fix typo and formatting errors

content/Math4201/Math4201_L16.md

@@ -119,7 +119,7 @@ Then $Z$ is a closed subset of $(\operatorname{Map}(X,Y),\rho)$.
 
 We need to show that $\overline{Z}=Z$.
 
-Since $\operatorname{Map}(X,Y)$ is a metric space, this is equivalent to showing that: $f_n:X\to Y\in Z$ continuous,
+Since $\operatorname{Map}(X,Y)$ is a metric space, this is equivalent to showing that: Let $f_n:X\to Y\in Z$ be a sequence of continuous maps,
 
 Which is to prove the uniform convergence,
 
@@ -127,28 +127,33 @@ $$
 f_n \to f \in \operatorname{Map}(X,Y)
 $$
 
-Then we want to show that $f$ is continuous.
+Then we want to show that $f$ is also continuous.
 
-Let $B_r(y)$ be an arbitrary ball in $Y$, it suffices to show that $f^{-1}(B_r(y))$ is open in $X$.
+It is to show that for any open subspace $V$ of $Y$, $f^{-1}(V)$ is open in $X$.
 
-Take $N$ to be large enough such that for $n\geq N$, we have
+Take $x_0\in f^{-1}(V)$, we'd like to show that there is an open neighborhood $U$ of $x_0$ such that $U\subseteq f^{-1}(V)$.
 
-$$
-\rho(f_n(x), f(x)) < \frac{r}{3}
-$$
+Since $x_0\in f^{-1}(V)$, then $f(x_0)\in V$. By metric definition, there is $r>0$ such that $B_r(f(x_0))\subseteq V$.
 
-In particular, this holds for $n=N$. So we have
+Take $N$ to be large enough such $\rho(f_N(x), f(x)) < \frac{r}{3}$
 
-$$
-d(f_N(x), f(x)) < \frac{r}{3},\forall x\in X
-$$
+So $\forall x\in X$, $d(f(x),f_N(x))<\frac{r}{3}$
 
-Take $x_0\in f^{-1}(B_r(y))$, we'd like to show that there is an open ball around $x_0$ in $f^{-1}(B_r(y))$.
+Since $f_N$ is continuous, $f_N^{-1}(B_{r/3}(f(x_0)))$ is an open set $U\subseteq X$ containing $x_0$.
 
-Since $f_N$ is continuous, $f^{-1}_N(B_{\frac{r}{3}}(y))$ is open in $X$.
+Take $x\in U$, $d(f(x),f(x_0))<d(f(x),f_N(x_0))+d(f_N(x),f_N(x_0))+d(f_N(x_0),f(x_0))$ using triangle inequality.
 
-$d(f(x_0), f(x_0))<\frac{r}{3}$
+Note that,
 
-continue the proof in bonus video
+$d(f(x),f_N(x))<\frac{r}{3}$ (using $N$ large enough),
 
+$d(f_N(x),f_N(x_0))<\frac{r}{3}$ (using $x\in U$, then $f_N(x)\in B_{r/3}(f_N(x_0))$, so $d(f_N(x),f_N(x_0))<\frac{r}{3}$),
+
+$d(f_N(x_0),f(x_0))<\frac{r}{3}$ (using $N$ large enough),
+
+So $d(f(x),f(x_0))<\frac{r}{3}+\frac{r}{3}+\frac{r}{3}=r$.
+
+So $f(x)\in B_r(f(x_0))\implies x\in f^{-1}(B_r(f(x_0)))\implies x\in f^{-1}(V)\implies U\subseteq f^{-1}(V)$.
+
+So $f^{-1}(V)$ is open in $X$.
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