Understanding the proof of $cl(cl(A))=cl(A)$

Question

I want to discuss a particular part of the proof of $cl(cl(A))=cl(A)$ in a topological space:
One way to proof this includes the following reasoning:
$xin cl(cl(A)) Rightarrow U_xcap cl(A)neq emptysetRightarrow exists y in U_x wedge yin cl(A) Rightarrow xin  cl(A)$
where $U_x$ is a neibeorhood of $x$.
What I dont understand here is the last Implication. OK: I know that there is an $y$ in $cl(A)$ and this $y$ is aswell in $U_x$, but how can I know that my $x$ from the beginning is in $cl(A)$?

Mark · Accepted Answer

$yin cl(A)$ implies that the intersection of every neighborhood of $y$ with the set $A$ is nonempty. So here $U_x$ is a neighborhood of $y$, and hence $U_xcap Aneemptyset$. Now, $U_x$ was just some random neighborhood of $x$. So we actually showed that if $U$ is any neighborhood of $x$ then $Ucap Aneemptyset$. Thus $xin cl(A)$.

Correct answer by Mark on December 27, 2020

Understanding the proof of $cl(cl(A))=cl(A)$

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