\section{Embeddings in Cubes}
\label{section:embeddings-in-cubes}

\begin{definition}[Completely Regular]
\label{definition:completely-regular}
    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 $f|_E = 0$. 
\end{definition}

\begin{definition}[Separation of Points and Closed Sets]
\label{definition:separate-points-closed-sets}
    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 $f(x) \not\in \ol{f(E)}$. 
\end{definition}

\begin{proposition}
\label{proposition:completely-regular-separate}
    Let $X$ be a $T_1$ space, then the following are equivalent:
    \begin{enumerate}
        \item $X$ is completely regular. 
        \item There exists $\cf \subset C(X; [0, 1])$ that separates points and closed sets.
    \end{enumerate}
\end{proposition}
\begin{proof}
    (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 $\phi(f(E)) = 0$. 
\end{proof}

\begin{definition}[Embedding in Cube]
\label{definition:embedding-in-cube}
    Let $X$ be a topological space, $\cf \subset C(X; [0, 1])$, and
    \[
    e: X \to [0, 1]^\cf \quad \pi_f(e(x)) = f(x)
    \]

    then:
    \begin{enumerate}
        \item $e \in C(X; [0, 1]^\cf)$. 
        \item If $\cf$ separates points, then $e$ is injective.
        \item If $X$ is $T_1$ and $\cf$ separates points and closed sets, then $e$ is an embedding. 
    \end{enumerate}

    The mapping $e$ is the \textbf{mapping of $X$ into the cube $[0, 1]^\cf$ associated with $\cf$}. 
\end{definition}
\begin{proof}[Proof, {{\cite[Proposition 4.53]{Folland}}}. ]
    (1): By (U) of the \hyperref[product topology]{definition:product-topology}. 

    (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 $y \in U$. 
\end{proof}

\begin{proposition}
\label{proposition:completely-regular-uniformisable}
    Let $X$ be a $T_1$ space, then the following are equivalent:
    \begin{enumerate}
        \item $X$ is completely regular.
        \item There exists a uniformity $\fU$ on $X$ that induces the topology on $X$. 
    \end{enumerate}
\end{proposition}
\begin{proof}
    (1) $\Rightarrow$ (2): By \autoref{definition:uniform-separated}. 

    (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 $X$. 
\end{proof}

\begin{theorem}[Uryson Metrisation Theorem]
\label{theorem:urysohn-metrisation}
    Let $X$ be a second countable regular space, then $X$ is metrisable. 
\end{theorem}
\begin{proof}
    By \autoref{proposition:second-countable-regular}, $X$ is normal. Let $\cb \subset 2^X$ be a countable base for $X$, and let
    \[
    \mathcal{S} = \bracsn{(E, F) \in \mathcal{B}^2 | \ol{E} \subset F}
    \]

    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
    \[
    \cf = \bracsn{f_{EF}|(E, F) \in \mathcal{S}} \subset C(X; [0, 1])
    \]
    
    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}. 
\end{proof}



