# The closure of a set is closed

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, $$0 If we choose $$0<\epsilon, 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.

• Look at this for additional reference math.stackexchange.com/questions/448468/… Jan 23, 2017 at 15:37
• 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. Jan 23, 2017 at 15:39
• @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 $A$. Jan 23, 2017 at 16:21
• You do not need any use of a metric.This is true in any topological space, whether metrizable or not. Apr 15, 2018 at 12:22
• To me, showing that the complement of a closure is open is much easier. Dec 25, 2021 at 13:12

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 $$A$$.