\section{Separable Normed Vector Spaces}
\label{section:separable-banach-space}

\begin{proposition}
\label{proposition:separable-dual}
    Let $K \in \RC$ and $\dpn{E, F}{\lambda}$ be a duality of normed vector spaces over $K$ with $E$ being separable, and $S = \bracsn{y \in F|\ \norm{y}_F \le 1}$ be the closed unit ball of $F$, then
    \begin{enumerate}
        \item $S$ is separable with respect to the $\sigma(F, E)$-topology. 
        \item $S$ is metrisable with respect to the $\sigma(F, E)$-topology. 
        \item For any $A \subset F$, $A$ is separable with respect to the $\sigma(F, E)$-topology. 
        \item If the duality is norming, then there exists $\seq{y_n} \subset F$ such that for each $x \in E$, $\norm{x}_E = \sup_{n \in \natp}|\dpn{x, y_n}{\lambda}|$. 
    \end{enumerate}
\end{proposition}
\begin{proof}
    (1), (2): Let $D \subset E$ be a countable dense subset. By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, $S$ is embedded as a subspace of $K^D$. By \autoref{theorem:uniform-metrisable}, $\real^D$ is metrisable. By \autoref{proposition:separable-product}, $K^D$ is separable. Thus $S$ is also metrisable and separable by \autoref{proposition:separable-metric-space}. 

    (3): For any $A \subset E$, $A = \bigcup_{n \in \natp}A \cap nS$. By \autoref{proposition:separable-metric-space}, $A \cap nS$ is separable for each $n \in \natp$. Therefore $A$ is also separable. 
\end{proof}

\begin{lemma}
\label{lemma:compact-embed}
    Let $E$ be a normed vector space over $K \in \RC$ and $A \subset [0, 1]$ be closed, then $C(A; E)$ embeds isometrically into $C([0, 1]; E)$. 
\end{lemma}
\begin{proof}
    First note that if $0 \not\in A$ or $1 \not\in A$, $C(A; E)$ embeds isometrically into $C(A \cup \bracs{0, 1}; E)$ through extension by $0$. Thus assume without loss of generality that $A$ contains the endpoints $0$ and $1$. 

    Let $U = [0, 1] \setminus A$, then there exists $\seq{(a_n, b_n)} \subset [0, 1]^2$ such that  $U = \bigsqcup_{n \in \natp}(a_n, b_n)$. For each $f \in C(A; E)$, let
    \[
    Tf: [0, 1] \to E \quad x \mapsto \begin{cases}
        f(x) &x \in A \\
         \frac{b_n - x}{b_n - a_n}f(a_n) + \frac{x - a_n}{b_n - a_n}f(b_n) &x \in (a_n, b_n) \subset [0, 1]
    \end{cases}
    \]

    then the mapping $f \mapsto Tf$ is an isometric embedding into $E^{[0, 1]}$ with respect to the uniform norm.
    
    Since $U$ is open and $Tf$ is affine on each component of $U$, $Tf$ is continuous on $U$. It remains to show that $Tf$ is continuous on $A$. Let $x \in A$ and $\eps > 0$, then there exists $\delta > 0$ such that $\norm{f(y) - f(x)}_E < \eps$ for all $y \in (x -\delta, x + \delta) \cap A$. Now, a case analysis: 
    \begin{enumerate}
        \item If there exists $y \in (x - \delta, x) \cap A$, then for any $z \in U \cap (y, x)$, there exists $n \in \natp$ such that $(a_n, b_n) \subset (y, x)$ and $z \in (a_n, b_n)$. In which case, since $\norm{f(a_n) - f(x)}_E < \eps$ and $\norm{f(b_n) - f(x)}_E < \eps$, $\norm{Tf(z) - Tf(x)}_E < \eps$. Thus $\norm{Tf(z) - Tf(x)}_E < \eps$ for all $z \in (y, x)$.
        \item Otherwise, $x = 0$ or $Tf|_{(x - \delta, x)}$ is an affine function. Either way, there exists $y \in (x - \delta, x)$ such that $\norm{Tf(z) - Tf(x)}_E < \eps$ for all $z \in (y, x) \cap [0, 1]$. 
    \end{enumerate}

    Thus there exists $y \in (x - \delta, x)$ with $\norm{Tf(z) - Tf(x)}_E < \eps$ for all $z \in (y, x) \cap [0, 1]$. Similarly, there exists $y' \in (x, x + \delta)$ with $\norm{Tf(z) - Tf(x)}_E < \eps$ for all $z \in (x, y') \cap [0, 1]$. Therefore $Tf$ is continuous at $x$. Since this holds for all $x \in U$ and $x \in A$, $Tf \in C([0, 1]; E)$. 
\end{proof}

\begin{theorem}[Banach-Mazur]
\label{theorem:banach-mazur}
    Let $E$ be a separable normed vector space over $K \in \RC$, then there exists an isometric embedding $\iota \in L(E; C([0, 1]; K))$. 
\end{theorem}
\begin{proof}
    Let $B$ be the closed unit ball of $E^*$, equipped with the weak* topology. By the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, the linear mapping
    \[
    E \to C(B; K) \quad x(\phi) = \dpn{x, \phi}{E}
    \]

    is an isometric embedding. By \autoref{proposition:separable-dual}, $B$ is a compact metric space. The \hyperref[Alexandroff-Hausdorff Theorem]{theorem:cantor-universality} then provides a continuous surjection $f: 2^{\natp} \to B$. Thus the composition map
    \[
    C(B; K) \to C(2^{\natp}; K) \quad g \mapsto g \circ f
    \]

    is a linear isometric embedding. Let $\mathcal{C} \subset [0, 1]$ be the Cantor set, then $\mathcal{C}$ is homeomorphic to $2^{\natp}$ through \autoref{proposition:cantor-space-embedding}. Hence $C(2^{\natp}; K)$ is isometrically isomorphic to $C(\mathcal{C}; K)$. 

    Finally, \autoref{lemma:compact-embed} provides yet another linear isometric embedding $C(\mathcal{C}; K) \to C([0, 1]; K)$. Composing the above maps as follows
    \[
    \xymatrix{
E \ar@{->}[r] & C(B; K) \ar@{->}[r] & C(2^{{\mathbb N}^+}; K) \ar@{->}[r] & C(\mathcal{C}; K) \ar@{->}[r] & C([0, 1]; K)
}
    \]

    yields the desired embedding. 
\end{proof}



\begin{proposition}
\label{proposition:separable-banach-borel-sigma-algebra}
    Let $E$ be a separable normed vector space, then the Borel $\sigma$-algebra on $E$ is generated by the following families of sets:
    \begin{enumerate}
        \item Open sets in $E$ with respect to the strong topology. 
        \item $\bracs{B(x, r)|x \in E, r > 0}$.
        \item $\bracsn{\ol{B(x, r)}|x \in E, r > 0}$. 
        \item Open sets in $E$ with respect to the weak topology. 
    \end{enumerate}

    
\end{proposition}
\begin{proof}
    (1) $\Leftrightarrow$ (2) $\Leftrightarrow$ (3): By \autoref{proposition:separable-metric-borel-sigma-algebra}.

    (4) $\subset$ (1): Every weakly open set is strongly open. 

    (2) $\subset$ (4): By \autoref{proposition:seminorm-lsc}, $\norm{\cdot}_E: E \to [0, \infty)$ is Borel measurable with respect to the weak topology. For any $x \in E$, let
    \[
    \phi_x: E \to [0, \infty) \quad y \mapsto \norm{x - y}_E
    \]

    then $\phi_x$ is Borel measurable with respect to the weak topology, so $B(x, r) = \bracs{\phi_x < r}$ is a Borel set with respect to the weak topology. 
\end{proof}