fix typos

content/Math4201/Math4201_L38.md

@@ -40,7 +40,7 @@ If $X$ is a normal (regular and second countable) topological space, then $X$ is
 
 We will show that there exists embedding $F:X\to \mathbb{R}^\omega$ such that $F$ is continuous, injective and if $Z=F(X)$, $F:X\to Z$ is a open map.
 
-Recall that [regular and second countable spaces are normal](./Math4201_L36.md/#theorem-for-constructing-normal-spaces)
+Recall that [regular and second countable spaces are normal](../Math4201_L36#theorem-for-constructing-normal-spaces)
 
 1. Since $X$ is regular, then 1 point sets in $X$ are closed.
 2. $X$ is regular if and only if $\forall x\in U\subseteq X$, $U$ is open in $X$. There exists $V$ open in $X$ such that $x\in V\subseteq\overline{V}\subseteq U$.
@@ -49,7 +49,7 @@ Let $\{B_n\}$ be a countable basis for $X$ (by second countability).
 
 Pass to $(n,m)$ such that $\overline{B_n}\subseteq B_m$.
 
-By [Urysohn lemma](./Math4201_L37.md/#urysohn-lemma), there exists continuous function $g_{m,n}: X\to [0,1]$ such that $g_{m,n}(\overline{B_n})=\{1\}$ and $g_{m,n}(B_m)=\{0\}$.
+By [Urysohn lemma](../Math4201_L37#urysohn-lemma), there exists continuous function $g_{m,n}: X\to [0,1]$ such that $g_{m,n}(\overline{B_n})=\{1\}$ and $g_{m,n}(B_m)=\{0\}$.
 
 Therefore, we have $\{g_{m,n}\}$ is a countable set of functions, where $\overline{B_n}\subseteq B_m$.