I was trying to prove the following statement:
Let $(X,\mathscr{T})$ be a topological space. If the set of accumulation points of $\{x\}$ is closed for every $x\in X$, then the set of accumulation points of each subset of $X$ is closed.

I've tried to start with, for a subset $S\subset X$, $x\in (S')'\setminus S'$, where $S'=\{\mbox{accumulation points of }S\}$, and get a contradiction, but all i've showed is that $x\in S$ and $\{x\}'\subset S'$ (but doesn't seem to help...): You can choose an open set $V$ containing $x$ and such that $V\cap \left(S\setminus\{x\}\right)=\emptyset$ (because $x\notin S'$). Then $\exists y\in V\cap S'$, so $V$ is a neighborhood of $y$, and then $V\cap (S\setminus\{y\})\ne\emptyset$. This implies necessarily that $V\cap (S\setminus\{y\})=\{x\}$ and $x\in S$.
If $y\in\{x\}'$, then every open neighborhood $U$ of $y$ is a open neighborhood of $x$ and, repeating the argument, $U\cap S\setminus\{y\}$ is non-void, so $y\in S'$.

Please, any hint? Thanks!

  • 1
    $\begingroup$ +1 for showing some work on the problem. To others: see how this gets a more useful answer to OP's issue, though it seems when Arturo answers it tends to be useful regardless. $\endgroup$ May 17 '12 at 5:52
  • $\begingroup$ Does $S'$ denote $S^c-\text{complement of} S$? $\endgroup$
    – Invisible
    Jan 21 '20 at 20:08

Let $S$ be an arbitrary set, and let $x\notin S'$. Then there exists an open set $U$ such that $x\in U$ and $ U\cap S \subseteq \{x\}$. If $U\cap S=\varnothing$, then $U\subseteq X-S'$, so $x$ is in the interior of $X-S'$. So we may assume that $x\in S$ and $U\cap S=\{x\}$.

Now, the set of accumulation points of $\{x\}$ is closed; call it $C$. Note that $x\notin C$, since there are no open sets $V$ that contain $x$ and such that $(V\cap\{x\})\setminus\{x\}\neq\varnothing$. Therefore, $X-C$ is an open neighborhood of $x$; let $W=U\cap (X-C)$, which is open and contains $x$.

Show that $W\subseteq X-S'$.

  • $\begingroup$ Completing: let $y\in W$. Then, you can find a open neighborhood $V$ of y such that $x\notin V$ and $V\subset U$, then $V\cap S=\emptyset$, and then, $y\notin S'$, in other words, $W\subset X\setminus S'$. Thanks Arturo Magidin! =p $\endgroup$
    – Yuki
    May 17 '12 at 4:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.