Definition: The closure of a set $A$ is $\bar A=A\cup A'$, where $A'$ is the set of all limit points of $A$.

Claim: $\bar A$ is a closed set.

Proof: (my attempt) If $\bar A$ is a closed set then that implies that it contains all its limit points. So suppose to the contrary that $\bar A$ is not a closed set. Then $\exists$ a limit point $p$ of $\bar A$ such that $p\not \in \bar A.$ Clearly, $p$ is not a limit point of $A$ because if it were then $p\in \bar A$. This means that $\exists$ a neighborhood $N_r(p)$ which does not contain any point of $A$. But $p$ is a limit point of $\bar A$ so it must contain an element $y\in \bar A-A$ in its neighborhood $N_r(p).$ Of course, $y$ is a limit point. Now, $0<d(p,y)=h<r.$ If we choose $0<\epsilon<r-h$, then $N_{\epsilon}(y)$ will contain no point $x\in A$, which is a contradiction since $y$ is a limit point.

I want to know if this proof is correct.

  • $\begingroup$ Look at this for additional reference math.stackexchange.com/questions/448468/… $\endgroup$
    – Juniven
    Jan 23, 2017 at 15:37
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    $\begingroup$ You may define the closure of $A$ as the smallest (with respect to $\subseteq$) closed set enclosing $A$. In such a way the claim is trivial. $\endgroup$ Jan 23, 2017 at 15:39
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    $\begingroup$ @JackD'Aurizio But then we have to prove that your closure is really $A\cup A'$. The point of this exercise is not really to show that the closure is closed (however it happens to be defined), but rather that the operation $A\mapsto A\cup A'$ (no matter what name it has) gives a closed set regardless of $A$. $\endgroup$
    – Arthur
    Jan 23, 2017 at 16:21
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    $\begingroup$ You do not need any use of a metric.This is true in any topological space, whether metrizable or not. $\endgroup$ Apr 15, 2018 at 12:22
  • $\begingroup$ To me, showing that the complement of a closure is open is much easier. $\endgroup$
    – Hermis14
    Dec 25, 2021 at 13:12

3 Answers 3


Your proof is correct, maybe we can make it slightly faster.

Let $z$ be a limit point of $\overline A$. Every open set $U$ containing $z$ must contain a point $x$ in $\overline A$. If this point $x$ is in $A^\circ$ then $U$ must intersect $A$ because it contains a limit point of $A$. If $x$ is in $A$ then $U$ obviously intersects $A$.

So every limit point of $\overline A$ is a limit point of $A$, and $\overline A$ contains all of its limit points.


The above proof given by Stella Biderman is in the setting of $X$ a metric space. We present here a more simple proof in the general case of $(X,\tau)$ a topological space (as given in the Armstrong's Book "Basic Topology", p. 30). To show that $X \setminus \overline{A}$ is open, let $x \in X \setminus \overline{A}= X \setminus (A \cup A')$. Then $x$ does not belong to $A$ and is also not a limit point of $A$. Hence there exists an open neighborhood $U(x)$ of $x$ which contains neither a point of $A$, nor a limit point of $A$, and so $U(x) \cap \overline{A} =\varnothing$. Consequently, $U(x) \subseteq X \setminus \overline{A}$. Thus $\overline{A}$ is a closed set.


If z is a limit point of $\bar{A}$ and $U$ is a open set containing $z$ then it must contain a point in $\bar{A}$. But as an open set containing a point in $\bar{A}$ it must contain a point in $A$.


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