As stated in the title the problem reads, show that if $A\subset B$ then $A^a\subset B^a$. I'm dealing with topologies here and they are relatively new to me so I'm trying to understand them better. In this case "a" refers to the accumulation/limit point.
So the definition I have for accumulation point is as follows:
$x\in A^a\iff\forall(G\in\tau,x\in G) G\cap A\nsubseteq \{x\}$
$G\cap A\nsubseteq \{x\}$ is the same as $\exists y\neq x\in G\cap A$.
I'm having trouble starting because I know my hypothesis is that $A\subset B$ however from there should I assume $x\in A^a$ and thus end up ultimately proving that $x\in B^a$?
Let $x\in A^a$ and let $G$ be an open set such that $x\in G$. Let $y\in A\cap G$, $y\ne x$. Then $y\in B$, so $y\in B\cap G$.
Thus every open set containing $x$ intersects $B$ at a point different from $x$, which is the condition for $x\in B^a$.
Alternatively, suppose $x\notin B^a$; then there exists an open set $G$ such that $x\in G$ and $B\cap G\subseteq\{x\}$. Then $A\cap G\subseteq B\cap G\subseteq\{x\}$, so $x\notin A^a$.