130 lines
6.6 KiB
TeX
130 lines
6.6 KiB
TeX
\section{Compactifications}
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\label{section:compactifications}
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\begin{definition}[Compactification]
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\label{definition:compactification}
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Let $X$ be a topological space, then a \textbf{compactification} of $X$ is a pair $(Y, f)$ where
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\begin{enumerate}
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\item $Y$ is a compact Hausdorff space.
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\item $f \in C(X; Y)$ is an embedding.
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\item $f(X)$ is dense in $Y$.
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\end{enumerate}
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\end{definition}
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\begin{definition}[Stone-Čech Compactification]
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\label{definition:stone-cech}
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Let $X$ be a completely regular space, then there exists a pair $(\beta X, e)$ such that:
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\begin{enumerate}
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\item $(\beta X, e)$ is a compactification of $X$.
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\item[(U1)] For any $f \in C(X; [0, 1])$, there exists a unique $\beta f \in C(\beta X; [0, 1])$ such that the following diagram commutes:
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\[
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\xymatrix{
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\beta X \ar@{->}[r]^{\beta f} & [0, 1] \\
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X \ar@{->}[u]^{e} \ar@{->}[ru]_{f} &
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}
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\]
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\end{enumerate}
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Moreover, if $(\beta X, e)$ is \textit{any} pair that satisfies (U1), then
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\begin{enumerate}
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\item[(U2)] For any pair $(Y, \varphi)$ satisfying (1), there exists a unique $\beta \varphi \in C(\beta X; Y)$ such that the following diagram commutes:
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\[
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\xymatrix{
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\beta X \ar@{->}[r]^{\beta \varphi} & Y \\
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X \ar@{->}[u]^{e} \ar@{->}[ru]_{\varphi} &
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}
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\]
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\end{enumerate}
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The pair $(\beta X, e)$ is the \textbf{Stone-Čech compactification} of $X$.
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\end{definition}
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\begin{proof}
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Let $e: X \to [0, 1]^{C(X; [0, 1])}$ be the embedding of $X$ into $[0, 1]^{C(X; [0, 1])}$ associated with $C(X; [0, 1])$ in \autoref{definition:embedding-in-cube}, and $\beta X = \ol{e(X)}$.
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(1): By \autoref{theorem:tychonoff} and \autoref{proposition:product-hausdorff}, $[0, 1]^{C(X; [0, 1])}$ is a compact Hausdorff space. By definition, $e(X)$ is dense in $\beta X$.
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(U1): For each $f \in C(X; [0, 1])$, $\pi_f \in C([0, 1]^{C(X; [0, 1])}; [0, 1])$ is an extension of $f$ to $e(x)$.
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(U2): Let $(Y, \varphi)$ be a compactification of $X$. For each $f \in C(Y; [0, 1])$, by (U1), there exists a unique $\beta(f \circ \varphi) \in C(X; [0, 1])$ such that the following diagram commutes:
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\[
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\xymatrix{
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X \ar@{->}[d]_{\varphi} \ar@{->}[r]^{e} & \beta X \ar@{->}[d]^{\beta (f \circ \varphi)} \\
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Y \ar@{->}[r]_{f} & [0, 1]
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}
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\]
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Let $e': Y \to [0, 1]^{C(Y; [0, 1])}$ be the embedding of $Y$ into $[0, 1]^{C(Y; [0, 1])}$ associated with $C(Y; [0, 1])$, then by (U) of the \hyperref[product topology]{definition:product-topology}, there exists $\beta(e' \circ \varphi) \in C(\beta X; [0, 1])$ such that the following diagram commutes:
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\[
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\xymatrix{
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X \ar@{->}[d]_{\varphi} \ar@{->}[r]^{e} & \beta X \ar@{->}[d]^{\beta (e' \circ \varphi)} \\
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Y \ar@{->}[r]_{e'} & [0, 1]^{C(Y; [0, 1])}
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}
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\]
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Since $Y$ is a compact Hausdorff space, $e'(Y)$ is closed by \autoref{proposition:compact-extensions} and \autoref{proposition:compact-closed}. As $e'$ is an embedding, identify $Y$ as a subspace of $[0, 1]^{C(Y; [0, 1])}$. Given that $e(X)$ is dense in $\beta X$, the the image of $\beta (e' \circ \varphi)$ lies in $Y$ by \autoref{proposition:closure-of-image}. Therefore under the identification, the following diagram commutes:
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\[
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\xymatrix{
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X \ar@{->}[d]_{\varphi} \ar@{->}[r]^{e} & \beta X \ar@{->}[ld]^{\beta (e' \circ \varphi)} \\
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Y &
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}
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\]
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\end{proof}
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\begin{lemma}
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\label{lemma:lch-compactification-open}
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Let $X$ be a LCH space and $(Y, \varphi)$ be a compactification of $X$, then $\varphi(X) \subset Y$ is open.
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\end{lemma}
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\begin{proof}
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For each $x \in X$, let $U \in \cn_X(x)$ be a compact neighbourhood. Since $Y$ is a compact Hausdorff space, $\varphi(U)$ is closed by \autoref{proposition:compact-closed}. As $\varphi \in C(X; Y)$ is an embedding, there exists $V \in \cn_Y(\varphi(x))$ such that $\varphi(U) = \varphi(X) \cap V$. Given that $\varphi(X)$ is dense in $Y$, $\varphi(U) = \ol{\varphi(X) \cap V} \supset V$. Therefore $\varphi(U) \in \cn_{Y}(\varphi(x))$, and $\varphi(X)$ is open in $Y$.
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\end{proof}
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\begin{definition}[One-Point Compactification]
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\label{definition:alexandroff-compactification}
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Let $(X, \mathcal{T})$ be a LCH space, then there exists a pair $(X^*, \iota)$ such that:
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\begin{enumerate}
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\item $(X^*, \iota)$ is a compactification of $X$.
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\item[(U)] For any pair $(Y, \varphi)$ satisfying (1), there exists a unique $\varphi^* \in C(Y; X^*)$ such that the following diagram commutes:
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\[
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\xymatrix{
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Y \ar@{->}[rd]^{\varphi^*} & \\
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X \ar@{->}[r]_{\iota} \ar@{->}[u]^{\varphi} & X^*
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}
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\]
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\end{enumerate}
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The pair $(X^*, \iota)$ is the \textbf{one-point compactification} of $X$.
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\end{definition}
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\begin{proof}
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Let $\infty$ be a point not in $X$, $X^* = X \sqcup \bracs{\infty}$, and $\mathcal{T}^* \subset 2^{X^*}$ such that for each $U \in \mathcal{T}^*$, one of the following holds:
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\begin{enumerate}[label=(\alph*)]
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\item $U \in \mathcal{T}$.
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\item $\infty \in U$ and $U^c \subset X$ is compact.
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\end{enumerate}
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Let $\seqi{U} \subset \mathcal{T}^*$ be an open cover of $X$, then there exists $i \in I$ such that $\infty \in U$. In which case, $U_i$ must satisfy (b), so there exists $J \subset I$ finite such that $\bigcup_{j \in J}U_j \supset U_i^c$, and $\bracsn{U_j|j \in J \cup \bracs{i}}$ is a finite subcover. Now, let $x \in X$, then since $X$ is locally compact, there exists a precompact neighbourhood $U \in \cn_X^o(x)$. In which case, $\ol{U}^c \in \cn_{X^*}(\infty)$ with $U \cap \ol{U}^c = \emptyset$. Therefore $X^*$ is a compact Hausdorff space.
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Let $\iota: X \to X^*$ be the inclusion map. For each $U \in \mathcal{T}^*$ satisfying (b), $\iota^{-1}(U) = U \cap X$. Since $U^c \subset X$ is compact, $U \cap X$ is open by \autoref{proposition:compact-closed}, so $\iota \in C(X; X^*)$. By (a), $\iota$ is an embedding.
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Finally, let
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\[
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\varphi^*: Y \to X^* \quad x \mapsto \begin{cases}
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\varphi^{-1}(x) &x \in \varphi(X) \\
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\infty &x \not\in \varphi(X)
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\end{cases}
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\]
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Let $U \subset X^*$ with $\infty \not\in U$, then $(\varphi^*)^{-1}(U) = \varphi(U)$ is open in $\varphi(X)$ because $\varphi$ is an embedding, and open in $Y$ by \autoref{lemma:lch-compactification-open}. On the other hand, for each $V \in \cn_{X^*}^o(\infty)$,
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\[
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(\varphi^*)^{-1}(V) = V \cup (Y \setminus \varphi(X))
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\]
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Since $\varphi \in C(X; Y)$ is an embedding, $V$ is relatively open in $\varphi(X)$, so $V \cup (Y \setminus \varphi(X))$ is open in $Y$.
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\end{proof}
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