fix errors and update news

content/Math4201/Math4201_L34.md

@@ -21,7 +21,7 @@ If $\mathbb{R}_l$ is second countable, then for any real number $x$, there is an
 
 Any such open sets is of the form $[x,x+\epsilon)\cap A$ with $\epsilon>0$ and any element of $A$ being larger than $\min(U_x)=x$.
 
-In summary, for any $x\in \mathbb{R}$, there is an element $U_x\in \mathcal{B}$ with $(U_x)=x$. In particular, if $x\neq y$, then $U_x\neq U_y$. SO there is an injective map $f:\mathbb{R}\rightarrow \mathcal{B}$ sending $x$ to $U_x$. This implies that $\mathbb{B}$ is uncountable.
+In summary, for any $x\in \mathbb{R}$, there is an element $U_x\in \mathcal{B}$ with $(U_x)=x$. In particular, if $x\neq y$, then $U_x\neq U_y$. So there is an injective map $f:\mathbb{R}\rightarrow \mathcal{B}$ sending $x$ to $U_x$. This implies that $\mathcal{B}$ is uncountable.
 
 </details>
 

content/Math4201/Math4201_L4.md

@@ -27,7 +27,7 @@ $$
 Let $(X,\mathcal{T})$ be a topological space. Let $\mathcal{C}\subseteq \mathcal{T}$ be a collection of subsets of $X$ satisfying the following property:
 
 $$
-\forall U\in \mathcal{T}, \exists C\in \mathcal{C} \text{ such that } U\subseteq C
+\forall U\in \mathcal{T}, \exists C\in \mathcal{C} \text{ such that } C\subseteq U
 $$
 
 Then $\mathcal{C}$ is a basis and the topology generated by $\mathcal{C}$ is $\mathcal{T}$.