Show that a subset of $\mathbb{R}$ is closed iff it contains all its accumulation points.
Well, the definition of accumulation point for a set S is that I have is that for all $\epsilon>0$, $B_\epsilon(x)\cap S\neq \varnothing$. Also the definition of an open set is that for all $\epsilon>0$, $B_\epsilon(x)\subset S$. So it follows that the closed set contains all the acc. points, just not sure how to formalize.