Typo fixes.

src/measure/measurable-maps/metric.tex

@@ -99,9 +99,9 @@
     k(N, x) = \min\bracs{n \in C(N, x) \bigg | d(f(x), y_n) = \min_{m \in C(N, x)}d(f(x), y_m)}
     \]
 
-    then for any $k \in \natp$,
+    then for any $k \in \natp$, $\bracs{x \in X|k(n, x) \le k}$ is equal to
     \[
-    \bracs{x \in X|k(n, x) \le k} = \bigcup_{j = 1}^k\bracs{x \in X \bigg |y_j \in C(n, x), d(f(x), y_j) = \min_{m \in C(N, x)}d(f(x), y_m)}
+     \bigcup_{j = 1}^k\bracs{x \in X \bigg |y_j \in C(n, x), d(f(x), y_j) = \min_{m \in C(N, x)}d(f(x), y_m)}
     \]
 
     For each $1 \le m \le N$, $y \mapsto d(y, y_m)$ is continuous. Thus $x \mapsto d(f(x), y_m)$ and $x \mapsto \min_{m \in C(N, x)}d(f(x), y_m)$ are $(\cm, \cb_\real)$-measurable by \autoref{proposition:limit-measurable} and assumption (c). By \autoref{proposition:metric-measurables},

src/topology/main/support.tex

@@ -13,5 +13,5 @@
 
 \begin{definition}
 \label{definition:compactly-supported-01}
-    Let $X$ be a topological space, $f \in C_c(X; [0, 1])$ and $U \subset X$ be open, then $f \prec U$ if $\supp{f} \subset U$
+    Let $X$ be a topological space, $f \in C_c(X; [0, 1])$ and $U \subset X$ be open, then $f \prec U$ if $\supp{f} \subset U$.
 \end{definition}