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I have been asked (by a book) to prove the following:

In a $T_1$ space, the set of accumulation points of a subset is closed.

My concern is that the proof I have written seems to work in any topological space. Please help me find my mistake, but please don't provide a proof of the statement.

My proof:

Let $X$ be a topological space. Let $A$ be a subset of $X$ and let $A'$ be the set of accumulation points of $A$. Let $x$ be an accumulation point of $A'$. If we let $N(x)$ be an open neighborhood of $x$ then we are guaranteed a point $a' \in N(x) \cap A'$. ($N(x)$ is open WLOG because every neighborhood contains an open neighborhood.) However, since $N(x)$ is open, this implies that $N(x)$ is also a neighborhood of $a'$. Since $a'$ is an accumulation point of $A$, this means that there is a point a of $A$ in $N(x)$, which implies that x was in fact an accumulation point of $A$, and therefore $x \in A'$. Since $A'$ contains all of its accumulation points, it must be closed.

Thank you.

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What if the only point of $A$ in $N(x)$ is $x$? –  Daniel Fischer Sep 18 '13 at 23:16
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For an example demonstrating @DanielFischer's comment take the $\{a,b\}$ with the indiscrete topology –  Stefan Hamcke Sep 18 '13 at 23:20
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It may help to look at a counterexample when $X$ is not $T_1$. Suppose that $X$ has the indiscrete topology and contains at least two points. Let $A=X\setminus\{x\}$ for some $x\in X$. Observe that the set of accum. pts. of $A$ is not closed. –  Brian M. Scott Sep 18 '13 at 23:20
    
You may take a look at my answer here where I give a counterexample for $T_0$ and show that for some $T_0$ spaces it still holds. –  Stefan Hamcke Sep 18 '13 at 23:23

1 Answer 1

You must ensure that such point in $A$ (let call it $a$) is distinct of $x$ because if it's the same point $x$ is not an accumulation point of $A$. So to find a point $a\neq x$ , you must use $T_{1}$'s axiom to separate $x$ from $a'$ then you can find an $a\neq x$ such that $a \in A$ .

Done it thanks to Daniel Fisher's idea.

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