I want to prove that, if $(X, \tau)$ is Hausdorff and $A\subseteq X$ is non empty then $A'$ the set of accumulation points is closed in $X$.
My reasoning:
$x \in X \setminus A' \rightarrow \exists\ U \subseteq X $ s.t. $x\in U$, $U$ is open and $U \cap A \subseteq \{x\} $
Now by Hausdorff condition, $\{x\}$ is closed and hence $(X\setminus \{x\}) \cap U = U\setminus \{x\}$ is open as well.
Let $y \in U$ if $y\not = x $ then $y\in U\setminus \{x\}$ and $U\setminus \{x\} \cap A = \emptyset$
so $y\in X\setminus A'$ hence, as $x \in X\setminus A'$,
$U\subseteq X\setminus A'$ and $X\setminus A'$ is open, giving the result.
Now, I am not completely sure, I do not think I used the Hausdorff condition fully as I only needed the first separation axiom (implied by Hausdorff) …
What do you think? thanks!
Best Answer
Your proof is correct. You only need the $T_1$ property.