Question on subset topology Let $A$ be a subset of $B$. Prove the closure of $A$ is a subset of the closure of $B$. 
Have tried multiple ways but have no idea how to proceed. Any help would be nice.
 A: Hint. If we take as definition of the closure
$$\overline{A}=\bigcap_{F\supset A \text{ closed}}F,$$
then it is almost immediate. If $A\subset B$ and $F\supset B$, then $F\supset A$. Can you proceed?
A: Let the closure be denoted $cl(A)$.  Then assume $cl(A)$ is not a subset of $cl(B)$.  This means that there is an $x\in cl(A)$ such that $x\not\in cl(B)$.  
by definition of closure, $x\in cl(A)$ implies the existence of a net $(x_\alpha)_{\alpha\in I} \subseteq A$ for some indexing set $I$, such that $\lim_{\alpha \in I} x_\alpha = x$.  
But since $A \subseteq B$, we have $(x_\alpha)_{\alpha\in I} \subseteq B$.  Which means $x$ is a limit point of elements of $B$. Which means $x \in cl(B)$
A: Here's a direct proof.
Let $a \in \bar{A}$. Then $a$ is a limit point of $A$. Let $U$ be a neighborhood of $a$ in $B$. By definition of a limit point, $U \cap A \not = \emptyset$. Since $A \subset B$, it follows $U \cap B \not = \emptyset$. Therefore $a$ is a limit point of $B$. It follows $a \in \bar{B}$. This holds for all limit points of $A$. Therefore $\bar{A} \subset \bar{B}$.
