8
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.