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Theorem 1.4.3 For every infinite set \( X \) ,\n\n\[ X \times \{ 0,1\} \equiv X \]
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Proof. Let\n\n\[ P = \{ \left( {A, f}\right) : A \subseteq X\\text{ and }f : A \times \{ 0,1\} \rightarrow A\\text{ a bijection }\} .\n\]\n\nSince \( X \) is infinite, it contains a countably infinite set, say \( D \) . By 1.1.3, \( D \times \{ 0,1\} \equiv D \) . Therefore, \( P \) is nonempty. Consider the partial order \( \\propto \) on \( P \) defined by\n\n\[ \\left( {A, f}\right) \\propto \\left( {B, g}\right) \\Leftrightarrow A \\subseteq B\\& f \\preccurlyeq g.\n\]\n\nFollowing the argument contained in the proof of 1.4.1, we see that the hypothesis of Zorn’s lemma is satisfied by \( P \) . So, \( P \) has a maximal element, say \( \\left( {A, f}\right) \) .\n\nTo complete the proof we show that \( A \\equiv X \) . Since \( X \) is infinite, by 1.3.1, it will be sufficient to show that \( X \\smallsetminus A \) is finite. Suppose not. By 1.3.2, there is a \( B \\subseteq X \\smallsetminus A \) such that \( B \\equiv \\mathbb{N} \) . So there is a one-to-one map \( g \) from \( B \times \{ 0,1\} \) onto \( B \) . Combining \( f \) and \( g \) we get a bijection\n\n\[ h : \\left( {A\\bigcup B}\right) \times \{ 0,1\} \\rightarrow A\\bigcup B\n\]\n\nthat extends \( f \) . This contradicts the maximality of \( \\left( {A, f}\right) \) . Hence, \( X \\smallsetminus A \) is finite. Therefore, \( A \\equiv X \) . The proof is complete.
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Yes
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Theorem 1.4.5 For every infinite set \( X \) , \[ X \times X \equiv X. \]
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Proof. Let \[ P = \{ \left( {A, f}\right) : A \subseteq X\text{ and }f : A \times A \rightarrow A\text{ a bijection }\} . \] Note that \( P \) is nonempty. Consider the partial order \( \propto \) on \( P \) defined by \[ \left( {A, f}\right) \propto \left( {B, g}\right) \Leftrightarrow A \subseteq B\& f \preccurlyeq g. \] By Zorn’s lemma, take a maximal element \( \left( {A, f}\right) \) of \( P \) as in the proof of 1.4.3. Note that \( A \) must be infinite. To complete the proof, we shall show that \( A \equiv X \) . Suppose not. Then \( A{ < }_{c}X \) . We first show that \( X \smallsetminus A \equiv X \) . Suppose \( X \smallsetminus A{ < }_{c}X \) . By 1.4.1, either \( A{ \leq }_{c}X \smallsetminus A \) or \( X \smallsetminus A{ \leq }_{c}A \) . Assume first \( X \smallsetminus A{ \leq }_{c}A \) . Using 1.4.3, take two disjoint sets \( {A}_{1},{A}_{2} \) of the same cardinality as \( A \) and \( {A}_{1}\bigcup {A}_{2} = A \) . Now, \[ X = A\bigcup \left( {X \smallsetminus A}\right) { \leq }_{c}{A}_{1}\bigcup {A}_{2} \equiv A{ < }_{c}X. \] This is a contradiction. Similarly we arrive at a contradiction from the other inequality. Thus, by 1.4.2, \( X \smallsetminus A \equiv X \) . Now choose \( B \subseteq X \smallsetminus A \) such that \( B \equiv A \) . By 1.4.4, write \( B \) as the union of three disjoint sets, say \( {B}_{1},{B}_{2} \), and \( {B}_{3} \), each of the same cardinality as \( A \) . Since there is a one-to-one map from \( A \times A \) onto \( A \), there exist bijections \( {f}_{1} : B \times A \rightarrow {B}_{1},{f}_{2} : B \times B \rightarrow {B}_{2} \), and \( {f}_{3} : A \times B \rightarrow {B}_{3} \) . Let \( C = \) \( A\bigcup B \) . Combining these four bijections, we get a bijection \( g : C \times C \rightarrow C \) that is a proper extension of \( f \) . This contradicts the maximality of \( \left( {A, f}\right) \) . Thus, \( A \equiv X \) . The proof is now complete.
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Yes
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Theorem 1.4.5 For every infinite set \( X \) , \[ X \times X \equiv X. \]
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Proof. Let \[ P = \{ \left( {A, f}\right) : A \subseteq X\text{ and }f : A \times A \rightarrow A\text{ a bijection }\} . \] Note that \( P \) is nonempty. Consider the partial order \( \propto \) on \( P \) defined by \[ \left( {A, f}\right) \propto \left( {B, g}\right) \Leftrightarrow A \subseteq B\& f \preccurlyeq g. \] By Zorn’s lemma, take a maximal element \( \left( {A, f}\right) \) of \( P \) as in the proof of 1.4.3. Note that \( A \) must be infinite. To complete the proof, we shall show that \( A \equiv X \) . Suppose not. Then \( A{ < }_{c}X \) . We first show that \( X \smallsetminus A \equiv X \) . Suppose \( X \smallsetminus A{ < }_{c}X \) . By 1.4.1, either \( A{ \leq }_{c}X \smallsetminus A \) or \( X \smallsetminus A{ \leq }_{c}A \) . Assume first \( X \smallsetminus A{ \leq }_{c}A \) . Using 1.4.3, take two disjoint sets \( {A}_{1},{A}_{2} \) of the same cardinality as \( A \) and \( {A}_{1}\bigcup {A}_{2} = A \) . Now, \[ X = A\bigcup \left( {X \smallsetminus A}\right) { \leq }_{c}{A}_{1}\bigcup {A}_{2} \equiv A{ < }_{c}X. \] This is a contradiction. Similarly we arrive at a contradiction from the other inequality. Thus, by 1.4.2, \( X \smallsetminus A \equiv X \) . Now choose \( B \subseteq X \smallsetminus A \) such that \( B \equiv A \) . By 1.4.4, write \( B \) as the union of three disjoint sets, say \( {B}_{1},{B}_{2} \), and \( {B}_{3} \), each of the same cardinality as \( A \) . Since there is a one-to-one map from \( A \times A \) onto \( A \), there exist bijections \( {f}_{1} : B \times A \rightarrow {B}_{1},{f}_{2} : B \times B \rightarrow {B}_{2} \), and \( {f}_{3} : A \times B \rightarrow {B}_{3} \) . Let \( C = \) \( A\bigcup B \) . Combining these four bijections, we get a bijection \( g : C \times C \rightarrow C \) that is a proper extension of \( f \) . This contradicts the maximality of \( \left( {A, f}\right) \) . Thus, \( A \equiv X \) . The proof is now complete.
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Yes
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Proposition 1.4.8 (J. König,[58]) Let \( \\left\\{ {{X}_{i} : i \\in I}\\right\\} \) and \( \\left\\{ {{Y}_{i} : i \\in I}\\right\\} \) be families of sets such that \( {X}_{i}{ < }_{c}{Y}_{i} \) for each \( i \\in I \) . Then there is no map \( f \) from \( \\mathop{\\bigcup }\\limits_{i}{X}_{i} \) onto \( {\\Pi }_{i}{Y}_{i} \) .
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Proof. Let \( f : \\mathop{\\bigcup }\\limits_{i}{X}_{i} \\rightarrow {\\Pi }_{i}{Y}_{i} \) be any map. For any \( i \\in I \\), let\n\n\[ \n{A}_{i} = {Y}_{i} \\smallsetminus {\\pi }_{i}\\left( {f\\left( {X}_{i}\\right) }\\right)\n\]\n\nwhere \( {\\pi }_{i} : \\mathop{\\prod }\\limits_{j}{Y}_{j} \\rightarrow {Y}_{i} \) is the projection map. Since for evry \( i,{X}_{i}{ < }_{c}{Y}_{i} \) , each \( {A}_{i} \) is nonempty. By \( \\mathbf{{AC}},{\\Pi }_{i}{A}_{i} \\neq \\varnothing \) . But\n\n\[ \n{\\Pi }_{i}{A}_{i}\\bigcap \\operatorname{range}\\left( f\\right) = \\varnothing .\n\]\n\nIt follows that \( f \) is not onto.
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Yes
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\[ {2}^{\mathfrak{c}} \leq {\aleph }_{0}^{\mathfrak{c}} \]
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\[ {2}^{\mathfrak{c}} \leq {\aleph }_{0}^{\mathfrak{c}}\;\text{ (since }2 \leq {\aleph }_{0}\text{ ) } \] \[ \leq {\mathfrak{c}}^{\mathfrak{c}}\;\text{ (since }{\aleph }_{0} \leq \mathfrak{c}\text{ ) } \] \[ = {\left( {2}^{{\aleph }_{0}}\right) }^{\mathfrak{c}}\;\left( {\text{since}\mathfrak{c} = {2}^{{\aleph }_{0}}}\right) \] \[ = {2}^{{\aleph }_{0} \cdot \mathfrak{c}}\;\text{(since for nonempty sets}X, Y, Z,{\left( {X}^{Y}\right) }^{Z} \equiv {X}^{Y \times Z}\text{)} \] \[ \leq {2}^{\mathfrak{c} \cdot \mathfrak{c}}\;\text{ (since }{\aleph }_{0} < \mathfrak{c}\text{ ) } \] \[ = {2}^{\mathfrak{c}}\;\text{(since}\mathfrak{c} \cdot \mathfrak{c} = \mathfrak{c}\text{).} \] So, by the Schröder - Bernstein theorem, \( {2}^{\mathfrak{c}} = {\aleph }_{0}^{\mathfrak{c}} = {\mathfrak{c}}^{\mathfrak{c}} \) . It follows that \[ \{ 0,1{\} }^{\mathbb{R}} \equiv {\mathbb{N}}^{\mathbb{R}} \equiv {\mathbb{R}}^{\mathbb{R}} \]
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Yes
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Proposition 1.6.2 A linearly ordered set \( \\left( {W, \\leq }\\right) \) is well-ordered if and only if there is no descending sequence \( {w}_{0} > {w}_{1} > {w}_{2} > \\cdots \) in \( W \) .
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Proof. Let \( W \) be not well-ordered. Then there is a nonempty subset \( A \) of \( W \) not having a least element. Choose any \( {w}_{0} \\in A \) . Since \( {w}_{0} \) is not the first element of \( A \), there is a \( {w}_{1} \\in A \) such that \( {w}_{1} < {w}_{0} \) . Since \( {w}_{1} \) is not the first element of \( A \), we get \( {w}_{2} < {w}_{1} \) in \( A \) . Proceeding similarly, we get a descending sequence \( \\left\\{ {{w}_{n} : n \\geq 0}\\right\\} \) in \( W \) . This completes the proof of the \
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Yes
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Proposition 1.6.5 No well-ordered set \( W \) is order isomorphic to an initial segment \( W\left( u\right) \) of itself.
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Proof. Let \( W \) be a well-ordered set and \( u \in W \) . Suppose \( W \) and \( W\left( u\right) \) are isomorphic. Let \( f : W \rightarrow W\left( u\right) \) be an order isomorphism. For \( n \in \mathbb{N} \) , let \( {w}_{n} = {f}^{n}\left( u\right) \) . Note that\n\n\[ \n{w}_{0} = {f}^{0}\left( u\right) = u > {f}^{1}\left( u\right) = f\left( u\right) = {w}_{1}.\n\]\n\nBy induction on \( n \), we see that \( {w}_{n} > {w}_{n + 1} \) for all \( n \), i.e., \( \left( {w}_{n}\right) \) is a descending sequence in \( W \) . By 1.6.2, \( W \) is not well-ordered. This contradiction proves our result.
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Yes
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Proposition 1.7.1 (Proof by induction) For each \( n \in \mathbb{N} \), let \( {P}_{n} \) be a mathematical proposition. Suppose \( {P}_{0} \) is true and for every \( n,{P}_{n + 1} \) is true whenever \( {P}_{n} \) is true. Then for every \( n,{P}_{n} \) is true. Symbolically, we can express this as follows.
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The proof of this proposition uses two basic properties of the set of natural numbers. First, it is well-ordered by the usual order, and second, every nonzero element in it is a successor. A repeated application of 1.7.1 gives us the following.
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No
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Theorem 1.7.3 (Proof by transfinite induction) Let \( \\left( {W, \\leq }\\right) \) be a well-ordered set, and for every \( w \\in W \), let \( {P}_{w} \) be a mathematical proposition. Suppose that for each \( w \\in W \), if \( {P}_{v} \) is true for each \( v < w \), then \( {P}_{w} \) is true. Then for every \( w \\in W,{P}_{w} \) is true. Symbolically, we express this as\n\n\[ \n\\left( {\\forall w \\in W}\\right) \\left( {\\left( {\\left( {\\forall v < w}\\right) {P}_{v}}\\right) \\Rightarrow {P}_{w}}\\right) \\Rightarrow \\left( {\\forall w \\in W}\\right) {P}_{w}.\n\]
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Proof. Let\n\n\[ \n\\left( {\\forall w \\in W}\\right) \\left( {\\left( {\\left( {\\forall v < w}\\right) {P}_{v}}\\right) \\Rightarrow {P}_{w}}\\right) .\n\]\n\n\( \\left( *\\right) \)\n\nSuppose \( {P}_{w} \) is false for some \( w \\in W \) . Consider\n\n\[ \nA = \\left\{ {w \\in W : {P}_{w}\\text{ does not hold }}\\right\} .\n\]\n\nBy our assumptions, \( A \\neq \\varnothing \) . Let \( {w}_{0} \) be the least element of \( A \) . Then for every \( v < {w}_{0},{P}_{v} \) holds. However, \( {P}_{{w}_{0}} \) does not hold. This contradicts \( \\left( \\star \\right) \). Therefore, for every \( w \\in W,{P}_{w} \) holds.
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Yes
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Theorem 1.7.4 (Definition by transfinite induction) Let \( \\left( {W, \\leq }\\right) \) be a well-ordered set, \( X \) a set, and \( \\mathcal{F} \) the set of all maps with domain an initial segment of \( W \) and range contained in \( X \) . If \( G : \\mathcal{F} \\rightarrow X \) is any map, then there is a unique map \( f : W \\rightarrow X \) such that for every \( u \\in W \) ,\n\n\[ f\\left( u\\right) = G\\left( {f \\mid W\\left( u\\right) }\\right) . \]
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Proof. For each \( w \\in W \), let \( {P}_{w} \) be the proposition \
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No
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Theorem 1.7.5 (Trichotomy theorem for well-ordered sets) For any two well-ordered sets \( W \) and \( {W}^{\prime } \), exactly one of\n\n\[ W \prec {W}^{\prime }, W \sim {W}^{\prime },\text{ and }{W}^{\prime } \prec W \]\n\nholds.
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Proof. It is easy to see that no two of these can hold simultaneously. For example, if \( W \sim {W}^{\prime } \) and \( {W}^{\prime } \prec W \), then \( W \) is isomorphic to an initial segment of itself. This is impossible by 1.6.5.\n\nTo show that at least one of these holds, take \( X = {W}^{\prime }\bigcup \{ \infty \} \), where \( \infty \) is a point outside \( {W}^{\prime } \) . Now define a map \( f : W \rightarrow X \) by transfinite induction as follows. Let \( w \in W \) and assume that \( f \) has been defined on \( W\left( w\right) \) . If \( {W}^{\prime } \smallsetminus f\left( {W\left( w\right) }\right) \neq \varnothing \), then we take \( f\left( w\right) \) to be the least element of \( {W}^{\prime } \smallsetminus f\left( {W\left( w\right) }\right) \) ; otherwise, \( f\left( w\right) = \infty \) . By 1.7.4, such a function exists.\n\nLet us assume that \( \infty \notin f\left( W\right) \) . Then\n\n(i) the map \( f \) is one-to-one and order preserving, and\n\n(ii) the range of \( f \) is either whole of \( {W}^{\prime } \) or an initial segment of \( {W}^{\prime } \) .\n\nSo, in this case at least one of \( W \sim {W}^{\prime } \) or \( W \prec {W}^{\prime } \) holds.\n\nIf \( \infty \in f\left( W\right) \), then let \( w \) be the first element of \( W \) such that \( f\left( w\right) = \infty \) . Then \( f \mid W\left( w\right) \) is an order isomorphism from \( W\left( w\right) \) onto \( {W}^{\prime } \) . Thus in this case \( {W}^{\prime } \prec W \) .
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Yes
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Corollary 1.7.6 Let \( \\left( {W, \\leq }\\right) ,\\left( {{W}^{\\prime },{ \\leq }^{\\prime }}\\right) \) be well-ordered sets. Then \( W \\preccurlyeq {W}^{\\prime } \) if and only if there is a one-to-one order-preserving map from \( W \) into \( {W}^{\\prime } \) .
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Proof. Suppose there is a one-to-one order-preserving map \( g \) from \( W \) into \( {W}^{\\prime } \) . Let \( X \) and \( f : W \\rightarrow X \) be as in the proof of 1.7.5. Then, by induction on \( w \), we easily show that for every \( w \\in W, f\\left( w\\right) { \\leq }^{\\prime }g\\left( w\\right) \) . Therefore, \( \\infty \\notin f\\left( W\\right) \) . Hence, \( W \\preccurlyeq {W}^{\\prime } \) . The converse is clear.
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Yes
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Theorem 1.7.7 Let \( \mathcal{W} = \left\{ {\left( {{W}_{i},{ \leq }_{i}}\right) : i \in I}\right\} \) be a family of pairwise non-isomorphic well-ordered sets. Then there is a \( W \in \mathcal{W} \) such that \( W \prec {W}^{\prime } \) for every \( {W}^{\prime } \in \mathcal{W} \) different from \( W \) .
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Proof. Suppose no such \( W \) exists. Then there is a descending sequence\n\n\[ \cdots \prec {W}_{n} \prec \cdots \prec {W}_{1} \prec {W}_{0} \]\n\nin \( \mathcal{W} \) . For \( n \in \mathbb{N} \), choose a \( {w}_{n}^{\prime } \in {W}_{n} \) such that \( {W}_{n + 1} \sim {W}_{n}\left( {w}_{n}^{\prime }\right) \) . Fix an order isomorphism \( {f}_{n} : {W}_{n + 1} \rightarrow {W}_{n}\left( {w}_{n}^{\prime }\right) \) . Let \( {w}_{0} = {w}_{0}^{\prime } \), and for \( n > 0 \), \n\n\[ {w}_{n} = {f}_{0}\left( {{f}_{1}\left( {\cdots {f}_{n - 1}\left( {w}_{n}^{\prime }\right) }\right) }\right) . \]\n\n(See Figure 1.2.) Then \( \left( {w}_{n}\right) \) is a descending sequence in \( {W}_{0} \) . This is a contradiction. The result follows.
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Yes
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Theorem 1.8.3 Every ordinal \( \alpha \) can be uniquely written as \[ \alpha = \beta + n \] where \( \beta \) is a limit ordinal and \( n \) finite.
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Proof. Let \( \alpha \) be an ordinal number. We first show that there exists a limit ordinal \( \beta \) and an \( n \in \omega \) such that \( \alpha = \beta + n \) . Choose a well-ordered set \( W \) such that \( t\left( W\right) = \alpha \) . If \( W \) has no last element, then we take \( \beta = \alpha \) and \( n = 0 \) . Suppose \( W \) has a last element, say \( {w}_{0} \) . If \( {w}_{0} \) has no immediate predecessor, then take \( \beta = t\left( {W\left( {w}_{0}\right) }\right) \) and \( n = 1 \) . Now suppose that \( {w}_{0} \) does have an immediate predecessor, say \( {w}_{1} \) . If \( {w}_{1} \) has no immediate predecessor, then we take \( \beta = t\left( {W\left( {w}_{1}\right) }\right) \) and \( n = 2 \) . Since \( W \) has no descending sequence, this process ends after finitely many steps. Thus we get \( {w}_{0},{w}_{1},\ldots ,{w}_{k - 1} \) such that \( {w}_{i} = {w}_{i - 1}^{ - } \) for all \( i > 0 \), and \( {w}_{k - 1} \) has no immediate predecessor. We take \( \beta = t\left( {W\left( {w}_{k - 1}\right) }\right) \) and \( n = k \) . We now show that \( \alpha \) has a unique representation of the type mentioned above. Let \( W,{W}^{\prime } \) be well-ordered sets with no last element, and \( {A}_{n},{B}_{m} \) finite well-ordered sets of cardinality \( n \) and \( m \) respectively such that \[ {A}_{n} \cap W = {B}_{m} \cap {W}^{\prime } = \varnothing . \] Let \( f : W + {A}_{n} \rightarrow {W}^{\prime } + {B}_{m} \) be an order isomorphism. It is easy to check that \( f\left( W\right) = {W}^{\prime } \) and \( f\left( {A}_{n}\right) = {B}_{m} \) . Uniqueness now follows.
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Yes
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Theorem 1.8.5 The set \( \Omega \) of all countable ordinals is uncountable.
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Proof. Suppose \( \Omega \) is countable. Fix an enumeration \( {\alpha }_{0},{\alpha }_{1},\ldots \) of \( \Omega \) . Then\n\n\[ \alpha = \mathop{\sum }\limits_{n}{\alpha }_{n} + 1 \]\n\n is a countable ordinal strictly larger than each \( {\alpha }_{n} \) . This is a contradiction. So, \( \Omega \) is uncountable.
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Yes
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Proposition 1.8.6 Let \( \alpha \) be a countable limit ordinal. Then there exist \( {\alpha }_{0} < {\alpha }_{1} < \cdots \) such that \( \sup \left\{ {{\alpha }_{n} : n \in \mathbb{N}}\right\} = \alpha \) .
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Proof. Since \( \alpha \) is countable, \( \{ \beta \in \mathbf{{ON}} : \beta < \alpha \} \) is countable. Fix an enumeration \( \left\{ {{\beta }_{n} : n \in \mathbb{N}}\right\} \) of all ordinals less than \( \alpha \) . We now define a sequence of ordinals \( \left( {\alpha }_{n}\right) \) by induction on \( n \) . Choose \( {\alpha }_{0} \) such that \( {\beta }_{0} < \) \( {\alpha }_{0} < \alpha \) . Since \( \alpha \) is a limit ordinal, such an ordinal exists. Suppose \( {\alpha }_{n} \) has been defined. Choose \( {\alpha }_{n + 1} \) greater than \( {\alpha }_{n} \) such that \( {\beta }_{n + 1} < {\alpha }_{n + 1} < \alpha \) . Clearly,\n\n\[ \n\alpha = \sup \left\{ {{\beta }_{n} : n \in \mathbb{N}}\right\} \leq \sup \left\{ {{\alpha }_{n} : n \in \mathbb{N}}\right\} \leq \alpha .\n\]\n\nSo, \( \sup \left\{ {{\alpha }_{n} : n \in \mathbb{N}}\right\} = \alpha \) .
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Yes
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Proposition 1.10.7 (König’s infinity lemma, [57]) Let \( T \) be a finitely splitting, infinite tree on \( A \) . Then \( T \) is ill-founded.
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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, \( \left( {a}_{0}\right) \) exists. By the same argument we get \( {a}_{1} \in A \) such that \( {s}_{1} = \left( {{a}_{0},{a}_{1}}\right) \) has infinitely many extensions in \( T \) . Proceeding similarly we get an \( \alpha = \left( {{a}_{0},{a}_{1},\ldots }\right) \) such that for all \( k,\alpha \mid k \) has infinitely many extensions in \( T \) . In particular, \( \alpha \in \left\lbrack T\right\rbrack \) , and the result is proved.
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Yes
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Proposition 1.10.7 (König’s infinity lemma, [57]) Let \( T \) be a finitely splitting, infinite tree on \( A \) . Then \( T \) is ill-founded.
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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, \( \left( {a}_{0}\right) \) exists. By the same argument we get \( {a}_{1} \in A \) such that \( {s}_{1} = \left( {{a}_{0},{a}_{1}}\right) \) has infinitely many extensions in \( T \) . Proceeding similarly we get an \( \alpha = \left( {{a}_{0},{a}_{1},\ldots }\right) \) such that for all \( k,\alpha \mid k \) has infinitely many extensions in \( T \) . In particular, \( \alpha \in \left\lbrack T\right\rbrack \) , and the result is proved.
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Yes
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Proposition 1.10.7 (König’s infinity lemma, [57]) Let \( T \) be a finitely splitting, infinite tree on \( A \) . Then \( T \) is ill-founded.
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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, \( \left( {a}_{0}\right) \) exists. By the same argument we get \( {a}_{1} \in A \) such that \( {s}_{1} = \left( {{a}_{0},{a}_{1}}\right) \) has infinitely many extensions in \( T \) . Proceeding similarly we get an \( \alpha = \left( {{a}_{0},{a}_{1},\ldots }\right) \) such that for all \( k,\alpha \mid k \) has infinitely many extensions in \( T \) . In particular, \( \alpha \in \left\lbrack T\right\rbrack \) , and the result is proved.
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Yes
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Proposition 1.10.8 Let \( T \) be a tree on a finite set \( A \) . Then\n\n\[ \left\lbrack T\right\rbrack \neq \varnothing \Leftrightarrow \left( {\forall k \in \mathbb{N}}\right) \left( {\exists u \in T}\right) \left( {\left| u\right| = k}\right) . \]
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Proof. \
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No
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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 \( T \) .
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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 \
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No
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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\[ \left( {\forall u \in T}\right) \left( {\left( {\left( {\forall v \in {T}_{u}\smallsetminus \{ e\} }\right) {P}_{u \land v}}\right) \Rightarrow {P}_{u}}\right) \Rightarrow \left( {\forall u \in T}\right) {P}_{u}. \]
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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 \( v \) of \( w \) in \( T,{P}_{v} \) holds. So, by the hypothesis, \( {P}_{w} \) holds. This is a contradiction. Therefore, \( {P}_{u} \) holds for all \( u \in T \) .
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Yes
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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 \( G : \mathcal{F} \rightarrow X \), there is a unique map \( f : T \rightarrow X \) such that for all \( v \in T \) , \[ f\left( v\right) = G\left( {f}_{v}\right) \] where \( {f}_{v} : {T}_{v} \smallsetminus \{ e\} \rightarrow X \) is the map defined by \[ {f}_{v}\left( u\right) = f\left( {v \hat{} u}\right), u \in {T}_{v} \smallsetminus \{ e\} .
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Proof. (Existence) For \( u \in T \), let \( {P}_{u} \) be the proposition \
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No
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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\[ \nB = \\mathop{\\bigcap }\\limits_{k}\\mathop{\\bigcup }\\limits_{{\\left| s\\right| = k}}\\left\\lbrack {{A}_{s} \\times \\sum \\left( s\\right) }\\right\\rbrack \n\] \n\nThen \( \\mathcal{A}\\left( \\left\\{ {A}_{s}\\right\\} \\right) = {\\pi }_{X}\\left( B\\right) \), where \( {\\pi }_{X} : X \\times {\\mathbb{N}}^{\\mathbb{N}} \\rightarrow X \) is the projection map.
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The proof of the above proposition is routine and is left as an exercise.
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No
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Proposition 1.12.2 For every family \( \mathcal{F} \) of subsets of \( X \) ,
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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) \( {\mathcal{F}}_{\sigma } \subseteq \mathcal{A}\left( \mathcal{F}\right) \) . Let \( \left( {A}_{n}\right) \) be a sequence in \( \mathcal{F} \) . For \( s = \)\n\n\[ \left( {{s}_{0},{s}_{1},\ldots ,{s}_{m - 1}}\right) \in {\mathbb{N}}^{ < \mathbb{N}} \], define \( {B}_{s} = {A}_{{s}_{0}} \) . Then \( \mathcal{A}\left( \left\{ {B}_{s}\right\} \right) = \bigcup {A}_{n} \).\n\n(iii) \( {\mathcal{F}}_{\delta } \subseteq \mathcal{A}\left( \mathcal{F}\right) \) . Let \( \left( {A}_{n}\right) \) be a sequence in \( \mathcal{F} \) . Take\n\n\[{C}_{s} = {A}_{\left| s\right| }, s \in {\mathbb{N}}^{ < \mathbb{N}}.\]\n\nClearly, \( \mathcal{A}\left( \left\{ {C}_{s}\right\} \right) = \bigcap {A}_{n} \).
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Yes
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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
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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 \( x \\in \\mathop{\\bigcap }\\limits_{n}\\mathop{\\bigcup }\\limits_{{\\left| s\\right| = n}}{A}_{s} \) . Conversely, let \( x \\in \) \( \\mathop{\\bigcap }\\limits_{n}\\mathop{\\bigcup }\\limits_{{\\left| s\\right| = n}}{A}_{s} \) . For each \( n \), choose \( {s}_{n} \\in {A}^{ < \\mathbb{N}} \) of length \( n \) such that \( x \\in {A}_{{s}_{n}} \) . Since \( {\\dot{A}}_{s} \\cap {A}_{t} = \\varnothing \) whenever \( s \\bot t \), the \( {s}_{n} \) ’s are compatible. Therefore, there is an \( \\alpha \\in {A}^{\\mathbb{N}} \) such that \( \\alpha \\mid n = {s}_{n} \) for all \( n \) . Thus \( x \\in {\\mathcal{A}}_{A}\\left( \\left\\{ {A}_{s}\\right\\} \\right) \) .
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Yes
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Proposition 1.12.4 If \( A \) is a finite set and \( \left\{ {{A}_{s} : s \in {A}^{ < \mathbb{N}}}\right\} \) regular, then
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\[ {\mathcal{A}}_{A}\left( \left\{ {A}_{s}\right\} \right) = \mathop{\bigcap }\limits_{n}\mathop{\bigcup }\limits_{{\left| s\right| = n}}{A}_{s} \] Proof. We have seen in the proof of 1.12.3 that \[ {\mathcal{A}}_{A}\left( \left\{ {A}_{s}\right\} \right) \subseteq \mathop{\bigcap }\limits_{n}\mathop{\bigcup }\limits_{{\left| s\right| = n}}{A}_{s} \] is always true. To prove the other inclusion, take any \( x \in \mathop{\bigcap }\limits_{n}\mathop{\bigcup }\limits_{{\left| s\right| = n}}{A}_{s} \) . Consider \[ T = \left\{ {s \in {A}^{ < \mathbb{N}} : x \in {A}_{s}}\right\} \] Since \( \left\{ {A}_{s}\right\} \) is regular, \( T \) is a tree. Since \( A \) is finite, the tree \( T \) is finitely splitting. By our hypothesis, it is infinite. Therefore, by König's infinity lemma (1.10.7), \( \left\lbrack T\right\rbrack \neq \varnothing \) . Let \( \alpha \in \left\lbrack T\right\rbrack \) . Then \( x \in {A}_{\alpha \mid n} \) for all \( n \) . Hence, \( x \in {\mathcal{A}}_{A}\left( \left\{ {A}_{s}\right\} \right) \]
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Yes
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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.
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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\[ \n{V}_{m}\bigcap {U}_{\beta }^{c} \neq \varnothing \& {V}_{m} \subseteq {U}_{\beta + 1} \n\]\n\nClearly, \( \beta \rightarrow n\left( \beta \right) \) is one-to-one and the result is proved.
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No
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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\[ u\left( x\right) = \left\{ \begin{array}{ll} 0 & \text{ if }x \in {A}_{0} \\ 1 & \text{ if }x \in {A}_{1} \end{array}\right. \]
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Proof. Let \( d \) be a compatible metric on \( X \) . Take\n\n\[ u\left( x\right) = \frac{d\left( {x,{A}_{0}}\right) }{d\left( {x,{A}_{0}}\right) + d\left( {x,{A}_{1}}\right) }.\]
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Yes
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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 \( A = \) \( {f}^{-1}\left( 0\right) \) .
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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\[ \n{f}_{n}\left( x\right) = \left\{ \begin{array}{ll} 0 & \text{ if }x \in A \\ 1 & \text{ if }x \in X \smallsetminus {U}_{n} \end{array}\right. \n\]\n\nTake \( f = \mathop{\sum }\limits_{0}^{\infty }\frac{1}{{2}^{n + 1}}{f}_{n} \) .
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Yes
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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 continuous maps \( {f}_{n} : X \rightarrow \mathbb{R} \) such that \( f\left( x\right) = \inf {f}_{n}\left( x\right) \) for all \( x \) .
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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 \( {U}_{r} = \mathop{\bigcup }\limits_{n}{F}_{n}^{r} \) . By the Tietze extension theorem, there is a continuous map \( {f}_{n}^{r} : X \rightarrow \lbrack r,\infty ) \) satisfying\n\n\[ \n{f}_{n}^{r}\left( x\right) = \left\{ \begin{array}{ll} r & \text{ if }x \in {F}_{n}^{r}, \\ g\left( x\right) & \text{ if }x \in X \smallsetminus {U}_{r}. \end{array}\right.\n\]\n\nWe claim that\n\n\[ \nf\left( x\right) = \inf \left\{ {{f}_{n}^{r}\left( x\right) : r \in \mathbb{Q}\text{ and }n \in \mathbb{N}}\right\}\n\]\n\nfor all \( x \) . Clearly, \( {f}_{n}^{r}\left( x\right) \geq f\left( x\right) \) for every \( x \in X \) . Fix any \( {x}_{0} \in X \) and \( \epsilon > 0 \) . To complete the proof, we show that for some \( r \) and for some \( n \),\n\n\[ \n{f}_{n}^{r}\left( {x}_{0}\right) < f\left( {x}_{0}\right) + \epsilon .\n\]\n\nTake any rational number \( r \) such that\n\n\[ \nf\left( {x}_{0}\right) < r < f\left( {x}_{0}\right) + \epsilon .\n\]\n\nTwo cases arise: \( g\left( {x}_{0}\right) \leq r \) or \( g\left( {x}_{0}\right) > r \) . If \( g\left( {x}_{0}\right) \leq r \), then \( {x}_{0} \in X \smallsetminus {U}_{r} \) . Hence,\n\n\[ \n{f}_{n}^{r}\left( {x}_{0}\right) = g\left( {x}_{0}\right) < f\left( {x}_{0}\right) + \epsilon \n\]\n\nfor all \( n \) . If \( g\left( {x}_{0}\right) > r \), then \( {x}_{0} \in {U}_{r} \) . Take any \( n \) such that \( {x}_{0} \in {F}_{n}^{r} \) . Then \( {f}_{n}^{r}\left( {x}_{0}\right) < f\left( {x}_{0}\right) + \epsilon \), and our result is proved.
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Yes
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Proposition 2.1.27 The product of countably many second countable (equivalently separable) metric spaces is second countable.
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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 topology, \( \left\{ {{\pi }_{i}^{-1}\left( {U}_{in}\right) : i, n \in \mathbb{N}}\right\} \) is a subbase for \( X \) . Since \( \left\{ {{\pi }_{i}^{-1}\left( {U}_{in}\right) : i, n \in }\right. \) \( \mathbb{N}\} \) is countable, the result follows from 2.1.8.
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Yes
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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 \( \\left( {F}_{n}\\right) \\rightarrow 0 \), the intersection \( \\mathop{\\bigcap }\\limits_{n}{F}_{n} \) is a singleton.
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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 and so convergent. It is easily seen that \( \\lim {x}_{n} \\in \\mathop{\\bigcap }\\limits_{n}{F}_{n} \) . Let \( x \\neq y \) . Then \( d\\left( {x, y}\\right) > 0 \) . Since diameter \( \\left( {F}_{n}\\right) \\rightarrow 0 \), there is an integer \( n \) such that both \( x \) and \( y \) cannot belong to \( {F}_{n} \) . It follows that both \( x \) and \( y \) cannot belong to \( \\bigcap {F}_{n} \) .\n\nTo show the converse, let \( \\left( {x}_{n}\\right) \) be a Cauchy sequence. Put\n\n\[ \n{F}_{n} = \\operatorname{cl}\\left( \\left\\{ {{x}_{m} : m \\geq n}\\right\\} \\right) .\n\]\n\nAs \( \\left( {x}_{n}\\right) \) is Cauchy, \( \\operatorname{diam}\\left( {F}_{n}\\right) \\rightarrow 0 \) . Take \( x \\in \\mathop{\\bigcap }\\limits_{n}{F}_{n} \) . Then \( \\lim {x}_{n} = x \) .
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Yes
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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.
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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 \( \alpha \left( k\right) \in {X}_{k} \) such that \( {\alpha }_{n}\left( k\right) \rightarrow \alpha \left( k\right) \) . By 2.1.26, the sequence \( \left( {\alpha }_{n}\right) \) converges to \( \alpha \) .
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Yes
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Theorem 2.1.32 Any second countable metrizable space \( X \) can be embedded in the Hilbert cube \( \mathbb{H} \) .
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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\[ \n{f}_{nm}\left( x\right) = \left\{ \begin{array}{ll} 0 & \text{ if }x \in \operatorname{cl}\left( {U}_{n}\right) \\ 1 & \text{ if }x \in X \smallsetminus {U}_{m} \end{array}\right.\n\]\n\nBy 2.1.18, such a function exists. Enumerate \( \left\{ {{f}_{nm} : m, n \in \mathbb{N}}\right\} \) as a sequence \( \left( {f}_{k}\right) \) . Define \( f \) on \( X \) by\n\n\[ \nf\left( x\right) = \left( {{f}_{0}\left( x\right) ,{f}_{1}\left( x\right) ,\ldots }\right) ,\;x \in X.\n\]\n\nWe can easily check that \( f \) embeds \( X \) in the Hilbert cube.
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Yes
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Proposition 2.1.34 Every nonempty open set \( U \) in \( \mathbb{R} \) is a countable union of pairwise disjoint nonempty open intervals.
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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 countable. Further, \( U = \mathop{\bigcup }\limits_{{x \in U}}{I}_{x} \) .
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Yes
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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 \( \\mathbb{R} \) .
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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 \( k \\in \\mathbb{N} \), we define integers \( {m}_{k},{n}_{k} \\in \\mathbb{N} \) and real numbers \( {a}_{k},{b}_{k} \\in \\left( {0,1}\\right) \) satisfying the following conditions.\n\n) \( {n}_{0} < {m}_{0} < \\cdots < {n}_{k} < {m}_{k} \), \n\n(ii) \( {a}_{0} < {a}_{1} < \\cdots < {a}_{k} < {b}_{k} < \\cdots {b}_{1} < {b}_{0}, \n\n\\)\n\n(iii)\( {a}_{k} \\in {A}_{{n}_{k}},{b}_{k} \\in {A}_{{m}_{k}}\\text{,}\n\n(iv) for every \( i \\leq {m}_{k},{A}_{i} \\cap \\left( {{a}_{k},{b}_{k}}\\right) = \\varnothing \) .\n\nAssume that we have done this. Take \( a = \\sup {a}_{k} \). Then \( {a}_{n} < a < {b}_{n} \) for all \( n \). Hence, \( a \\notin \\bigcup {A}_{i} \), which is a contradiction.\n\nWe define \( {m}_{k},{n}_{k},{a}_{k} \), and \( {b}_{k} \) by induction. Take \( {n}_{0} = 0 \) and \( {a}_{0} = \) \( \\sup {A}_{{n}_{0}} \). Let \( {m}_{0} \) be the first integer \( m \) such that \( {A}_{m} \\cap \\left( {{a}_{0},1}\\right) \\neq \\varnothing \). Put \( {b}_{0} = \\inf \\left\\lbrack {{A}_{{m}_{0}}\\bigcap \\left( {{a}_{0},1}\\right) }\\right\\rbrack \). Since \( {A}_{{n}_{0}} \) and \( {A}_{{m}_{0}} \) are disjoint and closed, \( {a}_{0} < {b}_{0} \). Note that \( {A}_{i} \\cap \\left( {{a}_{0},{b}_{0}}\\right) = \\varnothing \) for all \( i \\leq {m}_{0} \). Let \( k \\in \\mathbb{N} \) and suppose for every \( i \\leq k,{m}_{i},{n}_{i},{a}_{i} \), and \( {b}_{i} \) satisfying (i)-(iv) have been defined. Take \( {n}_{k + 1} \) to be the first integer \( n \) such that \( {A}_{n} \\cap \\left( {{a}_{k},{b}_{k}}\\right) \\neq \\varnothing \). Put \( {a}_{k + 1} = \) \( \\sup \\left\\lbrack {{A}_{{n}_{k + 1}}\\bigcap \\left( {{a}_{k},{b}_{k}}\\right) }\\right\\rbrack \). Clearly, \( {a}_{k + 1} < {b}_{k} \). Now, let \( {m}_{k + 1} \) be the first integer \( m \) such that \( {A}_{m} \\cap \\left( {{a}_{k + 1},{b}_{k}}\\right) \\neq \\varnothing \). Note that \( {m}_{k + 1} > {n}_{k + 1} \). Take \( {b}_{k + 1} = \) \( \\inf \\left\\lbrack {{A}_{{m}_{k + 1}} \\cap \\left( {{a}_{k + 1},{b}_{k}}\\right) }\\right\\rbrack \).
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Yes
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Theorem 2.2.1 (Alexandrov) Every \( {G}_{\delta } \) subset \( G \) of a completely metrizable (Polish) space \( X \) is completely metrizable (Polish).
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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 the following.\n\n(i) The function \( f \) is one-to-one.\n\n(ii) By 2.1.15 and 2.1.26, \( f \) is continuous.\n\n(iii) Since \( {f}^{-1} \) is \( {\pi }_{1} \mid f\left( G\right) \), it is continuous.\n\n(iv) The set \( f\left( G\right) \) is closed in \( X \times \mathbb{R} \) .\n\nTo see (iv), let \( \left( {x}_{n}\right) \) be a sequence in \( G \) and\n\n\[ f\left( {x}_{n}\right) = \left( {{x}_{n},1/d\left( {{x}_{n}, X \smallsetminus G}\right) }\right) \rightarrow \left( {x, y}\right) . \]\n\nThen, \( {x}_{n} \rightarrow x \) . Hence,\n\n\[ d\left( {{x}_{n}, X \smallsetminus G}\right) \rightarrow d\left( {x, X \smallsetminus G}\right) \]\n\nSince \( 1/d\left( {{x}_{n}, X \smallsetminus G}\right) \rightarrow y, y = 1/d\left( {x, X \smallsetminus G}\right) \) . Hence, \( d\left( {x, X \smallsetminus G}\right) \neq 0 \) . This\n\nimplies that \( x \in G \) and \( \left( {x, y}\right) = f\left( x\right) \in f\left( G\right) \) .\n\nSo, \( G \) is homeomorphic to \( f\left( G\right) \) . As \( f\left( G\right) \) is closed in the completely metrizable space \( X \times \mathbb{R} \), it is completely metrizable. Since \( f \) is a homeomorphism, \( G \) is completely metrizable.\n\nNow consider the case when \( G \) is a \( {G}_{\delta } \) set. Let \( G = \mathop{\bigcap }\limits_{n}{G}_{n} \), where the \( {G}_{n} \) ’s are open. Define \( f : G \rightarrow X \times {\mathbb{R}}^{\mathbb{N}} \) by\n\n\[ f\left( x\right) = \left( {x,\frac{1}{d\left( {x, X \smallsetminus {G}_{0}}\right) },\frac{1}{d\left( {x, X \smallsetminus {G}_{1}}\right) },\ldots }\right) ,\;x \in G. \]\n\nArguing as above, we see that \( f \) embeds \( G \) onto a closed subspace of \( X \times {\mathbb{R}}^{\mathbb{N}} \) , which completes the proof.
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Yes
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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 \( A \) .
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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 \( {O}_{f}\left( x\right) \) the oscillation of \( f \) at \( x \) . Put\n\n\[ \nB = \left\{ {x \in \operatorname{cl}\left( A\right) : {O}_{f}\left( x\right) = 0}\right\} .\n\]\n\nThe set \( B \) is \( {G}_{\delta } \) in \( W \) . To see this, take any \( t > 0 \) and note that for any \( x \in \operatorname{cl}\left( A\right) , \)\n\n\[ \n{O}_{f}\left( x\right) < t \Leftrightarrow \left( {\exists \text{ open }V \ni x}\right) \left( {\operatorname{diameter}\left( {f\left( {A \cap V}\right) }\right) < t}\right) .\n\]\n\nTherefore, the set\n\n\[ \n\left\{ {x \in \operatorname{cl}\left( A\right) : {O}_{f}\left( x\right) < t}\right\}\n\]\n\n\[ \n= \bigcup \{ V\bigcap \operatorname{cl}\left( A\right) : V\text{ open and diameter }\left( {f\left( {A \cap V}\right) }\right) < t\}\n\]\n\nand hence it is open in \( \operatorname{cl}\left( A\right) \) . Since\n\n\[ \nB = \mathop{\bigcap }\limits_{n}\left\{ {x \in \operatorname{cl}\left( A\right) : {O}_{f}\left( x\right) < \frac{1}{n + 1}}\right\}\n\]\n\nand \( \operatorname{cl}\left( A\right) \) is a \( {G}_{\delta } \) set in \( W, B \) is a \( {G}_{\delta } \) set. Since \( f \) is continuous on \( A \), the oscillation of \( f \) at every \( x \in A \) is 0 . Therefore, \( A \subseteq B \) .\n\nWe now define a continuous map \( g : B \rightarrow Z \) that extends \( f \) . Let \( x \in B \) . Take a sequence \( \left( {x}_{n}\right) \) in \( A \) converging to \( x \) . Since \( {O}_{f}\left( x\right) = 0,\left( {f\left( {x}_{n}\right) }\right) \) is a Cauchy sequence in \( \left( {Z,\rho }\right) \) . As \( \left( {Z,\rho }\right) \) is complete, \( \left( {f\left( {x}_{n}\right) }\right) \) is convergent. Put \( g\left( x\right) = \mathop{\lim }\limits_{n}f\left( {x}_{n}\right) \) . The following statements are easy to prove.\n\n(i) The map \( g \) is well-defined.\n\n(ii) It is continuous.\n\n(iii) It extends \( f \) .
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Yes
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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 \( B \) .
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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 \( {g}^{\prime } : {B}^{\prime } \rightarrow X \) of \( g \) . Let\n\n\[ H = \left\{ {\left( {x, y}\right) \in {A}^{\prime } \times Y : y = {f}^{\prime }\left( x\right) }\right\} = \operatorname{graph}\left( {f}^{\prime }\right) \]\n\nand\n\n\[ K = \left\{ {\left( {x, y}\right) \in X \times {B}^{\prime } : x = {g}^{\prime }\left( y\right) }\right\} = \operatorname{graph}\left( {g}^{\prime }\right) . \]\n\nLet \( {A}^{ * } = {\pi }_{1}\left( {H \cap K}\right) \) and \( {B}^{ * } = {\pi }_{2}\left( {H \cap K}\right) \), where \( {\pi }_{1} \) and \( {\pi }_{2} \) are the two projection functions. Note that\n\n\[ \left. {{A}^{ * } = \left\{ {x \in {A}^{\prime } : \left( {x,{f}^{\prime }\left( x\right) }\right) \in K}\right) }\right\} \]\n\nand\n\n\[ \left. \left. {{B}^{ * } = \left\{ {y \in {B}^{\prime } : \left( {{g}^{\prime }\left( y\right), y}\right) }\right) \in H}\right) \right\} . \]\n\nSince \( K \) is closed in \( X \times {B}^{\prime } \) and \( {B}^{\prime } \) is a \( {G}_{\delta }, K \) is a \( {G}_{\delta } \) set. As \( {f}^{\prime } \) is continuous on the \( {G}_{\delta } \) set \( {A}^{\prime },{A}^{ * } \) is a \( {G}_{\delta } \) set. Similarly, we can show that \( {B}^{ * } \) is a \( {G}_{\delta } \) set. It is easy to check that \( {f}^{ * } = {f}^{\prime } \mid {A}^{ * } \) is a homeomorphism from \( {A}^{ * } \) onto \( {B}^{ * } \) that extends \( f \) .
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Yes
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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 \( X \) .
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Proof. The result follows from 2.2.6 by taking \( A = B = Y \) and \( f : A \rightarrow \) \( B \) the identity map.
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No
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Lemma 2.2.9 Every second countable, zero-dimensional metrizable space \( X \) can be embedded in \( \mathcal{C} \) .
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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\[ f\left( x\right) = \left( {{\chi }_{{U}_{0}}\left( x\right) ,{\chi }_{{U}_{1}}\left( x\right) ,{\chi }_{{U}_{2}}\left( x\right) ,\ldots }\right) ,\;x \in X. \]\n\nSince the characteristic function of a clopen set is continuous and since a map into a product space is continuous if its composition with the projection to each of its coordinate spaces is continuous, \( f \) is continuous. Since \( \left\{ {{U}_{n} : n \in \mathbb{N}}\right\} \) is a base for \( X, f \) is one-to-one. Further,\n\n\[ f\left( {U}_{n}\right) = f\left( X\right) \bigcap \{ \alpha \in \mathcal{C} : \alpha \left( n\right) = 1\} . \]\n\nTherefore, \( {f}^{-1} : f\left( X\right) \rightarrow X \) is also continuous. Thus, \( f \) is an embedding of \( X \) in \( \mathcal{C} \) .
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Yes
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Proposition 2.2.11 Every zero-dimensional Polish space is homeomorphic to a \( {G}_{\delta } \) subset of \( \mathcal{C} \) .
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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 \( {\mathbb{N}}^{\mathbb{N}} \) .
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No
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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 \( A \) .
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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, \( \sum \left( {\alpha \mid k}\right) \subseteq {A}^{\mathbb{N}} \smallsetminus \left\lbrack T\right\rbrack \), whence \( {A}^{\mathbb{N}} \smallsetminus \left\lbrack T\right\rbrack \) is open.\n\nConversely, let \( C \) be closed in \( {A}^{\mathbb{N}} \) . Let\n\n\[ T = \{ \alpha \mid k : \alpha \in C\text{ and }k \in \mathbb{N}\} .\n\]\n\nClearly, \( C \subseteq \left\lbrack T\right\rbrack \) . Take any \( \alpha \notin C \) . Since \( C \) is closed, choose a \( k \in \mathbb{N} \) such that \( \sum \left( {\alpha \mid k}\right) \subseteq {A}^{\mathbb{N}} \smallsetminus C \) . Thus \( \alpha \mid k \notin T \) . Hence \( \alpha \notin \left\lbrack T\right\rbrack \) .
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Yes
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Proposition 2.3.6 A continuous image of a compact space is compact.
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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 \( {f}^{-1}\left( {U}_{1}\right) ,{f}^{-1}\left( {U}_{2}\right) ,\ldots ,{f}^{-1}\left( {U}_{n}\right) \) . Hence \( {U}_{1},{U}_{2},\ldots ,{U}_{n} \) cover \( f\left( X\right) \) .
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Yes
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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.
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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.
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Yes
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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.
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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.
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Yes
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Proposition 2.3.11 Every compact metric space \( \left( {X,\rho }\right) \) is complete.
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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.
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Yes
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Proposition 2.3.14 Every compact metrizable space \( X \) is separable and hence second countable.
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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 \( X \) .
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Yes
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Corollary 2.3.15 Every zero-dimensional compact metrizable space \( X \) is homeomorphic to a closed subset of \( \mathcal{C} \) .
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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.
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Yes
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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 \( U \in \mathcal{U} \) .
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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 convergent, \( \left( {x}_{n}\right) \) has a convergent subsequence, converging to \( x \), say. Choose \( U \in \mathcal{U} \) containing \( x \) . Fix \( r > 0 \) such that \( B\left( {x, r}\right) \subseteq U \) . Note that \( {x}_{n} \in B\left( {x, r/2}\right) \) for infinitely many \( n \) . Choose \( {n}_{0} \) such that \( 1/{n}_{0} < r/2 \) and \( {x}_{{n}_{0}} \in B\left( {x, r/2}\right) \) . As diameter \( \left( {A}_{{n}_{0}}\right) < \) \( 1/{n}_{0} < r/2 \n\n\[ \n{A}_{{n}_{0}} \subseteq B\left( {x, r}\right) \subseteq U. \n\]\n\nThis contradiction proves the result.
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Yes
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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.
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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 \( d\left( {x, y}\right) < \delta \) .
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Yes
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Proposition 2.3.18 Every sequentially compact metric space \( \left( {X, d}\right) \) is totally bounded.
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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 \( n > 0,{x}_{n} \notin \mathop{\bigcup }\limits_{{i < n}}B\left( {{x}_{i},\epsilon }\right) \) . Thus for any \( m \neq n, d\left( {{x}_{m},{x}_{n}}\right) \geq \epsilon \) . Such a sequence \( \left( {x}_{n}\right) \) has no convergent subsequence.
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Yes
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Proposition 2.3.19 Every sequentially compact metric space is compact.
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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 \( {U}_{k} \in \mathcal{U} \) containing \( B\left( {{x}_{k},\delta /3}\right) \) . Plainly, \( \left\{ {{U}_{1},{U}_{2},\ldots ,{U}_{n}}\right\} \) is a finite subcover of \( \mathcal{U} \) .
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Yes
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Theorem 2.3.22 A metric space is compact if and only if it is complete and totally bounded.
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Proof. We have already proved the \
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No
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Theorem 2.3.23 The product of a sequence of compact metric spaces is compact.
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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 \( \left( {x}_{n}\right) \) has a convergent subsequence.\n\nSince \( {X}_{0} \) is compact, there is a convergent subsequence \( \left( {{x}_{{n}_{k}^{0}}\left( 0\right) }\right) \) of \( \left( {{x}_{n}\left( 0\right) }\right) \) . Similarly, as \( {X}_{1} \) is compact, there is a convergent subsequence \( \left( {{x}_{{n}_{k}^{1}}\left( 1\right) }\right) \) of \( \left( {{x}_{{n}_{k}^{0}}\left( 1\right) }\right) \) . Proceeding similarly we obtain a double sequence \( \left( {x}_{{n}_{k}^{i}}\right) \) such that\n\n(i) \( {\left( {x}_{{n}_{k}^{i}}\left( i\right) \right) }_{k \in \mathbb{N}} \) is convergent for each \( i \), and\n\n(ii) \( {\left( {x}_{{n}_{k}^{i + 1}}\right) }_{k \in \mathbb{N}} \) is a subsequence of \( {\left( {x}_{{n}_{k}^{i}}\right) }_{k \in \mathbb{N}} \) .\n\nDefine \( {y}_{i} = {x}_{{n}_{i}^{i}}, i \in \mathbb{N} \) . As \( {y}_{i}\left( k\right) \) is convergent for each \( k,\left( {y}_{i}\right) \) is a convergent subsequence of \( \left( {x}_{n}\right) \) .
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Yes
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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\[ \mathop{\limsup }\limits_{n}{f}_{n}\left( {x}_{n}\right) \leq f\left( x\right) \]
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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-semicontinuous, and \( \left( {h}_{n}\right) \) decreases to \( h \) . By 2.3.27, \( {h}_{n} \rightarrow h \) uniformly on \( X \) . Hence,\n\n\[ \mathop{\limsup }\limits_{n}{f}_{n}\left( {x}_{n}\right) \leq \mathop{\lim }\limits_{n}{h}_{n}\left( {x}_{n}\right) = h\left( x\right) \leq f\left( x\right) + \epsilon . \]\n\nSince \( \epsilon > 0 \) was arbitrary, our result is proved.
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Yes
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Theorem 2.3.30 Every locally compact metrizable space \( X \) is completely metrizable.
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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 \( \left( {X, d}\right) \) . Then \( X \) is a locally compact dense subspace of \( \widehat{X} \) . By 2.3.31, \( X \) is open in \( \widehat{X} \) . By 2.2.1, \( X \) is completely metrizable.\n\nThe 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 \( \operatorname{cl}\left( U\right) \subseteq \operatorname{cl}\left( U\right) \bigcap Y \subseteq Y \) . Choose an open set \( V \) in \( X \) such that \( U = V \cap Y \) . Since \( Y \) is dense and \( V \) open, \( \operatorname{cl}\left( V\right) = \operatorname{cl}\left( {V \cap Y}\right) \) . Thus we have\n\n\[ \n x \in V \subseteq \operatorname{cl}\left( V\right) = \operatorname{cl}\left( {V\bigcap Y}\right) = \operatorname{cl}\left( U\right) \subseteq Y.\n\]\n\nWe have shown that for every \( x \in Y \) there is an open set \( V \) in \( X \) such that \( x \in V \subseteq Y \) . Therefore, \( Y \) is open.
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Yes
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Lemma 2.3.31 Let \( Y \) be a locally compact dense subspace of a metrizable space \( X \) . Then \( Y \) is open in \( X \) .
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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 \( \operatorname{cl}\left( U\right) \subseteq \operatorname{cl}\left( U\right) \bigcap Y \subseteq Y \) . Choose an open set \( V \) in \( X \) such that \( U = V \cap Y \) . Since \( Y \) is dense and \( V \) open, \( \operatorname{cl}\left( V\right) = \operatorname{cl}\left( {V \cap Y}\right) \) . Thus we have\n\n\[ \n x \in V \subseteq \operatorname{cl}\left( V\right) = \operatorname{cl}\left( {V \bigcap Y}\right) = \operatorname{cl}\left( U\right) \subseteq Y. \n\]\n\nWe have shown that for every \( x \in Y \) there is an open set \( V \) in \( X \) such that \( x \in V \subseteq Y \) . Therefore, \( Y \) is open.
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Yes
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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.
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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 cover \( {\mathcal{U}}_{l} = \left\{ {{U}_{0},{U}_{1},\ldots }\right\} \) such that diameter \( \left( {U}_{i}\right) < 1/l \) for each \( i \) . Fix such an \( {\mathcal{U}}_{l} \) for each \( l \) . Put\n\n\[ \n{C}_{m, n} = \{ f \in C\left( {X, Y}\right) : \forall x, y\left( {d\left( {x, y}\right) < 1/m \Rightarrow \rho \left( {f\left( x\right), f\left( y\right) }\right) < 1/n}\right) \} .\n\]\n\nFor each \( k \) -tuple \( s = \left( {{i}_{1},{i}_{2},\ldots ,{i}_{k}}\right) \), whenever possible, choose an \( {f}_{s} \in \) \( {C}_{m, n} \) such that \( {f}_{s}\left( {x}_{j}\right) \in {U}_{{i}_{j}} \) for all \( 1 \leq j \leq k \) . Let \( {D}_{m, n, l} \) be the collection of all these \( {f}_{s} \) and set \( {D}_{m, n} = \mathop{\bigcup }\limits_{{l > 0}}{D}_{m, n, l} \) .\n\nWe claim that for all \( f \in {C}_{m, n} \) and all \( \epsilon > 0 \) there is a \( g \in {D}_{m, n} \) such that \( \rho \left( {f\left( y\right), g\left( y\right) }\right) < \epsilon \) for every \( y \in {X}_{m} \) . To see this, take \( l > 1/\epsilon \) and choose \( {i}_{1},{i}_{2},\ldots ,{i}_{k} \) such that \( f\left( {x}_{j}\right) \in {U}_{{i}_{j}} \) for all \( 1 \leq j \leq k \) . Thus \( {f}_{s} \) exists for \( s = \left( {{i}_{1},{i}_{2},\ldots ,{i}_{k}}\right) \) . Take \( g = {f}_{s} \) .\n\nSet \( D = \mathop{\bigcup }\limits_{{m, n}}{D}_{m, n} \) . Note that \( D \) is countable. We show that \( D \) is dense in \( C\left( {X, Y}\right) \) . Take \( f \in C\left( {X, Y}\right) \) and \( \epsilon > 0 \) . Take any \( n > 3/\epsilon \) . Since \( f \) is uniformly continuous, \( f \in {C}_{m, n} \) for some \( m \) . We choose \( g \in {D}_{m, n} \) such that \( \rho \left( {f\left( y\right), g\left( y\right) }\right) < \epsilon /3 \) for \( y \in {X}_{m} \) . Since \( {X}_{m} \) is a \( 1/m \) -net, by the triangle inequality we see that \( \rho \left( {f\left( x\right), g\left( x\right) }\right) < \epsilon \) for all \( x \in X \) . So, \( D \) is dense, and our theorem is proved.
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Yes
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Proposition 2.4.6 \( \operatorname{irr}\left( n\right) \) is Polish.
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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\[ A\text{ is irreducible } \Leftrightarrow \forall m\exists k\left| {{A}_{0} - {P}_{k}\left( {A,{A}^{ * }}\right) }\right| < {2}^{-m}. \]\n\nSo,\n\n\[ \operatorname{irr}\left( n\right) = \mathop{\bigcap }\limits_{m}{G}_{m} \]\n\nwhere\n\n\[ {G}_{m} = \left\{ {A \in {M}_{n} : \left| {{A}_{0} - {P}_{k}\left( {A,{A}^{ * }}\right) }\right| < {2}^{-m}}\right\} \text{ for some }k\} . \]\n\nClearly, \( {G}_{m} \) is open. Hence, \( \operatorname{irr}\left( n\right) \) is a \( {G}_{\delta } \) set.
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Yes
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Proposition 2.4.6 \( \operatorname{irr}\left( n\right) \) is Polish.
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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\[ A\text{ is irreducible } \Leftrightarrow \forall m\exists k\left| {{A}_{0} - {P}_{k}\left( {A,{A}^{ * }}\right) }\right| < {2}^{-m}. \]\n\nSo,\n\n\[ \operatorname{irr}\left( n\right) = \mathop{\bigcap }\limits_{m}{G}_{m} \]\n\nwhere\n\n\[ {G}_{m} = \left\{ {A \in {M}_{n} : \left| {{A}_{0} - {P}_{k}\left( {A,{A}^{ * }}\right) }\right| < {2}^{-m}}\right\} \text{ for some }k\} . \]\n\nClearly, \( {G}_{m} \) is open. Hence, \( \operatorname{irr}\left( n\right) \) is a \( {G}_{\delta } \) set.
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Yes
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Proposition 2.4.15 If \( \left( {X, d}\right) \) is a complete metric space, so is \( \left( {K\left( X\right) ,{\delta }_{H}}\right) \) .
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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 compact. Further, they have the finite intersection property. Therefore, \( K \) is nonempty and compact. We show that \( {\delta }_{H}\left( {{K}_{n}, K}\right) \rightarrow 0 \) as \( n \rightarrow \infty \) .\n\nFix \( \epsilon > 0 \) . Choose \( N \) such that for \( m, n \geq N,{\delta }_{H}\left( {{K}_{m},{K}_{n}}\right) < \epsilon /2 \) . We show that \( {\delta }_{H}\left( {{K}_{n}, K}\right) < \epsilon \) for every \( n \geq N \) . Fix \( n \geq N \) .\n\n(i) Let \( x \in K \) . As \( x \in \operatorname{cl}\left( {\mathop{\bigcup }\limits_{{i > n}}{K}_{i}}\right) \), there exist \( i \geq n \) and \( {x}_{i} \in {K}_{i} \) such that \( d\left( {x,{x}_{i}}\right) < \epsilon /2 \) . Since \( {\delta }_{H}\left( {{K}_{i},{K}_{n}}\right) < \epsilon /2 \), take \( y \in {K}_{n} \) such that \( d\left( {y,{x}_{i}}\right) < \epsilon /2 \) . By the triangle inequality \( d\left( {x, y}\right) < \epsilon \) . Thus, \( d\left( {x,{K}_{n}}\right) < \epsilon \) for every \( x \in K \) . So, \( K \subseteq B\left( {{K}_{n},\epsilon }\right) \).\n\n(ii) Let \( x \in {K}_{n} \) . We prove that \( d\left( {x, K}\right) < \epsilon \) . This would show that \( {K}_{n} \subseteq B\left( {K,\epsilon }\right) \) . For each \( i \geq N,{\delta }_{H}\left( {{K}_{i},{K}_{n}}\right) < \epsilon /2 \) . Choose \( {x}_{i} \in {K}_{i} \) such that \( d\left( {x,{x}_{i}}\right) < \epsilon /2 \) . Since \( \operatorname{cl}\left( {\mathop{\bigcup }\limits_{{i > N}}{K}_{i}}\right) \) is compact, \( \left( {x}_{i}\right) \) has a convergent subsequence converging to \( y \), say. Clearly, \( y \in K \), and \( d\left( {x, y}\right) \leq \epsilon /2 < \epsilon \) . ∎
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Yes
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Proposition 2.4.17 If \( X \) is compact metrizable, so is \( K\left( X\right) \) .
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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.
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No
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Proposition 2.4.17 If \( X \) is compact metrizable, so is \( K\left( X\right) \) .
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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.
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No
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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 \( X \) .
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Proof. For the \
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No
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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.
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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, \( {U}_{0} \cap V \) is nonempty. Choose an open ball \( {B}_{0} \) of diameter \( \mathrm{j}1 \) such that \( \operatorname{cl}\left( {B}_{0}\right) \subseteq {U}_{0} \cap V \) . Since \( {U}_{1} \) is dense, by the same argument we get an open ball \( {B}_{1} \) of diameter \( < 1/2 \) such that \( \operatorname{cl}\left( {B}_{1}\right) \subseteq {U}_{1}\bigcap {B}_{0} \) . Proceeding similarly, we define a sequence \( \left( {B}_{n}\right) \) of open balls in \( X \) such that for each \( n \) ,\n\n(i) \( \operatorname{diameter}\left( {B}_{n}\right) < 1/{2}^{n} \) ,\n\n(ii) \( \operatorname{cl}\left( {B}_{0}\right) \subseteq {U}_{0} \cap V \), and\n\n(iii) \( \operatorname{cl}\left( {B}_{n + 1}\right) \subseteq {U}_{n + 1} \cap {B}_{n} \) .\n\nSince \( \left( {X, d}\right) \) is a complete metric space, by 2.1.29, \( \mathop{\bigcap }\limits_{n}{B}_{n} = \mathop{\bigcap }\limits_{n}\operatorname{cl}\left( {B}_{n}\right) \) is a singleton, say \( \{ x\} \) . Clearly, \( x \in \mathop{\bigcap }\limits_{n}{U}_{n} \cap V \) .
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Yes
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Corollary 2.5.6 Every completely metrizable space is of second category in itself.
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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 \( \mathop{\bigcap }\limits_{n}{U}_{n} = \varnothing \) . This contradicts the Baire category theorem.
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Yes
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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.
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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 \( n \). Fix \( z \\in \\operatorname{int}\\left( {K}_{n}\\right) \). For any \( x \\in G \), \( \\left( {x \\cdot {z}^{-1}}\\right) {K}_{n} \) is a compact neighborhood of \( x \) where, for \( A \\subseteq G \) and \( g \\in G \), \( {gA} = \\{ g \\cdot h : h \\in A\\} \). So, \( G \) is locally compact.\n\nThe converse follows from 2.3.33.
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Yes
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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.
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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 \( n \) . Fix \( z \\in \\operatorname{int}\\left( {K}_{n}\\right) \) . For any \( x \\in G \) , \( \\left( {x \\cdot {z}^{-1}}\\right) {K}_{n} \) is a compact neighborhood of \( x \) where, for \( A \\subseteq G \) and \( g \\in G \) , \( {gA} = \\{ g \\cdot h : h \\in A\\} \) . So, \( G \) is locally compact.\n\nThe converse follows from 2.3.33.
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Yes
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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 \( G \) .
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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 \( {G}^{\prime } \) . As it is also dense in \( {G}^{\prime } \), it is comeager in \( {G}^{\prime } \) . Suppose \( H \neq {G}^{\prime } \) . Take any \( x \in {G}^{\prime } \smallsetminus H \) . Then the coset \( {xH} \) is comeager in \( {G}^{\prime } \) and disjoint from \( H \) . By the Baire category theorem, \( {G}^{\prime } \) cannot have two disjoint comeager subsets. This contradiction shows that \( H = {G}^{\prime } \) .
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Yes
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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 \( U \). (ii) If each \( {U}_{i} \) is meager in \( X \), so is \( U \).
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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( {\mathop{\bigcup }\limits_{i}\operatorname{int}\left( {\operatorname{cl}\left( {U}_{i}\right) }\right) }\right) .\n\nProof of the lemma. Since \( {U}_{i} \subseteq U,\left( {i \in I}\right) \),\n\n\[ \operatorname{int}\left( {\operatorname{cl}\left( {U}_{i}\right) }\right) \subseteq \operatorname{int}\left( {\operatorname{cl}\left( U\right) }\right) .\n\nTherefore,\n\n\[ \operatorname{cl}\left( {\operatorname{int}\left( {\operatorname{cl}\left( U\right) }\right) }\right) \supseteq \operatorname{cl}\left( {\mathop{\bigcup }\limits_{i}\operatorname{int}\left( {\operatorname{cl}\left( {U}_{i}\right) }\right) }\right)\n\nThe reverse inclusion follows from\n\n\[ \operatorname{int}\left( {\operatorname{cl}\left( U\right) }\right) \subseteq \operatorname{cl}\left( {\mathop{\bigcup }\limits_{i}\operatorname{int}\left( {\operatorname{cl}\left( {U}_{i}\right) }\right) }\right)\n\nwhich we show now. We make two observations first.\n\n(i) Take any \( i \in I \) . Since \( {U}_{i} \) is open in \( U \),\n\n\[ {U}_{i} = U \smallsetminus \operatorname{cl}\left( {U \smallsetminus {U}_{i}}\right) \subseteq X \smallsetminus \operatorname{cl}\left( {U \smallsetminus {U}_{i}}\right)\n\nTherefore,\n\n\[ U \subseteq \mathop{\bigcup }\limits_{i}\left( {X \smallsetminus \operatorname{cl}\left( {U \smallsetminus {U}_{i}}\right) }\right)\n\n(ii) Since \( \operatorname{int}\left( {\operatorname{cl}\left( U\right) }\right) \smallsetminus \operatorname{cl}\left( {U \smallsetminus {U}_{i}}\right) \subseteq \operatorname{cl}\left( U\right) \smallsetminus \operatorname{cl}\left( {U \smallsetminus {U}_{i}}\right) \subseteq \operatorname{cl}\left( {U}_{i}\right) \),\n\n\[ \operatorname{int}\left( {\operatorname{cl}\left( U\right) }\right) \smallsetminus \operatorname{cl}\left( {U \smallsetminus {U}_{i}}\right) \subseteq \operatorname{int}\left( {\operatorname{cl}\left( {U}_{i}\right) }\right) .\n\nNow,\n\n\[ \operatorname{int}\left( {\operatorname{cl}\left( U\right) }\right) = \operatorname{int}\left( {\operatorname{cl}\left( U\right) }\right) \cap \operatorname{cl}\left( U\right)\n\n\[ \subseteq \operatorname{cl}
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Yes
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Lemma 2.5.17 Let \( X,{U}_{i}\left( {i \in I}\right) \), and \( U \) satisfy the hypothesis of the theorem. Then \[ \operatorname{cl}\left( {\operatorname{int}\left( {\operatorname{cl}\left( U\right) }\right) }\right) = \operatorname{cl}\left( {\mathop{\bigcup }\limits_{i}\operatorname{int}\left( {\operatorname{cl}\left( {U}_{i}\right) }\right) }\right) . \]
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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, \[ \operatorname{cl}\left( {\operatorname{int}\left( {\operatorname{cl}\left( U\right) }\right) }\right) \supseteq \operatorname{cl}\left( {\mathop{\bigcup }\limits_{i}\operatorname{int}\left( {\operatorname{cl}\left( {U}_{i}\right) }\right) }\right) \] The reverse inclusion follows from \[ \operatorname{int}\left( {\operatorname{cl}\left( U\right) }\right) \subseteq \operatorname{cl}\left( {\mathop{\bigcup }\limits_{i}\operatorname{int}\left( {\operatorname{cl}\left( {U}_{i}\right) }\right) }\right) \] which we show now. We make two observations first. (i) Take any \( i \in I \) . Since \( {U}_{i} \) is open in \( U \) , \[ {U}_{i} = U \smallsetminus \operatorname{cl}\left( {U \smallsetminus {U}_{i}}\right) \subseteq X \smallsetminus \operatorname{cl}\left( {U \smallsetminus {U}_{i}}\right) \] Therefore, \[ U \subseteq \mathop{\bigcup }\limits_{i}\left( {X \smallsetminus \operatorname{cl}\left( {U \smallsetminus {U}_{i}}\right) }\right) \] (ii) Since \( \operatorname{int}\left( {\operatorname{cl}\left( U\right) }\right) \smallsetminus \operatorname{cl}\left( {U \smallsetminus {U}_{i}}\right) \subseteq \operatorname{cl}\left( U\right) \smallsetminus \operatorname{cl}\left( {U \smallsetminus {U}_{i}}\right) \subseteq \operatorname{cl}\left( {U}_{i}\right) \), \[ \operatorname{int}\left( {\operatorname{cl}\left( U\right) }\right) \smallsetminus \operatorname{cl}\left( {U \smallsetminus {U}_{i}}\right) \subseteq \operatorname{int}\left( {\operatorname{cl}\left( {U}_{i}\right) }\right) . \] Now, \[ \operatorname{int}\left( {\operatorname{cl}\left( U\right) }\right) = \operatorname{int}\left( {\operatorname{cl}\left( U\right) }\right) \cap \operatorname{cl}\left( U\right) \] \[ \subseteq \operatorname{cl}\left( {\operatorname{int}\left( {\operatorname{cl}\left( U\right) }\right) \bigcap U}\right) \] \[ \subseteq \;\mathrm{{cl}}\left( {\mathrm{{int}}\left( {\mathrm{{cl}}\left( U\right) }\right) \bigcap \mathop{\bigcup }\limits_{i}\left( {X \smallsetminus \mathrm{{cl}}\left( {U \smallsetminus {U}_{i}}\right) }\right) }\right) \] (by (i)) \[ = \mathrm{{cl}}\left( {\mathop{\bigcup }\limits_{i}(\mathrm{{int}}\left( {\mathrm{{cl}}\left( U\right) }\right) \smallsetminus \mathrm{{cl}}\left( {U \smallsetminus {U}_{i}}\right) }\right) \] \[ \subseteq \mathrm{{cl}}\left( {\mathop{\bigcup }\limits_{i}\mathrm{{int}}\left( {\mathrm{{cl}}\left( {U}_{i}\right) }\right) }\right) \] (by (ii)) The proof of the lemma is complete.
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Yes
|
Proposition 2.6.1 Every dense-in-itself Polish space \( X \) contains a homeomorph of \( \mathcal{C} \) .
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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\[ s \bot t \Rightarrow \operatorname{cl}\left( {U}_{s}\right) \bigcap \operatorname{cl}\left( {U}_{t}\right) = \varnothing . \]\n\nAssuming that such a system of sets \( \left\{ {{U}_{s} : s \in {2}^{ < \mathbb{N}}}\right\} \) exists, define \( {F}_{s} = \operatorname{cl}\left( {U}_{s}\right), s \in {2}^{ < \mathbb{N}} \) . Then \( \left\{ {{F}_{s} : s \in {2}^{ < \mathbb{N}}}\right\} \) is a Cantor scheme on \( X \) , and so \( X \) contains a homeomorph of the Cantor set by (iv).\n\nWe define \( \left\{ {{U}_{s} : s \in {2}^{ < \mathbb{N}}}\right\} \) by induction on the length of \( s \) . Take \( {U}_{e} = X \) . Suppose for some \( s \in {2}^{ < \mathbb{N}},{U}_{s} \) has been defined and is a nonempty open set. Since \( X \) is dense-in-itself, there exist two distinct points \( {x}_{0},{x}_{1} \) in \( {U}_{s} \) . Choose open sets \( {U}_{{s}^{ \frown }0},{U}_{{s}^{ \frown }1} \), containing \( {x}_{0},{x}_{1} \) respectively, of diameters \( \leq {2}^{-\left( {\left| s\right| + 1}\right) } \) whose closures are disjoint and contained in \( {U}_{s} \) .
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Yes
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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 \( Y \cap Z = \varnothing \) .
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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 \( Y = X \smallsetminus Z \) .
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No
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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.
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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 set \( U \) in \( Y \) has uncountably many inequivalent elements. If necessary, we replace \( X \) by \( Y \) and assume that every nonempty open set has uncountably many inequivalent elements.\n\nWe now define a system \( \left\{ {{U}_{s} : s \in {2}^{ < \mathbb{N}}}\right\} \) of nonempty open sets such that\n\n(i) \( \operatorname{diameter}\left( {\operatorname{cl}\left( {U}_{s}\right) }\right) \leq \frac{1}{{2}^{\left| s\right| }} \) ;\n\n(ii) \( \operatorname{cl}\left( {U}_{s \cap \epsilon }\right) \subseteq {U}_{s} \) for \( \epsilon = 0 \) or 1 ; and\n\n(iii) \( s \bot t \Rightarrow E \cap \left( {{F}_{s} \times {F}_{t}}\right) = \varnothing \), where, \( {F}_{s} = \operatorname{cl}\left( {U}_{s}\right) .\n\nSuppose such a system has been defined. Take \( D = {\mathcal{A}}_{2}\left( \left\{ {F}_{s}\right\} \right) \) . Then \( D \) is a homeomorph of the Cantor set. Let \( \alpha \neq \beta \) be two elements of \( D \) . So there exists an \( n \) such that \( \alpha \left| {n \neq \beta }\right| n \) . As \( \alpha \in {F}_{\alpha \mid n} \) and \( \beta \in {F}_{\beta \mid n} \), they are inequivalent by (iii).\n\nThe definition of \( \left\{ {{U}_{s} : s \in {2}^{ < \mathbb{N}}}\right\} \) . Put \( {U}_{e} = X \) . Take two inequivalent elements \( {x}_{0} \) and \( {x}_{1} \) . Then \( \left( {{x}_{0},{x}_{1}}\right) \notin E \) . Since \( E \) is closed, we get open sets \( {U}_{0} \ni {x}_{0} \) and \( {U}_{1} \ni {x}_{1} \) of diameters less than \( 1/2 \) such that\n\n\[ \left( {\operatorname{cl}\left( {U}_{0}\right) \times \operatorname{cl}\left( {U}_{1}\right) }\right) \bigcap E = \varnothing . \]\n\nSuppose \( {U}_{s} \) has been defined for all \( s \) of length less than or equal to \( n \) satisfying conditions (i) to (iii). Fix an \( s \) of length \( n \) . Choose inequivalent elements \( {y}_{0} \) and \( {y}_{1} \) in \( {U}_{s} \) . Using the same arguments, choose open sets \( {U}_{s}{ \hat{} }_{0} \) and \( {U}_{s \cap 1} \) of diameters less than \( 1/{2}^{n + 1} \) such that\n\n\[ {y}_{\epsilon } \in {U}_{s \hat{} \epsilon } \subseteq \operatorname{cl}\left( {U}_{s \hat{} \epsilon }\right) \subseteq {U}_{s} \]\n\n\( \epsilon = 0 \) or 1, and\n\n\[ \left( {\operatorname{cl}\left( {U}_{s \cap 0}\right) \times \operatorname{cl}\left( {U}_{s \cap 1}\right) }\right) \cap E = \varnothing . \]\n\nOur construction of the system \( \left\{ {{U}_{s} : s \in {2}^{ < \mathbb{N}}}\right\} \) is complete.
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Yes
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Theorem 2.6.9 Every Polish space \( X \) is a one-to-one, continuous image of a closed subset \( D \) of \( {\mathbb{N}}^{\mathbb{N}} \) .
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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 family \( \left\{ {{F}_{s} : s \in {\mathbb{N}}^{ < \mathbb{N}}}\right\} \) by induction on \( \left| s\right| \) such that each \( {F}_{s} \) is an \( {F}_{\sigma } \) set. Suppose \( {F}_{s} \) has been defined. Write \( {F}_{s} = \mathop{\bigcup }\limits_{i}{C}_{i} \) where \( \left\{ {C}_{i}\right\} \) is a sequence of closed sets of diameter less than \( {2}^{-\left( {\left| s\right| + 1}\right) } \) . Put \( {F}_{{s}^{ \frown }i} = {C}_{i} \smallsetminus {C}_{i - 1} \) . (We take \( {C}_{-1} = \varnothing \) .) Since an open set in a metrizable space is an \( {F}_{\sigma } \) set, so is \( {F}_{s \cap i} \) . The proof is complete.
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Yes
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Theorem 2.6.10 Every compact metric space \( X \) is a continuous image of a zero-dimensional compact metric space \( Z \) .
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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 \( {F}_{s} \) such that\n\n(i) \( {F}_{e} = X \) ;\n\n(ii) \( {F}_{s} = \mathop{\bigcup }\limits_{{i \leq {n}_{\left| s\right| }}}{F}_{{s}^{ \frown }i} \) ;\n\n(iii) \( \operatorname{diameter}\left( {F}_{s}\right) \leq {2}^{-\left| s\right| } \) .\n\nTo define such a family we proceed by induction. As \( X \) is compact, there is an \( {n}_{0} \in \mathbb{N} \) and a finite open cover \( \left\{ {{U}_{0}^{e},{U}_{1}^{e},\ldots ,{U}_{{n}_{0}}^{e}}\right\} \) of \( X \) such that the diameter of each \( {U}_{i}^{e} \) is less than 1 . Take\n\n\[ {F}_{i} = \operatorname{cl}\left( {U}_{i}^{e}\right) ,\;1 \leq i \leq {n}_{0}. \]\n\nLet \( k \in \mathbb{N} \) . Suppose \( {n}_{0},{n}_{1},\ldots ,{n}_{k} \) and sets \( {F}_{s} \) for \( s \in \left\{ {0,1,\ldots ,{n}_{0}}\right\} \times \) \( \cdots \times \left\{ {0,1,\ldots ,{n}_{k}}\right\} \) satisfying conditions (i)-(iii) have been defined. Fix \( s \in \) \( \left\{ {0,1,\ldots ,{n}_{0}}\right\} \times \cdots \times \left\{ {0,1,\ldots ,{n}_{k}}\right\} \) . As \( {F}_{s} \) is compact, we obtain a finite open cover \( \left\{ {{U}_{i}^{s} : i \leq {n}_{s}}\right\} \) of \( {F}_{s} \) such that diameter \( \left( {U}_{i}^{s}\right) < {2}^{-\left( {k + 1}\right) } \) . Since there are only finitely many sequences of length \( k \), we can assume that there exist \( {n}_{k + 1} \) such that \( {n}_{s} = {n}_{k + 1} \) for all \( s \) . Put \( {F}_{{s}^{ \land }i} = \operatorname{cl}\left( {U}_{i}^{s}\right) \cap {F}_{s} \) .\n\nTo complete the proof, take\n\n\[ Z = \left\{ {0,1,\ldots ,{n}_{0}}\right\} \times \left\{ {0,1,\ldots ,{n}_{1}}\right\} \times \cdots \]\n\nwith the product of discrete topologies. For \( \alpha \in Z \), take \( f\left( \alpha \right) \) to be the unique element of \( \mathop{\bigcap }\limits_{n}{F}_{\alpha \mid n} \) . As before, we see that \( f : Z \rightarrow X \) is continuous and onto.
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Yes
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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 \( X \) .
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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 \( \alpha \in C \) . Suppose \( \alpha \notin C \) . As \( C \) is closed, there is an integer \( k \) such that \( \sum \left( {\alpha \mid k}\right) \cap C = \varnothing \) . Let \( k \) be the largest natural number such that \( C\bigcap \sum \left( {\alpha \mid k}\right) \neq \varnothing \) . Define \( f\left( \alpha \right) = {x}_{\alpha \mid k} \) .\n\nWe now show that \( f \) is continuous at every \( \alpha \in X \) . Let \( \alpha \notin C \) and \( f\left( \alpha \right) = \beta \) . Let \( k \) be the least natural number such that \( C\bigcap \sum \left( {\alpha \mid k}\right) = \varnothing \) . Then \( f \equiv \beta \) on \( \sum \left( {\alpha \mid k}\right) \) . So, \( f \) is continuous at \( \alpha \) .\n\nNow assume that \( \alpha \in C \) . Then \( f\left( \alpha \right) = \alpha \) and \( f\left( {\sum \left( {\alpha \mid k}\right) }\right) \subseteq \sum \left( {\alpha \mid k}\right) \) for all \( k \) . So, \( f \) is continuous at \( \alpha \), and our result is proved.
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Yes
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Theorem 2.6.14 Every zero-dimensional compact, dense-in-itself metric space is homeomorphic to \( \mathcal{C} \) .
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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 \( \left\{ {{C}_{s} : s \in {2}^{ < \mathbb{N}}}\right\} \) . Since \( X \) is perfect and zero-dimensional, we can write \( X = {X}_{1} \cup \cdots \cup {X}_{n} \), where \( n > 1 \) and the \( {X}_{i} \) are pairwise-disjoint nonempty clopen sets of diameter less than \( 1/2 \) . Put \( {C}_{e} = X,{C}_{0} = \mathop{\bigcup }\limits_{{i > 1}}{X}_{i},{C}_{1} = {X}_{1},{C}_{00} = \mathop{\bigcup }\limits_{{i > 2}}{X}_{i},{C}_{01} = {X}_{2} \), etc. Thus we\n\nhave\n\[ \n{C}_{s} = \left\{ \begin{array}{ll} \mathop{\bigcup }\limits_{{i > j}}{X}_{i} & \text{ if }s = {0}^{j}\& j < n, \\ {X}_{j + 1} & \text{ if }s = {0}^{j}{}^{ \frown }1\& j < n - 1. \end{array}\right. \n\]\n\nFor the next stage of construction, fix \( i,1 \leq i \leq n \) . Let \( {C}_{s} = {X}_{i} \) . Note that \( {C}_{s} \) is perfect and zero-dimensional. Write \( {C}_{s} \) as a finite union of pairwise-disjoint nonempty clopen sets \( {Y}_{1},{Y}_{2},\ldots ,{Y}_{m} \) of diameter less than \( 1/3 \) . Repeat the above process replacing \( X \) by \( {C}_{s} \) and \( {X}_{1},{X}_{2},\ldots ,{X}_{n} \) by \( {Y}_{1},{Y}_{2},\ldots ,{Y}_{m} \) to get \( {C}_{{s}^{ \frown }t} \) for \( t = {0}^{j},1 \leq j \leq m - 1 \), or \( t = {0}^{j}{}^{ \frown }1,0 \leq j \leq \) \( m - 2 \) . Continuing this process, we get the required Cantor scheme.
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Yes
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Theorem 2.6.14 Every zero-dimensional compact, dense-in-itself metric space is homeomorphic to \( \mathcal{C} \) .
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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 \( \left\{ {{C}_{s} : s \in {2}^{ < \mathbb{N}}}\right\} \) . Since \( X \) is perfect and zero-dimensional, we can write \( X = {X}_{1} \cup \cdots \cup {X}_{n} \), where \( n > 1 \) and the \( {X}_{i} \) are pairwise-disjoint nonempty clopen sets of diameter less than \( 1/2 \) . Put \( {C}_{e} = X,{C}_{0} = \mathop{\bigcup }\limits_{{i > 1}}{X}_{i},{C}_{1} = {X}_{1},{C}_{00} = \mathop{\bigcup }\limits_{{i > 2}}{X}_{i},{C}_{01} = {X}_{2} \), etc. Thus we\n\nhave\n\[ \n{C}_{s} = \left\{ \begin{array}{ll} \mathop{\bigcup }\limits_{{i > j}}{X}_{i} & \text{ if }s = {0}^{j}\& j < n, \\ {X}_{j + 1} & \text{ if }s = {0}^{j}{}^{ \frown }1\& j < n - 1. \end{array}\right. \n\]\n\nFor the next stage of construction, fix \( i,1 \leq i \leq n \) . Let \( {C}_{s} = {X}_{i} \) . Note that \( {C}_{s} \) is perfect and zero-dimensional. Write \( {C}_{s} \) as a finite union of pairwise-disjoint nonempty clopen sets \( {Y}_{1},{Y}_{2},\ldots ,{Y}_{m} \) of diameter less than \( 1/3 \) . Repeat the above process replacing \( X \) by \( {C}_{s} \) and \( {X}_{1},{X}_{2},\ldots ,{X}_{n} \) by \( {Y}_{1},{Y}_{2},\ldots ,{Y}_{m} \) to get \( {C}_{{s}^{ \frown }t} \) for \( t = {0}^{j},1 \leq j \leq m - 1 \), or \( t = {0}^{j}{}^{ \frown }1,0 \leq j \leq \) \( m - 2 \) . Continuing this process, we get the required Cantor scheme.
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Yes
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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 \( y \in A \) .
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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.
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Yes
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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 \( y \in A \) .
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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.
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Yes
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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 \( A \\in \\sigma \\left( {\\mathcal{G}}^{\\prime }\\right) \) .
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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 \( \\mathcal{G} \\subseteq \\mathcal{A} \) .\n\nLet \( {A}_{0},{A}_{1},{A}_{2},\\ldots \\in \\mathcal{A} \) . Choose countable \( {\\mathcal{G}}_{n} \\subseteq \\mathcal{G} \) such that \( {A}_{n} \\in \\sigma \\left( {\\mathcal{G}}_{n}\\right) \) . Set \( {\\mathcal{G}}^{\\prime } = \\mathop{\\bigcup }\\limits_{n}{\\mathcal{G}}_{n} \) . Then \( {\\mathcal{G}}^{\\prime } \) is countable, and \( \\mathop{\\bigcup }\\limits_{n}{A}_{n} \\in \\sigma \\left( {\\mathcal{G}}^{\\prime }\\right) \) . Thus \( \\mathcal{A} \) is closed under countable unions. The proof is complete.
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Yes
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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.
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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 \( \mathcal{B} \) is closed under complementation. Towards proving this, consider\n\n\[ \mathcal{D} = \left\{ {A \in \mathcal{B} : {A}^{c} \in \mathcal{B}}\right\} \]\n\nWe need to show that \( \mathcal{B} \subseteq \mathcal{D} \) . Since every closed set in a metrizable space is a \( {G}_{\delta } \) set, open sets are in \( \mathcal{D} \) . Now suppose \( {A}_{0},{A}_{1},{A}_{2},\ldots \) are in \( \mathcal{D} \) . Then \( {A}_{i},{A}_{i}^{c} \in \mathcal{B} \) for all \( i \) . As\n\n\[ {\left( \mathop{\bigcup }\limits_{i}{A}_{i}\right) }^{c} = \mathop{\bigcap }\limits_{i}{A}_{i}^{c}\text{ and }{\left( \mathop{\bigcap }\limits_{i}{A}_{i}\right) }^{c} = \mathop{\bigcup }\limits_{i}{A}_{i}^{c} \]\n\n\( \mathop{\bigcup }\limits_{i}{A}_{i} \) and \( \mathop{\bigcap }\limits_{i}{A}_{i} \) belong to \( \mathcal{D} \) . Thus \( \mathcal{D} \) contains open sets and is closed under countable unions and countable intersections. Since \( \mathcal{B} \) is the smallest such family, \( \mathcal{B} \subseteq \mathcal{D} \) .
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Yes
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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.
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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 \( \mathcal{D} \) . We now show that \( \mathcal{D} \) is closed under countable disjoint unions and countable intersections.\n\nFix \( {A}_{0},{A}_{1},{A}_{2},\ldots \in \mathcal{D} \) . Then \( {A}_{i},{A}_{i}^{c} \in \mathcal{B} \) for all \( i \) . We have \( \mathop{\bigcap }\limits_{i}{A}_{i} \in \mathcal{B} \) . Note that the sets \( {B}_{0} = {A}_{0}^{c},{B}_{1} = {A}_{1}^{c} \cap {A}_{0},{B}_{2} = {A}_{2}^{c} \cap {A}_{0} \cap {A}_{1},\ldots \) are pairwise disjoint and belong to \( \mathcal{B} \) . Further, \( {\left( \mathop{\bigcap }\limits_{i}{A}_{i}\right) }^{c} = \mathop{\bigcup }\limits_{i}{B}_{i} \in \mathcal{B} \) . Thus \( \mathcal{D} \) is closed under countable intersections. Similarly, we show that \( \mathcal{D} \) is closed under countable disjoint unions. As before, we conclude that \( \mathcal{B} \subseteq \mathcal{D} \) ; i.e., \( \mathcal{B} \) is closed under complementation.
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Yes
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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.
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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 family \( \mathcal{F} \), let \( {\mathcal{F}}_{ + } \) denote the family of countable disjoint unions of sets in \( \mathcal{F} \) .\n\nLemma 3.1.13 Let \( \mathcal{F} \) be the set of closed subsets of \( \mathbb{R} \) . Then \( (0,1\rbrack \in {\mathcal{F}}_{+\delta + } \).\n\nAssuming the lemma, we complete the proof as follows. Given an open set \( U \subseteq X \), by 2.1.19, choose a continuous map \( f : X \rightarrow \left\lbrack {0,1}\right\rbrack \) such that \( U = {f}^{-1}(\left( {0,1\rbrack }\right) \) . The lemma immediately implies that \( U \in \mathcal{B} \) .\n\nProof 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, \( D = \left\{ {\frac{1}{3},\frac{2}{3},\frac{1}{9},\frac{2}{9},\frac{7}{9},\frac{8}{9},\ldots }\right\} \) . Let \( E = D\bigcup \{ 0\} \) and \( P = \mathbf{C} \smallsetminus E \).\n\nNote that\n\n\[ (0,1\rbrack \smallsetminus P = \left\lbrack {\frac{1}{3},\frac{2}{3}}\right\rbrack \cup \left\lbrack {\frac{1}{9},\frac{2}{9}}\right\rbrack \cup \left\lbrack {\frac{7}{9},\frac{8}{9}}\right\rbrack \cup \cdots ; \]\n\ni.e., \( (0,1\rbrack \smallsetminus P \) is the union of the closures of the middle-third intervals removed to form \( \mathbf{C} \) . These interval are, of course, disjoint. Therefore, the lemma will be proved if we show that \( P \) is in \( {\mathcal{F}}_{+\delta } \) . Now,\n\n\[ P = \mathop{\bigcap }\limits_{{x \in E}}\left( {\mathbf{C}\smallsetminus \{ x\} }\right) \]\n\n\( \left( \star \right) \)\n\nSince \( \mathbf{C} \) is a zero-dimensional compact metric space, each \( \mathbf{C} \smallsetminus \{ x\} \) is a countable disjoint union of clopen subsets of \( \mathbf{C} \), which, being compact, are closed in \( \mathbb{R} \) ; i.e., \( C \smallsetminus \{ x\} \in {\mathcal{F}}_{ + } \) . Since \( E \) is countable, \( P \in {\mathcal{F}}_{+\delta } \) by \( \left( \star \right) \) .
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Yes
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Lemma 3.1.13 Let \( \mathcal{F} \) be the set of closed subsets of \( \mathbb{R} \) . Then \( (0,1\rbrack \in {\mathcal{F}}_{+\delta + } \) .
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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, \( D = \left\{ {\frac{1}{3},\frac{2}{3},\frac{1}{9},\frac{2}{9},\frac{7}{9},\frac{8}{9},\ldots }\right\} \) . Let \( E = D\bigcup \{ 0\} \) and \( P = \mathbf{C} \smallsetminus E \) .\n\nNote that\n\n\[ \n(0,1\rbrack \smallsetminus P = \left\lbrack {\frac{1}{3},\frac{2}{3}}\right\rbrack \cup \left\lbrack {\frac{1}{9},\frac{2}{9}}\right\rbrack \cup \left\lbrack {\frac{7}{9},\frac{8}{9}}\right\rbrack \cup \cdots ; \n\] \n\ni.e., \( (0,1\rbrack \smallsetminus P \) is the union of the closures of the middle-third intervals removed to form \( \mathbf{C} \) . These interval are, of course, disjoint. Therefore, the lemma will be proved if we show that \( P \) is in \( {\mathcal{F}}_{+\delta } \) . Now, \n\n\[ \nP = \mathop{\bigcap }\limits_{{x \in E}}\left( {\mathbf{C}\smallsetminus \{ x\} }\right) \n\] \n\n\( \left( \star \right) \)\n\nSince \( \mathbf{C} \) is a zero-dimensional compact metric space, each \( \mathbf{C} \smallsetminus \{ x\} \) is a countable disjoint union of clopen subsets of \( \mathbf{C} \), which, being compact, are closed in \( \mathbb{R} \) ; i.e., \( C \smallsetminus \{ x\} \in {\mathcal{F}}_{ + } \) . Since \( E \) is countable, \( P \in {\mathcal{F}}_{+\delta } \) by \( \left( \star \right) \) .
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Yes
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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 \( \mathcal{A} \) .
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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 \[ \mathcal{M}\left( A\right) = \{ B \in \mathcal{M} : A\bigcap B \in \mathcal{M}\} . \] As \( \mathcal{M} \) is a monotone class, \( \mathcal{M}\left( A\right) \) is a monotone class. As \( \mathcal{A} \) is an algebra, \( \mathcal{A} \subseteq \mathcal{M}\left( A\right) \) for every \( A \in \mathcal{A} \) . Therefore, \( \mathcal{M} \subseteq \mathcal{M}\left( A\right) \) for \( A \in \mathcal{A} \) . Thus for every \( A \in \mathcal{A} \) and every \( B \in \mathcal{M}, A \cap B \in \mathcal{M} \) . Using this and following the above argument, we see that for every \( A \in \mathcal{M},\mathcal{M}\left( A\right) \) is a monotone class containing \( \mathcal{A} \) . So, \( \mathcal{M} \subseteq \mathcal{M}\left( A\right) \) . This proves our claim. As \( \mathcal{M} \) is a monotone class closed under finite intersections, it is closed under countable intersections. Our proof will be complete if we show that \( \mathcal{M} \) is closed under complementation. Consider \[ \mathcal{D} = \left\{ {A \in \mathcal{M} : {A}^{c} \in \mathcal{M}}\right\} \] It is routine to show that \( \mathcal{D} \) is a monotone class. Clearly, \( \mathcal{D} \supseteq \mathcal{A} \) . So, \( \mathcal{M} \subseteq \mathcal{D} \) ; i.e., \( \mathcal{M} \) is closed under complementation.
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Yes
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Proposition 3.1.15 Every countably generated measurable space is atomic.
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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 \( \alpha = \left( {{\epsilon }_{0},{\epsilon }_{1},{\epsilon }_{2},\ldots }\right) \) of 0 ’s and 1’s, define\n\n\[ A\left( \alpha \right) = \mathop{\bigcap }\limits_{n}{A}_{n}^{{\epsilon }_{n}} \]\n\nEach \( A\left( \alpha \right) \) is clearly measurable. Let \( x \in X \) . Put \( {\epsilon }_{n} = {\chi }_{{A}_{n}}\left( x\right) \) and \( \alpha = \) \( \left( {{\epsilon }_{0},{\epsilon }_{1},{\epsilon }_{2},\ldots }\right) \) . Then \( x \in A\left( \alpha \right) \) . Thus \( X \) is the union of \( A\left( \alpha \right) \) ’s. Note that \( x, y \) belong to the same \( A\left( \alpha \right) \) if and only if for every \( n \), either both \( x \) and \( y \) belong to \( {A}_{n} \) or neither does.\n\nWe now show that each \( A\left( \alpha \right) \) is an atom of \( \mathcal{A} \) . Suppose this is not the case. Thus, there is an \( \alpha = \left( {{\epsilon }_{0},{\epsilon }_{1},{\epsilon }_{2},\ldots }\right) \) such that \( A\left( \alpha \right) \) contains a nonempty, proper, measurable subset, say \( B \) . Choose \( x \in B \) and \( y \in A\left( \alpha \right) \smallsetminus B \) . By 3.1.6, there is an \( n \) such that \( {A}_{n} \) contains exactly one of \( x \) and \( y \) . Since both \( x, y \in A\left( \alpha \right) \), for every \( m, x \in {A}_{m} \) if and only if \( y \in {A}_{m} \) . This contradicts 3.1.6.
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Yes
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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 \( \\mathcal{A} \\otimes {\\mathcal{B}}_{Y} \) .
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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\[ \\operatorname{graph}\\left( f\\right) = {\\left\\lbrack \\mathop{\\bigcup }\\limits_{n}\\left( {f}^{-1}\\left( {U}_{n}\\right) \\times {U}_{n}^{c}\\right) \\right\\rbrack }^{c},\n\]\n\nand the result follows.
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Yes
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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.
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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 measurable. By 3.1.21, \( \\operatorname{graph}\\left( f\\right) \\in \\mathcal{A}\\bigotimes {\\mathcal{B}}_{Y} \\subseteq \\mathcal{A}\\bigotimes \\mathcal{P}\\left( Y\\right) \) .
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Yes
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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\[ \n{\mathcal{B}}_{X} = {\bigotimes }_{i}{\mathcal{B}}_{{X}_{i}} \n\]
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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, since \( \mathcal{G} \) is a subbase for the topology on \( X \), by 3.1.8, it generates \( {\mathcal{B}}_{X} \).
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Yes
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Theorem 3.1.24 \( \mathcal{P}\left( {\omega }_{1}\right) \bigotimes \mathcal{P}\left( {\omega }_{1}\right) = \mathcal{P}\left( {{\omega }_{1} \times {\omega }_{1}}\right) \) .
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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\[ C = A\bigcap \left\{ {\left( {\alpha ,\beta }\right) \in {\omega }_{1} \times {\omega }_{1} : \alpha \leq \beta }\right\} \]\n\nWe shall show that \( B \) is in the product \( \sigma \) -algebra. By symmetry it will follow that \( C \) is in the product \( \sigma \) -algebra. The result will then follow.\n\nFor each \( \alpha < {\omega }_{1},{B}_{\alpha } \) is countable, say \( {B}_{\alpha } = \left\{ {{\alpha }_{0},{\alpha }_{1},{\alpha }_{2},\ldots }\right\} \) . By 3.1.22,\n\n\[ {G}_{n} = \left\{ {\left( {\alpha ,{\alpha }_{n}}\right) : \alpha \in {\omega }_{1}}\right\} \]\n\nis in the product of discrete \( \sigma \) -algebras. Now note that \( B = \mathop{\bigcup }\limits_{n}{G}_{n} \) .
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Yes
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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.
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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\[ \nU = \\mathop{\\bigcup }\\limits_{k}{U}_{k} = \\mathop{\\bigcup }\\limits_{k}\\operatorname{cl}\\left( {U}_{k}\\right)\n\]\n\nNote that for every \( x \\in X \), we have\n\n\[ \nf\\left( x\\right) \\in U \\Rightarrow \\exists k\\mathop{\\lim }\\limits_{n}{f}_{n}\\left( x\\right) \\in {U}_{k}\n\]\n\n\[ \n\\Rightarrow \\;\\exists k\\exists N\\forall n \\geq N{f}_{n}\\left( x\\right) \\in {U}_{k}\n\]\n\n\[ \n\\Rightarrow \\;\\exists {kf}\\left( x\\right) \\in \\operatorname{cl}\\left( {U}_{k}\\right)\n\]\n\n\[ \n\\Rightarrow f\\left( x\\right) \\in U\\text{.}\n\]\n\nThus,\n\n\[ \n{f}^{-1}\\left( U\\right) = \\mathop{\\bigcup }\\limits_{k}\\mathop{\\liminf }\\limits_{n}{f}_{n}^{-1}\\left( {U}_{k}\\right)\n\]\n\nSince each \( {f}_{n} \) is measurable, it follows from the above observation that \( f \) is measurable.
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Yes
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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.
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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\[ \nU = \mathop{\bigcup }\limits_{k}{U}_{k} = \mathop{\bigcup }\limits_{k}\operatorname{cl}\left( {U}_{k}\right) \n\]\n\nNote that for every \( x \in X \), we have\n\n\[ \nf\left( x\right) \in U \Rightarrow \exists k\mathop{\lim }\limits_{n}{f}_{n}\left( x\right) \in {U}_{k} \n\]\n\n\[ \n\Rightarrow \;\exists k\exists N\forall n \geq N{f}_{n}\left( x\right) \in {U}_{k} \n\]\n\n\[ \n\Rightarrow \;\exists {kf}\left( x\right) \in \operatorname{cl}\left( {U}_{k}\right) \n\]\n\n\[ \n\Rightarrow f\left( x\right) \in U\text{.} \n\]\n\nThus,\n\n\[ \n{f}^{-1}\left( U\right) = \mathop{\bigcup }\limits_{k}\mathop{\liminf }\limits_{n}{f}_{n}^{-1}\left( {U}_{k}\right) \n\]\n\nSince each \( {f}_{n} \) is measurable, it follows from the above observation that \( f \) is measurable.
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Yes
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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.
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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\[ \n{f}_{n} = \mathop{\sum }\limits_{{j = - n{2}^{n}}}^{{\left( {n - 1}\right) {2}^{n}}}\frac{j}{{2}^{n}}{\chi }_{{B}_{j}^{n}}\n\]\n\nClearly, \( {f}_{n} \) is a simple Borel function. It is easy to check that \( {f}_{n} \rightarrow f \) pointwise.
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Yes
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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 \( Y \) to \( Z \) is continuous for every \( x \in X \) ;\n\n(ii) \( x \rightarrow f\left( {x, y}\right) \) is measurable for all \( y \in D \) .\n\nThen \( f : X \times Y \rightarrow Z \) is measurable.
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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\[ f\left( {x, y}\right) \in C \Leftrightarrow \left( {\forall n \geq 1}\right) \left( {\exists {y}^{\prime } \in D}\right) \left\lbrack {d\left( {y,{y}^{\prime }}\right) \leq \frac{1}{n}\& \rho \left( {f\left( {x,{y}^{\prime }}\right), C}\right) \leq \frac{1}{n}}\right\rbrack .\n\]\n\nTherefore,\n\n\[ {f}^{-1}\left( C\right) = \mathop{\bigcap }\limits_{n}\mathop{\bigcup }\limits_{{{y}^{\prime } \in D}}\left\lbrack {\left\{ {x \in X : \rho \left( {f\left( {x,{y}^{\prime }}\right), C}\right) \leq \frac{1}{n}}\right\} \times \{ y \in Y : d\left( {y,{y}^{\prime }}\right) \leq \frac{1}{n}\} }\right\rbrack .\n\]\n\nBy our hypothesis, \( {f}^{-1}\left( C\right) \in \mathcal{A} \otimes {\mathcal{B}}_{Y} \) .
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Yes
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Proposition 3.1.32 Let \( X \) and \( Y \) be metrizable spaces. Then every Baire function \( f : X \rightarrow Y \) is Borel.
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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.
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Yes
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Proposition 3.1.32 Let \( X \) and \( Y \) be metrizable spaces. Then every Baire function \( f : X \rightarrow Y \) is Borel.
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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.
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Yes
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