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?

  • $\begingroup$ yes, the proof is very good. $\endgroup$ – Jorge Fernández Hidalgo Jan 23 '17 at 15:37
  • $\begingroup$ Look at this for additional reference math.stackexchange.com/questions/448468/… $\endgroup$ – Juniven Jan 23 '17 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. $\endgroup$ – Jack D'Aurizio Jan 23 '17 at 15:39
  • $\begingroup$ @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 '17 at 16:21
  • $\begingroup$ You do not need any use of a metric.This is true in any topological space, whether metrizable or not. $\endgroup$ – DanielWainfleet Apr 15 '18 at 12:22

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

Let $z$ be a limit point of $\overline A$. Every open set $U$ 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.


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