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In a topological space $X$, call $x\in X$ an accumulation point if $\forall$ open set $U\ni x$, $U \cap A \neq \emptyset$, and $y\in X$ a cluster point if $\forall$ open set $U\ni y$, $U\cap A\setminus \{y\} \neq \emptyset$. (These are the terminologies used by my lecturer. I'm aware that different ones exist.)

Call a set $A\subseteq X$ closed if its complement is open.

My lecturer gave us a proof that $A$ is closed iff $A$ contains all of its accumulation points (see below). However, I managed to modify it to show that $A$ is closed iff $A$ contains all of its cluster points (see below, marked with []). What went wrong here? If the latter is false in general, in what special cases is it true (I heard it's true in metric spaces)?

The proof:

($\Rightarrow$): Suppose $A$ is closed and $x_0 \in X \setminus A$. Take $U:= X\setminus A$, an open set containing $x_0$. Now $U\cap A =\emptyset$, so $x_0$ is not an accumulation point. [$x_0$ is not an accumulation point and so it is not a cluster point either.]

($\Leftarrow$): Suppose $A$ is not closed, then $X\setminus A$ is not open. $\exists x_0 \in X\setminus A$ such that no open set $U\ni x_0$ is contained in $X\setminus A$, i.e. any open set $U\ni x_0$ satisfies $U\cap A \neq \emptyset$. So $x_0$ is an accumulation point of $A$ but not in $A$. [For this $x_0$, note that $x_0 \notin U\cap A$ because $x_0 \notin A$. So any open set $U\ni x_0$ satisfies $U\cap A \setminus \{x_0\} \neq \emptyset$, i.e. $x_0$ is a cluster point of $A$ but not in $A$.]

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In the french literature one uses the term accumulation point ("point d'accumalation" see…) for what your lecturer calls cluster point, and the term limit point (or "point adherent" see for what your lecturer calls accumulation point. The statement you are talking about is usually stated as: $A$ is closed iff $A$ contains all its limit points, meaning that $A=\bar{A}$. – Mercy King Jun 23 '12 at 13:02
What your lecturer stated can be translated as: $A$ is closed iff $\partial A \subset A$ since $\bar{A}=A\cup\partial A$. – Mercy King Jun 23 '12 at 13:06
Your modification isn't a new result, it's known to be equivalent to what your lecturer stated! – Mercy King Jun 23 '12 at 13:12
@Mercy: The OP isn’t claiming that it’s a new result, but rather asking whether it’s correct and whether the proof given is correct. The answer to both questions is yes. – Brian M. Scott Jun 23 '12 at 13:18
up vote 10 down vote accepted

Your result is correct, as is your argument. You can even prove directly that if $A$ contains all of its cluster points, then it contains all of its accumulation points. Suppose that a set $A$ contains all of its cluster points but fails to contain its accumulation point $x$. Then $x$ is not a cluster point, so $x$ has an open nbhd $U$ such that $U\cap A\subseteq\{x\}$. But $x\notin A$, so $U\cap A=\varnothing$, contradicting the assumption that $x$ was an accumulation point of $A$.

Added: Your lecturer could have proved a stronger result. Let $\operatorname{cl}A$ be the set of accumulation points of $A$; then $A$ is closed iff $A=\operatorname{cl}A$. Suppose first that $A$ is closed. You’ve already proved that $A\supseteq\operatorname{cl}A$, and it’s clear that every point of $A$ is an accumulation point of $A$, so $A=\operatorname{cl}A$. Conversely, if $A$ is not closed, you already know that it fails to contain some accumulation point, so $A\ne\operatorname{cl}A$.

This stronger result fails for cluster points. Let $X$ be any $T_1$-space with at least two points, and let $x\in X$. Then $\{x\}$ is closed, but it has no cluster points, so it can’t be equal to the set of its cluster points.

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Thanks, especially for the edit – hwhm Jun 23 '12 at 13:25

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