82 lines
4.1 KiB
TeX
82 lines
4.1 KiB
TeX
\section{Embeddings in Cubes}
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\label{section:embeddings-in-cubes}
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\begin{definition}[Completely Regular]
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\label{definition:completely-regular}
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Let $X$ be a topological space, then $X$ is \textbf{completely regular} if for any $E \subset X$ closed and $x \in X \setminus E$, there exists $f \in C(X; [0, 1])$ such that $f(x) = 1$ and $f|_E = 0$.
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\end{definition}
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\begin{definition}[Separation of Points and Closed Sets]
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\label{definition:separate-points-closed-sets}
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Let $X$ be a topological space and $\cf \subset C(X; [0, 1])$, then $\cf$ \textbf{separates points and closed sets} if for any $E \subset X$ closed and $x \in X \setminus E$, there exists $f \in \cf$ such that $f(x) \not\in \ol{f(E)}$.
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\end{definition}
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\begin{proposition}
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\label{proposition:completely-regular-separate}
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Let $X$ be a $T_1$ space, then the following are equivalent:
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\begin{enumerate}
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\item $X$ is completely regular.
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\item There exists $\cf \subset C(X; [0, 1])$ that separates points and closed sets.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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(2) $\Rightarrow$ (1): Let $E \subset X$ closed and $x \in X \setminus E$, then there exists $f \in \cf$ such that $x \not\in \ol{f(E)}$. By \hyperref[Urysohn's lemma]{lemma:urysohn}, there exists $\phi \in C([0, 1]; [0, 1])$ such that $\phi(f(x)) = 1$ and $\phi(f(E)) = 0$.
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\end{proof}
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\begin{definition}[Embedding in Cube]
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\label{definition:embedding-in-cube}
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Let $X$ be a topological space, $\cf \subset C(X; [0, 1])$, and
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\[
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e: X \to [0, 1]^\cf \quad \pi_f(e(x)) = f(x)
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\]
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then:
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\begin{enumerate}
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\item $e \in C(X; [0, 1]^\cf)$.
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\item If $\cf$ separates points, then $e$ is injective.
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\item If $X$ is $T_1$ and $\cf$ separates points and closed sets, then $e$ is an embedding.
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\end{enumerate}
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The mapping $e$ is the \textbf{mapping of $X$ into the cube $[0, 1]^\cf$ associated with $\cf$}.
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\end{definition}
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\begin{proof}[Proof, {{\cite[Proposition 4.53]{Folland}}}. ]
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(1): By (U) of the \hyperref[product topology]{definition:product-topology}.
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(3): Since $X$ is $T_1$, $e$ is injective by (2). Let $x \in X$ and $U \in \cn_X^o(x)$, then there exists $f \in \cf$ such that $f(x) \not\in \ol{f(U^c)}$. In which case, there exists $V \in \cn_{[0, 1]}^o(f(x))$ such that $V \cap f(U^c) = \emptyset$. Thus for any $y \in X$ with $\pi_f(e(y)) \in V$, $f(y) \not\in f(U^c)$, so $y \in U$.
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\end{proof}
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\begin{proposition}
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\label{proposition:completely-regular-uniformisable}
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Let $X$ be a $T_1$ space, then the following are equivalent:
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\begin{enumerate}
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\item $X$ is completely regular.
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\item There exists a uniformity $\fU$ on $X$ that induces the topology on $X$.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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(1) $\Rightarrow$ (2): By \autoref{definition:uniform-separated}.
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(2) $\Rightarrow$ (1): By \autoref{definition:embedding-in-cube}, $X$ embeds into $[0, 1]^{C(X; [0, 1])}$, which is a uniform space. The subspace uniformity on $X$ then induces the topology on $X$.
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\end{proof}
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\begin{theorem}[Uryson Metrisation Theorem]
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\label{theorem:urysohn-metrisation}
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Let $X$ be a second countable regular space, then $X$ is metrisable.
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\end{theorem}
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\begin{proof}
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By \autoref{proposition:second-countable-regular}, $X$ is normal. Let $\cb \subset 2^X$ be a countable base for $X$, and let
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\[
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\mathcal{S} = \bracsn{(E, F) \in \mathcal{B}^2 | \ol{E} \subset F}
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\]
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By \hyperref[Urysohn's Lemma]{lemma:urysohn}, for each $(E, F) \in \mathcal{S}$, there exists $f_{EF} \in C(X; [0, 1])$ such that $f|_E = 1$ and $f|_{F^c} = 0$. For any $x \in X$ and $U \in \cn^o_X(x)$, there exists $E, F \in \mathcal{B}$ such that $x \in E \subset \ol{E} \subset F \subset U$. Thus $f_{EF}(x) = 1$ and $f_{EF}|_{U^c} = 0$. Therefore
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\[
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\cf = \bracsn{f_{EF}|(E, F) \in \mathcal{S}} \subset C(X; [0, 1])
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\]
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is a countable family of continuous functions that separate points and closed sets. By \autoref{definition:embedding-in-cube}, $X$ embeds into $[0, 1]^{\cf}$, which is metrisable by \autoref{proposition:countable-metric}.
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\end{proof}
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