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This may be a very simple question about "written-math" but as a non-english speaker it maker the following proposition difficult to understand

Proposition. Let $X$ be a topological space. If $U$ is an open set of $X$ then the map $Y\mapsto\overline{Y}$, restricted to closed irreducible sets $Y$ in $U$ is injective

The sets $Y$ are $X$-closed or $U$-closed (very different things since $U$ is open)?. Is there an established "semantic" difference between closed sets in $U$ and closed sets of $U$?

I assume the closure is the $X$-closure because there is no indication to do otherwise. Then the sets $Y$ should be $U$-closed for the closure to be a non-trivial map. Then why use $U$ open set of $X$ but $Y$ closed in $U$?

So let $Y,Z\subseteq U$ be closed sets (of/in? $U$) such that $\overline{Y}=\overline{Z}$. Then $Y=Y_1\cap U$ and $Z=Z_1\cap U$ for $Y_1,Z_1$ closed sets of $X$ $$Z\subseteq\overline{Z}=\overline{Y}=\overline{Y_1\cap U}\subseteq Y_1\cap\overline{U}\subseteq Y_1$$ $$Z=Z\cap U\subseteq Y_1\cap U=Y$$ And $Y\subseteq Z$ for the same reason so $Y=Z$. Is this proof correct?

Thanks

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  • $\begingroup$ What is $V$? $Z$? $\endgroup$ – Hagen von Eitzen Jun 11 at 21:23
  • $\begingroup$ @HagenvonEitzen sorry they are the same $\endgroup$ – Pedro Jun 11 at 21:25
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    $\begingroup$ Where do you use that $Y$ and $Z$ are closed irreducible? Btw, what does it mean? $\endgroup$ – Berci Jun 11 at 22:05
  • $\begingroup$ @Berci good point i do not use it. A set is irreducible if it cannot be written as the union of two proper closed subsets of it $\endgroup$ – Pedro Jun 11 at 22:15
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    $\begingroup$ And your interpretation about closures is also correct. $\endgroup$ – Berci Jun 11 at 22:23

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