbrack T\right\rbrack \) is open. Let \( \alpha \notin \left\lbrack T\right\rbrack \) . Then there exists a \( k \in \mathbb{N} \) such that \( \alpha \mid k \notin T \) . So, \( \sum \left( {\alpha \mid k}\right) \subseteq ... | Yes |
Define a sequence \( \left( {C}_{n}\right) \) of subsets of \( \left\lbrack {0,1}\right\rbrack \) inductively as follows. Take\n\n\[ \n{C}_{0} = \left\lbrack {0,1}\right\rbrack \n\] \n\nSuppose \( {C}_{n} \) has been defined and is a union of \( {2}^{n} \) pairwise disjoint closed intervals \( \left\{ {{I}_{j} : 1 \leq... | Null | No |
Proposition 2.3.6 A continuous image of a compact space is compact. | Proof. Let \( X \) be compact and \( f : X \rightarrow Y \) continuous. Suppose \( \mathcal{U} \) is an open cover for \( f\left( X\right) \) . Then \( \left\{ {{f}^{-1}\left( U\right) : U \in \mathcal{U}}\right\} \) is a cover of \( X \) . As \( X \) is compact, there is a finite subcover of \( X \), say \( {f}^{-1}\l... | Yes |
Proposition 2.3.10 If \( \left( {X, d}\right) \) is a compact metric space, then every sequence in \( \left( {X, d}\right) \) has a convergent subsequence. | Proof. Suppose \( \left( {X, d}\right) \) is compact but that there is a sequence \( \left( {x}_{n}\right) \) in \( X \) with no convergent subsequence. Then \( \left\{ {{x}_{n} : n \in \mathbb{N}}\right\} \) is a closed and infinite discrete subspace of \( X \) . This contradicts the fact that \( X \) is compact. | Yes |
Proposition 2.3.10 If \( \left( {X, d}\right) \) is a compact metric space, then every sequence in \( \left( {X, d}\right) \) has a convergent subsequence. | Proof. Suppose \( \left( {X, d}\right) \) is compact but that there is a sequence \( \left( {x}_{n}\right) \) in \( X \) with no convergent subsequence. Then \( \left\{ {{x}_{n} : n \in \mathbb{N}}\right\} \) is a closed and infinite discrete subspace of \( X \) . This contradicts the fact that \( X \) is compact. | Yes |
Proposition 2.3.11 Every compact metric space \( \left( {X,\rho }\right) \) is complete. | Proof. By 2.3.10, every Cauchy sequence in \( \left( {X,\rho }\right) \) has a convergent subsequence. So, every Cauchy sequence in \( \left( {X,\rho }\right) \) is convergent. | Yes |
Proposition 2.3.12 Every compact metric space is totally bounded. | Null | No |
Proposition 2.3.14 Every compact metrizable space \( X \) is separable and hence second countable. | Proof. Let \( d \) be a compatible metric on \( X \) . For any \( n > 0 \), choose a \( \frac{1}{n} \) -net \( {A}_{n} \) in \( X \) . Then \( \mathop{\bigcup }\limits_{n}{A}_{n} \) is a countable, dense set in \( X \) . | Yes |
Corollary 2.3.15 Every zero-dimensional compact metrizable space \( X \) is homeomorphic to a closed subset of \( \mathcal{C} \) . | Proof. By 2.3.14, \( X \) is second countable. Therefore, by 2.2.9, there is an embedding \( f \) of \( X \) into \( \mathcal{C} \) . By 2.3.6, the range of \( f \) is compact and therefore closed. | Yes |
Proposition 2.3.16 Let \( \left( {X, d}\right) \) be sequentially compact and \( \mathcal{U} \) an open cover of \( X \) . Then there is a \( \delta > 0 \) such that every \( A \subseteq X \) of diameter less than \( \delta \) is contained in some \( U \in \mathcal{U} \) . | Proof. Suppose such a \( \delta \) does not exist. For every \( n > 0 \), choose \( {A}_{n} \subseteq X \) such that \( \operatorname{diameter}\left( {A}_{n}\right) < \frac{1}{n} \) and \( {A}_{n} \) is not a contained in any \( U \in \mathcal{U} \) .\n\nChoose \( {x}_{n} \in {A}_{n} \) . Since \( X \) is sequentially ... | Yes |
Proposition 2.3.17 Suppose \( \left( {X, d}\right) \) and \( \left( {Y,\rho }\right) \) are metric spaces with \( X \) sequentially compact. Then every continuous \( f : X \rightarrow Y \) is uniformly continuous. | Proof. Fix \( \epsilon > 0 \) . Let\n\n\[ \mathcal{U} = \left\{ {{f}^{-1}\left( B\right) : B\text{ an open ball of radius } < \epsilon /2}\right\} . \]\n\nLet \( \delta \) be a Lebesgue number of \( \mathcal{U} \) . Plainly, \( \rho \left( {f\left( x\right), f\left( y\right) }\right) < \epsilon \) whenever \( d\left( {... | Yes |
Proposition 2.3.18 Every sequentially compact metric space \( \left( {X, d}\right) \) is totally bounded. | Proof. Let \( X \) be not totally bounded. Choose \( \epsilon > 0 \) such that no finite family of open balls of radius \( \epsilon \) cover \( X \) . Then, by induction on \( n \), we can define a sequence \( \left( {x}_{n}\right) \) in \( X \) such that for all \( n > 0,{x}_{n} \notin \mathop{\bigcup }\limits_{{i < n... | Yes |
Proposition 2.3.19 Every sequentially compact metric space is compact. | Proof. Let \( \left( {X, d}\right) \) be sequentially compact and \( \mathcal{U} \) an open cover for \( X \) . Let \( \delta > 0 \) be a Lebesgue number of \( \mathcal{U} \) and \( \left\{ {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\} \) a \( \delta /3 \) - net in \( X \) . For each \( k \leq n \), choose \( {U}_{k} \in \... | Yes |
Theorem 2.3.22 A metric space is compact if and only if it is complete and totally bounded. | Proof. We have already proved the \ | No |
Theorem 2.3.23 The product of a sequence of compact metric spaces is compact. | Proof. Let \( \left( {{X}_{0},{d}_{0}}\right) ,\left( {{X}_{1},{d}_{1}}\right) ,\left( {{X}_{2},{d}_{2}}\right) ,\ldots \) be a sequence of compact metric spaces, \( X = \mathop{\prod }\limits_{n}{X}_{n} \), and \( d \) the product metric on \( X \) . Fix a sequence \( \left( {x}_{n}\right) \) in \( X \) . We show that... | Yes |
Lemma 2.3.28 Let \( X \) be a compact metric space. Suppose \( f,{f}_{n} : X \rightarrow \mathbb{R} \) are upper-semicontinuous and \( {f}_{n} \) decreases pointwise to \( f \) . If \( {x}_{n} \rightarrow x \) in \( X \), then\n\n\[ \mathop{\limsup }\limits_{n}{f}_{n}\left( {x}_{n}\right) \leq f\left( x\right) \] | Proof. Let \( \epsilon > 0 \) . By 2.3.25 and 2.1.25, there is a continuous \( h : \mathbb{R} \rightarrow \mathbb{R} \) such that \( f \leq h \) and \( h\left( x\right) \leq f\left( x\right) + \epsilon \) . Set\n\n\[ {h}_{n} = \max \left( {{f}_{n}, h}\right) ,\;n \in \mathbb{N}. \]\n\nThen \( {h}_{n} \) is upper-semico... | Yes |
Theorem 2.3.30 Every locally compact metrizable space \( X \) is completely metrizable. | Proof. We need a lemma.\n\nLemma 2.3.31 Let \( Y \) be a locally compact dense subspace of a metrizable space \( X \) . Then \( Y \) is open in \( X \) .\n\nAssuming the lemma, the proof is completed as follows. Let \( d \) be a metric on \( X \) inducing its topology and \( \widehat{X} \) the completion of \( \left( {... | Yes |
Lemma 2.3.31 Let \( Y \) be a locally compact dense subspace of a metrizable space \( X \) . Then \( Y \) is open in \( X \) . | The proof of lemma 2.3.31. Fix \( x \in Y \) and choose an open set \( U \) in \( Y \) containing \( x \) such that \( \operatorname{cl}\left( U\right) \cap Y \) is compact, and hence closed in \( X \) . Since\n\n\( U \subseteq \operatorname{cl}\left( U\right) \bigcap Y \), we have \( \operatorname{cl}\left( U\right) \... | Yes |
Theorem 2.4.3 If \( \left( {X, d}\right) \) is a compact metrc space and \( \left( {Y,\rho }\right) \) Polish, then \( C\left( {X, Y}\right) \), equipped with the topology of uniform convergence, is Polish. | Proof. We only need to check that \( C\left( {X, Y}\right) \) is separable. Let \( l, m \) , and \( n \) be positive integers. As \( X \) is compact, there is a \( 1/\mathrm{m} \) -net \( {X}_{m} = \) \( \left\{ {{x}_{1},{x}_{2},\ldots ,{x}_{k}}\right\} \) in \( X \) . As \( Y \) is separable, there is a countable open... | Yes |
Theorem 2.4.4 (Jacobson density theorem) A matrix \( A \in {M}_{n} \) is irreducible if and only if the \( {C}^{ * } \) -algebra generated by \( A \) is the whole of \( {M}_{n} \) . (See [4] for the definition of \( {C}^{ * } - \) algebra.) | Null | No |
Proposition 2.4.6 \( \operatorname{irr}\left( n\right) \) is Polish. | Proof. By 2.2.1 it is sufficient to show that \( \operatorname{irr}\left( n\right) \) is a \( {G}_{\delta } \) set in \( {M}_{n} \) . Towards showing this, fix any irreducible matrix \( {A}_{0} \) . For any matrix \( A \), by 2.4.5 we have\n\n\[ A\text{ is irreducible } \Leftrightarrow \forall m\exists k\left| {{A}_{0}... | Yes |
Proposition 2.4.6 \( \operatorname{irr}\left( n\right) \) is Polish. | Proof. By 2.2.1 it is sufficient to show that \( \operatorname{irr}\left( n\right) \) is a \( {G}_{\delta } \) set in \( {M}_{n} \) . Towards showing this, fix any irreducible matrix \( {A}_{0} \) . For any matrix \( A \), by 2.4.5 we have\n\n\[ A\text{ is irreducible } \Leftrightarrow \forall m\exists k\left| {{A}_{0}... | Yes |
Proposition 2.4.15 If \( \left( {X, d}\right) \) is a complete metric space, so is \( \left( {K\left( X\right) ,{\delta }_{H}}\right) \) . | Proof. Let \( \left( {K}_{n}\right) \) be a Cauchy sequence in \( K\left( X\right) \) . Let\n\n\[ K = \mathop{\bigcap }\limits_{n}\operatorname{cl}\left( {\mathop{\bigcup }\limits_{{i \geq n}}{K}_{i}}\right) \]\n\nBy Observation \( 1,\operatorname{cl}\left( {\mathop{\bigcup }\limits_{{i > n}}{K}_{i}}\right) \) are comp... | Yes |
Proposition 2.4.17 If \( X \) is compact metrizable, so is \( K\left( X\right) \) . | Proof. Let \( d \) be a compatible metric on \( X \) . By 2.4.15, \( \left( {K\left( X\right) ,{\delta }_{H}}\right) \) is completely metrizable. By Observation 2, it is also totally bounded. The result follows. | Yes |
Proposition 2.4.17 If \( X \) is compact metrizable, so is \( K\left( X\right) \) . | Proof. Let \( d \) be a compatible metric on \( X \) . By 2.4.15, \( \left( {K\left( X\right) ,{\delta }_{H}}\right) \) is completely metrizable. By Observation 2, it is also totally bounded. The result follows. | Yes |
Proposition 2.5.4 Let \( X \) be a topological space, \( U \) open in \( X \), and \( A \subseteq U \) . Then \( A \) is meager in \( U \) if and only if it is meager in \( X \) . | Proof. For the \ | No |
Theorem 2.5.5 (The Baire category theorem) Let \( X \) be a completely metrizable space. Then the intersection of countably many dense open sets in \( X \) is dense. | Proof. Fix a compatible complete metric \( d \) on \( X \) . Take any sequence \( \left( {U}_{n}\right) \) of dense open sets in \( X \) . Let \( \mathrm{V} \) be a nonempty open set in \( X \) . We show that \( \mathop{\bigcap }\limits_{n}{U}_{n} \cap V \neq \varnothing \) . Since \( {U}_{0} \) is dense, \( {U}_{0} \c... | Yes |
Corollary 2.5.6 Every completely metrizable space is of second category in itself. | Proof. Let \( X \) be a completely metrizable space. Suppose \( X \) is of the first category in itself. Choose a sequence \( \left( {F}_{n}\right) \) of closed and nowhere dense sets such that \( X = \mathop{\bigcup }\limits_{n}{F}_{n} \) . Then the sets \( {U}_{n} = X \smallsetminus {F}_{n} \) are dense and open, and... | Yes |
Corollary 2.5.7 The set of rationals \( \mathbb{Q} \) with the usual topology is not completely metrizable. More generally, no countable dense-in-itself space is completely metrizable. | Null | No |
Corollary 2.5.9 Let \( \\left( {G, \\cdot }\\right) \) be a Polish group. Then \( G \) is locally compact if and only if it is a \( {K}_{\\sigma } \) set. | Proof. Let \( G \) be a Polish space that is a \( {K}_{\\sigma } \) set. Choose a sequence \( \\left( {K}_{n}\\right) \) of compact subsets of \( G \) such that \( G = \\mathop{\\bigcup }\\limits_{n}{K}_{n} \). By the Baire category theorem, \( \\operatorname{int}\\left( {K}_{n}\\right) \\neq \\varnothing \) for some \... | Yes |
Corollary 2.5.9 Let \( \\left( {G, \\cdot }\\right) \) be a Polish group. Then \( G \) is locally compact if and only if it is a \( {K}_{\\sigma } \) set. | Proof. Let \( G \) be a Polish space that is a \( {K}_{\\sigma } \) set. Choose a sequence \( \\left( {K}_{n}\\right) \) of compact subsets of \( G \) such that \( G = \\mathop{\\bigcup }\\limits_{n}{K}_{n} \) . By the Baire category theorem, \( \\operatorname{int}\\left( {K}_{n}\\right) \\neq \\varnothing \) for some ... | Yes |
Corollary 2.5.10 Let \( \left( {G, \cdot }\right) \) be a completely metrizable group and \( H \) any subgroup. Then \( H \) is completely metrizable if and only if it is closed in \( G \) . | Proof. Let \( H \) be completely metrizable. Consider \( {G}^{\prime } = \operatorname{cl}\left( H\right) \) . By 2.4.8, \( {G}^{\prime } \) is a topological group. It is clearly completely metrizable. We show that \( {G}^{\prime } = H \), which will complete the proof. By 2.2.7, \( H \) is a \( {G}_{\delta } \) set in... | Yes |
Theorem 2.5.16 (The Banach category theorem) Let \( X \) be a topological space, \( \mathcal{U} = \left\{ {{U}_{i} : i \in I}\right\} \), and \( U = \bigcup \left\{ {{U}_{i} : i \in I}\right\} \). Assume that each \( {U}_{i} \) is open in \( U \). (i) If each \( {U}_{i} \) is nowhere dense in \( X \), so is \( U \). (i... | Proof. Assertion (i) immediately follows from the following lemma.\n\nLemma 2.5.17 Let \( X,{U}_{i}\left( {i \in I}\right) \), and \( U \) satisfy the hypothesis of the theorem. Then\n\n\[ \operatorname{cl}\left( {\operatorname{int}\left( {\operatorname{cl}\left( U\right) }\right) }\right) = \operatorname{cl}\left( {\m... | Yes |
Lemma 2.5.17 Let \( X,{U}_{i}\left( {i \in I}\right) \), and \( U \) satisfy the hypothesis of the theorem. Then \[ \operatorname{cl}\left( {\operatorname{int}\left( {\operatorname{cl}\left( U\right) }\right) }\right) = \operatorname{cl}\left( {\mathop{\bigcup }\limits_{i}\operatorname{int}\left( {\operatorname{cl}\lef... | Proof of the lemma. Since \( {U}_{i} \subseteq U,\left( {i \in I}\right) \), \[ \operatorname{int}\left( {\operatorname{cl}\left( {U}_{i}\right) }\right) \subseteq \operatorname{int}\left( {\operatorname{cl}\left( U\right) }\right) . \] Therefore, \[ \operatorname{cl}\left( {\operatorname{int}\left( {\operatorname{cl}\... | Yes |
Proposition 2.6.1 Every dense-in-itself Polish space \( X \) contains a homeomorph of \( \mathcal{C} \) . | Proof. Let \( d \leq 1 \) be a compatible complete metric on \( X \) . We show that there is a Souslin scheme \( \left\{ {{U}_{s} : s \in {2}^{ < \mathbb{N}}}\right\} \) of nonempty open sets such that\n\n\[ s \bot t \Rightarrow \operatorname{cl}\left( {U}_{s}\right) \bigcap \operatorname{cl}\left( {U}_{t}\right) = \va... | Yes |
Proposition 2.6.2 (Cantor - Bendixson theorem) Every separable metric space \( X \) can be written as \( X = Y\bigcup Z \) where \( Z \) is countable, \( Y \) closed with no isolated point, and \( Y \cap Z = \varnothing \) . | Proof. Let \( \left( {U}_{n}\right) \) be a countable base for \( X \) . Take\n\n\[ Z = \bigcup \left\{ {{U}_{n} : {U}_{n}\text{ countable }}\right\} \]\n\nand \( Y = X \smallsetminus Z \) . | No |
Theorem 2.6.3 Every uncountable Polish space contains a homeomorph of \( \mathcal{C} \), and hence is of cardinality \( \mathfrak{c} \) . | Null | No |
Theorem 2.6.7 Let \( E \) be a closed equivalence relation on a Polish space \( X \) with uncountably many equivalence classes. Then there is a homeomorph \( D \) of the Cantor set in \( X \) consisting of pairwise inequivalent elements. In particular, there are exactly \( \mathfrak{c} \) equivalence classes. | Proof. Fix a compatible complete metric \( d \leq 1 \) on \( X \) and a countable base \( \left( {V}_{n}\right) \) for \( X \) . Let\n\n\[ Z = \bigcup \left\{ {{V}_{n} : E \mid {V}_{n}}\right. \text{has countably many equivalence classes}\} \text{,} \]\n\nand \( Y = X \smallsetminus Z \) . Note that every nonempty open... | Yes |
Theorem 2.6.9 Every Polish space \( X \) is a one-to-one, continuous image of a closed subset \( D \) of \( {\mathbb{N}}^{\mathbb{N}} \) . | Proof. Fix a complete metric \( d \leq 1 \) on \( X \) compatible with its topology. It is enough to define a Lusin scheme \( \left\{ {{F}_{s} : s \in {\mathbb{N}}^{ < \mathbb{N}}}\right\} \) on \( X \) such that\n\n\[ \n{F}_{e} = X\& {F}_{s} = \mathop{\bigcup }\limits_{i}{F}_{s \cap i}.\n\]\n\nWe construct such a fami... | No |
Theorem 2.6.10 Every compact metric space \( X \) is a continuous image of a zero-dimensional compact metric space \( Z \) . | Proof. Fix a metric \( d \leq 1 \) on \( X \) compatible with its topology. We define a sequence \( \left( {n}_{i}\right) \) of positive integers and for each \( k \) and for each \( s \in \left\{ {0,1,\ldots ,{n}_{0}}\right\} \times \cdots \times \left\{ {0,1,\ldots ,{n}_{k}}\right\} \), a nonempty closed set \( {F}_{... | Yes |
Proposition 2.6.11 Let \( A \) be a discrete space and \( X = {A}^{\mathbb{N}} \) . Then every nonempty closed subset of \( X \) is a retract of \( X \) . | Proof. Let \( C \) be a nonempty closed set in \( X \) . For each \( s \in {A}^{ < \mathbb{N}} \) such that \( C\bigcap \sum \left( s\right) \neq \varnothing \), choose and fix \( {x}_{s} \in C\bigcap \sum \left( s\right) \) . Let \( \alpha \in X \) . Define \( f\left( \alpha \right) = \alpha \) for \( \alpha \in C \) ... | Yes |
Theorem 2.6.12 Every Polish space \( X \) is a continuous image of \( {\mathbb{N}}^{\mathbb{N}} \) . | Null | No |
Theorem 2.6.14 Every zero-dimensional compact, dense-in-itself metric space is homeomorphic to \( \mathcal{C} \) . | Proof. It is sufficient to show that there is a Cantor scheme \( \left\{ {{C}_{s} : s \in }\right. \) \( \left. {2}^{ < \mathbb{N}}\right\} \) on \( X \) of clopen sets such that \( {C}_{e} = X \) and \( {C}_{s} = {C}_{{s}^{ \frown }0}\bigcup {C}_{{s}^{ \frown }1} \) for all \( s \) .\n\nConstruction of \( \left\{ {{C}... | Yes |
Theorem 2.6.14 Every zero-dimensional compact, dense-in-itself metric space is homeomorphic to \( \mathcal{C} \) . | Proof. It is sufficient to show that there is a Cantor scheme \( \left\{ {{C}_{s} : s \in }\right. \) \( \left. {2}^{ < \mathbb{N}}\right\} \) on \( X \) of clopen sets such that \( {C}_{e} = X \) and \( {C}_{s} = {C}_{{s}^{ \frown }0}\bigcup {C}_{{s}^{ \frown }1} \) for all \( s \) .\n\nConstruction of \( \left\{ {{C}... | Yes |
Lemma 3.1.6 Let \( \left( {X,\mathcal{A}}\right) \) be a measurable space, where \( \mathcal{A} = \sigma \left( \mathcal{G}\right) \) . Suppose \( x, y \in X \) are such that for every \( G \in \mathcal{G}, x \in G \) if and only if \( y \in G \) . Then for all \( A \in \mathcal{A}, x \in A \) if and only if \( y \in A... | Proof. Let\n\n\[ \mathcal{B} = \{ A \subseteq X : x \in A \Leftrightarrow y \in A\} . \]\n\nIt is easy to see that \( \mathcal{B} \) is a \( \sigma \) -algebra. By our assumption, it contains \( \mathcal{G} \) . The result follows. | Yes |
Example 3.1.2 Let \( X \) be an infinite set and\n\n\[ \mathcal{A} = \left\{ {A \subseteq X : \text{ either }A\text{ or }{A}^{c}\text{ is finite }}\right\} .\n\]\nThen \( \mathcal{A} \) is an algebra that is not a \( \sigma \) -algebra. | Null | No |
Example 3.1.4 The family of finite disjoint unions of nondegenerate intervals including the empty set is an algebra on \( \mathbb{R} \) . | Null | No |
Lemma 3.1.6 Let \( \left( {X,\mathcal{A}}\right) \) be a measurable space, where \( \mathcal{A} = \sigma \left( \mathcal{G}\right) \) . Suppose \( x, y \in X \) are such that for every \( G \in \mathcal{G}, x \in G \) if and only if \( y \in G \) . Then for all \( A \in \mathcal{A}, x \in A \) if and only if \( y \in A... | Proof. Let\n\n\[ \mathcal{B} = \{ A \subseteq X : x \in A \Leftrightarrow y \in A\} . \]\n\nIt is easy to see that \( \mathcal{B} \) is a \( \sigma \) -algebra. By our assumption, it contains \( \mathcal{G} \) . The result follows. | Yes |
Proposition 3.1.7 Let \( \\left( {X,\\mathcal{B}}\\right) \) be a measurable space, \( \\mathcal{G} \) a generator of \( \\mathcal{B} \) , and \( A \\in \\mathcal{B} \) . Then there exists a countable \( {\\mathcal{G}}^{\\prime } \\subseteq \\mathcal{G} \) such that \( A \\in \\sigma \\left( {\\mathcal{G}}^{\\prime }\\... | Proof. Let \( \\mathcal{A} \) be the collection of all subsets \( A \) of \( X \) such that \( A \\in \\sigma \\left( {\\mathcal{G}}^{\\prime }\\right) \) for some countable \( {\\mathcal{G}}^{\\prime } \\subseteq \\mathcal{G} \) .\n\nClearly, \( \\mathcal{A} \) is closed under complementation, and \( \\mathcal{G} \\su... | Yes |
Proposition 3.1.9 The Borel \( \sigma \) -algebra \( {\mathcal{B}}_{X} \) of a metrizable space \( X \) equals the smallest family \( \mathcal{B} \) of subsets of \( X \) that contains all open sets and that is closed under countable intersections and countable unions. | Proof. Since \( \mathcal{B} \) is the smallest family of subsets of \( X \) containing all open sets, closed under countable intersections and countable unions, and \( {\mathcal{B}}_{X} \) is one such family, \( \mathcal{B} \subseteq {\mathcal{B}}_{X} \) . The reverse inclusion will be shown if we show that \( \mathcal... | Yes |
Proposition 3.1.10 The Borel \( \sigma \) -algebra \( {\mathcal{B}}_{X} \) of a metrizable space \( X \) equals the smallest family \( \mathcal{B} \) of subsets of \( X \) that contains all closed sets and that is closed under countable intersections and countable unions. | Null | No |
The Borel \( \sigma \) -algebra \( {\mathcal{B}}_{X} \) of a metrizable space \( X \) equals the smallest family \( \mathcal{B} \) that contains all open subsets of \( X \) and that is closed under countable intersections and countable disjoint unions. | By the argument contained in the proof of 3.1.9, it is sufficient to prove that \( \mathcal{B} \) is closed under complementation. Let\n\n\[ \mathcal{D} = \left\{ {B \in \mathcal{B} : {B}^{c} \in \mathcal{B}}\right\} \]\n\nSince every closed set in \( X \) is a \( {G}_{\delta } \) set, all open sets belong to \( \mathc... | Yes |
Proposition 3.1.12 (Sierpiński) The Borel \( \sigma \) -algebra \( {\mathcal{B}}_{X} \) of a metrizable space \( X \) equals the smallest family \( \mathcal{B} \) that contains all closed subsets of \( X \) and that is closed under countable intersections and countable disjoint unions. | Proof. By 3.1.11, it is sufficient to show that every open set belongs to \( \mathcal{B} \) . The main difficulty lies here. Recall that in 2.1.35 we showed that \( \left( {0,1}\right) \) cannot be expressed as a countable disjoint union of closed subsets of \( \mathbb{R} \) . We need a lemma.\n\nNotation. For any fami... | Yes |
Lemma 3.1.13 Let \( \mathcal{F} \) be the set of closed subsets of \( \mathbb{R} \) . Then \( (0,1\rbrack \in {\mathcal{F}}_{+\delta + } \) . | Proof of the lemma. Let \( D \) be the set of all endpoints of the middle-third intervals removed from \( \left\lbrack {0,1}\right\rbrack \) to construct the Cantor ternary set \( \mathbf{C} \n\n(2.3.3). So, \( D = \left\{ {\frac{1}{3},\frac{2}{3},\frac{1}{9},\frac{2}{9},\frac{7}{9},\frac{8}{9},\ldots }\right\} \) . Le... | Yes |
Proposition 3.1.14 (The monotone class theorem) The smallest monotone class \( \mathcal{M} \) containing an algebra \( \mathcal{A} \) on a set \( X \) equals \( \sigma \left( \mathcal{A}\right) \), the \( \sigma \) -algebra generated by \( \mathcal{A} \) . | Proof. Since every \( \sigma \) -algebra is a monotone class, \( \mathcal{M} \subseteq \sigma \left( \mathcal{A}\right) \) . To show the other inclusion, we first show that \( \mathcal{M} \) is closed under finite intersections. For \( A \subset X \), let \[ \mathcal{M}\left( A\right) = \{ B \in \mathcal{M} : A\bigcap ... | Yes |
Proposition 3.1.15 Every countably generated measurable space is atomic. | Proof. Let \( \mathcal{A} \) be a countably generated \( \sigma \) -algebra on \( X \) . Fix a countable generator \( \mathcal{G} = \left\{ {{A}_{n} : n \in \mathbb{N}}\right\} \) for \( \mathcal{A} \) . For any \( B \subseteq X \), set \( {B}^{0} = B \) and \( {B}^{1} = X \smallsetminus B \) . For every sequence \( \a... | Yes |
Proposition 3.1.21 Let \( \\left( {X,\\mathcal{A}}\\right) \) be a measurable space and \( Y \) a second countable metrizable space. If \( f : X \\rightarrow Y \) is a measurable function, then \( \\operatorname{graph}\\left( f\\right) \) is in \( \\mathcal{A} \\otimes {\\mathcal{B}}_{Y} \) . | Proof. Let \( \\left( {U}_{n}\\right) \) be a countable base for \( Y \). Note that\n\n\[ y \\neq f\\left( x\\right) \\Leftrightarrow \\exists n\\left( {f\\left( x\\right) \\in {U}_{n}\\& y \\notin {U}_{n}}\\right) .\n\]\n\nTherefore,\n\n\[ \\operatorname{graph}\\left( f\\right) = {\\left\\lbrack \\mathop{\\bigcup }\\l... | Yes |
Corollary 3.1.22 Let \( \\left( {X,\\mathcal{A}}\\right) \) be a measurable space and \( Y \) a discrete measurable space of cardinality at most \( \\mathfrak{c} \) . Then the graph of every measurable function \( f : X \\rightarrow Y \) is measurable. | Proof. Without loss of generality, assume \( Y \\subseteq \\mathbb{R} \) . Let \( f : \\left( {X,\\mathcal{A}}\\right) \\rightarrow \\left( {Y,\\mathcal{P}\\left( Y\\right) }\\right) \) be measurable. In particular, \( f : \\left( {X,\\mathcal{A}}\\right) \\rightarrow \\left( {Y,{\\mathcal{B}}_{Y}}\\right) \) is also m... | Yes |
Proposition 3.1.23 Let \( {X}_{i}, i = 0,1,\ldots \), be a sequence of second countable metrizable spaces and \( X = \mathop{\prod }\limits_{i}{X}_{i} \). Then\n\n\[ \n{\mathcal{B}}_{X} = {\bigotimes }_{i}{\mathcal{B}}_{{X}_{i}} \n\] | Proof. Fix a countable base \( {\mathcal{B}}_{i} \) for \( {X}_{i}, i \in \mathbb{N} \), and put\n\n\[ \n\mathcal{G} = \left\{ {{\pi }_{i}^{-1}\left( B\right) : B \in {\mathcal{B}}_{i}, i \in \mathbb{N}}\right\} \n\]\n\nThen \( \mathcal{G} \) generates \( {\bigotimes }_{i}{\mathcal{B}}_{{X}_{i}} \). On the other hand, ... | Yes |
Theorem 3.1.24 \( \mathcal{P}\left( {\omega }_{1}\right) \bigotimes \mathcal{P}\left( {\omega }_{1}\right) = \mathcal{P}\left( {{\omega }_{1} \times {\omega }_{1}}\right) \) . | Proof. Let \( A \subseteq {\omega }_{1} \times {\omega }_{1} \) . Write \( A = B \cup C \), where\n\n\[ B = A\bigcap \left\{ {\left( {\alpha ,\beta }\right) \in {\omega }_{1} \times {\omega }_{1} : \alpha \geq \beta }\right\} \]\n\nand\n\n\[ C = A\bigcap \left\{ {\left( {\alpha ,\beta }\right) \in {\omega }_{1} \times ... | Yes |
Proposition 3.1.27 Let \( \\left( {f}_{n}\\right) \) be a sequence of measurable maps from a measurable space \( X \) to a metrizable space \( Y \) converging pointwise to \( f \) . Then \( f : X \\rightarrow Y \) is measurable. | Proof. Let \( d \) be a compatible metric on \( Y \) . Fix any open set \( U \) in \( Y \) . For each positive integer \( k \), set\n\n\[ \n{U}_{k} = \\left\{ {x \\in U : d\\left( {x,{U}^{c}\\right) > 1/k}\\right\} .\n\]\n\nSince \( U \) is open,\n\n\[ \nU = \\mathop{\\bigcup }\\limits_{k}{U}_{k} = \\mathop{\\bigcup }\... | Yes |
Proposition 3.1.27 Let \( \left( {f}_{n}\right) \) be a sequence of measurable maps from a measurable space \( X \) to a metrizable space \( Y \) converging pointwise to \( f \) . Then \( f : X \rightarrow Y \) is measurable. | Proof. Let \( d \) be a compatible metric on \( Y \) . Fix any open set \( U \) in \( Y \) . For each positive integer \( k \), set\n\n\[ \n{U}_{k} = \left\{ {x \in U : d\left( {x,{U}^{c}}\right) > 1/k}\right\} . \n\]\n\nSince \( U \) is open,\n\n\[ \nU = \mathop{\bigcup }\limits_{k}{U}_{k} = \mathop{\bigcup }\limits_{... | Yes |
Proposition 3.1.28 Let \( X \) be metrizable. Then every Borel function \( f \) : \( X \rightarrow \mathbb{R} \) is the pointwise limit of a sequence of simple Borel functions. | Proof. Fix \( n \geq 1 \) . For \( - n{2}^{n} \leq j < n{2}^{n} \), let\n\n\[ \n{B}_{j}^{n} = {f}^{-1}\left( \left\lbrack {j/{2}^{n},\left( {j + 1}\right) /{2}^{n}}\right) \right) .\n\]\n\nAs \( f \) is Borel, each \( {B}_{j}^{n} \) is Borel. Set\n\n\[ \n{f}_{n} = \mathop{\sum }\limits_{{j = - n{2}^{n}}}^{{\left( {n - ... | Yes |
Proposition 3.1.29 (i) If \( f : \left( {X,\mathcal{A}}\right) \rightarrow \left( {Y,\mathcal{B}}\right) \) and \( g : \left( {Y,\mathcal{B}}\right) \rightarrow \left( {Z,\mathcal{C}}\right) \) are measurable, then so is \( g \circ f : \left( {X,\mathcal{A}}\right) \rightarrow \left( {Z,\mathcal{C}}\right) \) . | Null | No |
Theorem 3.1.30 Let \( \left( {X,\mathcal{A}}\right) \) be a measurable space, \( Y \) and \( Z \) metrizable spaces with \( Y \) second countable. Suppose \( D \) is a countable dense set in \( Y \) and \( f : X \times Y \rightarrow Z \) a map such that\n\n(i) the map \( y \rightarrow f\left( {x, y}\right) \) from \( Y... | Proof. Fix compatible metrics \( d \) and \( \rho \) on \( Y \) and \( Z \) respectively. Take any closed set \( C \) in \( Z \) . For \( \left( {x, y}\right) \in X \times Y \), it is routine to check that\n\n\[ f\left( {x, y}\right) \in C \Leftrightarrow \left( {\forall n \geq 1}\right) \left( {\exists {y}^{\prime } \... | Yes |
Proposition 3.1.32 Let \( X \) and \( Y \) be metrizable spaces. Then every Baire function \( f : X \rightarrow Y \) is Borel. | Proof. Since every continuous function is Borel and since the limit of a pointwise convergent sequence of Borel functions is Borel (3.1.27), Baire functions are Borel. | Yes |
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