As the title says, I'm trying to answer a question which asks me to show that, for a non-empty subset $A$ of a Hausdorff topological space, a point $x$ is an accumulation point of $A$ (i.e. $x \in A'$) if and only if $x$ is in $\overline{A\setminus{\{x}\}}$.
I don't know how to go about doing this, so I'd really appreciate any help you could offer.
It seems that Hausdorffness is redundant. I'm using the definition that $x$ is an accumulation point of $A$ if and only if every neighborhood of $x$ contains an element of $A$ different from $x$.
Consider $x \in A'$ and suppose that $x \not\in\overline{A\setminus{\{x}\}}$. Then, we can find a closed set $F$ such that $A\setminus\{x\} \subseteq F$ but $x \not\in F$. So, $F^c$ is an open set containing $x$ such that $F^c \cap \left( A\setminus\{x\}\right)=\emptyset$. Contradiction.
Conversely, suppose that $x \in \overline{A\setminus{\{x}\}}$ but $x\not\in A'$. Hence, there exists an open set $U$ such that $x \in U$ but $U \cap \left( A\setminus\{x\}\right) = \emptyset$. This means that we have found a closed set $U^c$ such that $A-\{x\}\subseteq U^c$. Hence, $\overline{A\setminus{\{x}\}} \subseteq U^c$ by definition of closure. But since $x \in \overline{A\setminus{\{x}\}}$, we have that $x\in U^c$. Contradiction.