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

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 $$cl(A)$$?

Mathematics Asked by Averroes2 on December 27, 2020

$$yin cl(A)$$ 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

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