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!

  • $\begingroup$ It looks fine; all you need is $T_1$. $\endgroup$ – Brian M. Scott Nov 23 '12 at 15:36
  • $\begingroup$ ah ok thank you! it seemed strange that they gave me more than i needed! $\endgroup$ – Moritzplatz Nov 23 '12 at 15:51
  • 1
    $\begingroup$ It’s quite common to deal only with Hausdorff spaces, even when less separation is sufficient. $\endgroup$ – Brian M. Scott Nov 23 '12 at 15:54

Your proof is correct. You only need the $T_1$ property.


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