# [Proof Verification]: 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, $0<d(p,y)=h<r.$ 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 or not?

• yes, the proof is very good. – Jorge Fernández-Hidalgo Jan 23 '17 at 15:37
• Look at this for additional reference math.stackexchange.com/questions/448468/… – Juniven Jan 23 '17 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. – Jack D'Aurizio Jan 23 '17 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 $A$. – Arthur Jan 23 '17 at 16:21
• You do not need any use of a metric.This is true in any topological space, whether metrizable or not. – DanielWainfleet Apr 15 '18 at 12:22

Let $$z$$ be a limit point of $$\overline A$$. Every open set $$U$$ containing $$z$$ must contain a point $$x$$ in $$\overline A$$, if the point is in $$A^\circ$$ then $$U$$ must intersect $$A$$ because it contains a limit point of $$A$$. if $$x$$ is in $$A$$ then $$U$$ also intersects $$A$$.
So every limit point of $$\overline A$$ is a limit point of $$A$$, and $$\overline A$$ contains all of its limit points.
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.
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 $$X \setminus A^-$$ is open, let $$x \in X \setminus A^-= X \setminus (A \cup A′)$$. 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 $$U(x) \cap A^- =\varnothing$$. Consequently, $$U(x) \subseteq X \setminus A^-$$. Thus $$A^-$$ is a closed set.