diff --git a/src/fa/norm/separable.tex b/src/fa/norm/separable.tex
index 979cc64..6a47c36 100644
--- a/src/fa/norm/separable.tex
+++ b/src/fa/norm/separable.tex
@@ -31,31 +31,6 @@
     (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{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}
-
 \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 $C([0, 1]; E)$. 
@@ -82,5 +57,59 @@
     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}
+
 
 
diff --git a/src/measure/radon/c0.tex b/src/measure/radon/c0.tex
index 42cf621..a769f6d 100644
--- a/src/measure/radon/c0.tex
+++ b/src/measure/radon/c0.tex
@@ -125,7 +125,7 @@
     
     As such a $\phi \in C_0(X; E)$ exists for all $\eps > 0$ and $\seqf{A_j}$, $\norm{I_\mu}_{C_0(X; E)^*} \ge \norm{\mu}_{\text{var}}$. Therefore the map $\mu \mapsto I_\mu$ is isometric. 
 
-    (Surjective): Let $B = \bracsn{\phi \in E^*|\norm{\phi}_{E^*} \le 1}$ and equip it with the weak*-topology and
+    (Surjective): Let $B = \bracsn{\phi \in E^*|\norm{\phi}_{E^*} \le 1}$ and equip it with the weak* topology and
     \[
     T: C_0(X; E) \to C_0(X \times B; K) \quad (Tf)(x, \phi) = \dpn{f(x), \phi}{E}
     \]
diff --git a/src/topology/dst/zero.tex b/src/topology/dst/zero.tex
index 914ce6f..746e1b7 100644
--- a/src/topology/dst/zero.tex
+++ b/src/topology/dst/zero.tex
@@ -22,6 +22,17 @@
     Let $2 = \bracs{0, 1}$ be equipped with the discrete topology, then $2^{\natp}$ is the \textbf{Cantor space}. 
 \end{definition}
 
+\begin{proposition}
+\label{proposition:cantor-space-embedding}
+    The mapping
+    \[
+    2^{\natp} \to [0, 1] \quad \seq{x_n} \mapsto 2\sum_{n \in \natp} \frac{x_n}{3^n}
+    \]
+
+    is an embedding. 
+\end{proposition}
+
+
 \begin{definition}[Baire Space]
 \label{definition:the-baire-space}
     Let $\natp$ be equipped with the discrete topology, then $\mathscr{N} = (\natp)^{\natp}$ is the \textbf{Baire space}. 
@@ -57,11 +68,11 @@
     Finally, suppose that $\bracsn{C(n_1, \cdots, n_N)| \bracsn{n_k}_1^N \subset \natp, N \in \natp} \subset 2^X$ has been constructed to satisfy (i)-(iv). Since $X$ is complete, for each $\seq{n_k} \in \mathscr{N}$, there exists a unique $f(\seq{n_k}) \in X$ such that $\bracs{f(\seq{n_k})} = \bigcap_{N \in \natp}C(n_1, \cdots, n_N)$. For each $x \in X$, (iii) and (iv) imply that there exists $\seq{n_k} \in \mathscr{N}$ such that $x = f(\seq{n_k})$. As such, $f: \mathscr{N} \to X$ is a surjective mapping. For any $\seq{m_k}, \seq{n_k} \in \mathscr{N}$ and $N \in \natp$ with $m_k = n_k$ for each $1 \le k \le N$, $d(f(\seq{m_k}), f(\seq{n_k})) \le 1/N$. Therefore $f \in C(\mathscr{N}; X)$. 
 \end{proof}
 
-\begin{proposition}
-\label{proposition:cantor-universality}
+\begin{theorem}[Alexandroff-Hausdorff]
+\label{theorem:cantor-universality}
     Let $X$ be a non-empty compact metrisable space, then there exists a surjective mapping $f \in C(2^{\natp}; X)$. 
-\end{proposition}
-\begin{proof}
+\end{theorem}
+\begin{proof}[Proof, adapted from {{\cite[Proposition 8.2.7]{CohnMeasure}}}. ]
     Let $d: X \times X \to [0, \infty)$ be a metric on $X$. Since $X$ is compact, for each $N \in \natp$, there exists $K_N \in \natp$ and $\seqf{x_{N, k}|1 \le k \le K_N} \subset X$ such that $X = \bigcup_{k = 1}^{K_N}B(x_{N, k}, 1/N)$.
 
     To construct the desired map, it is sufficient to construct