# $A$ closed in $Y$ and $Y$ closed in $X$ , then $A$ is closed in $X$

This sounds really simple and I'm struggling with it.

I first tried to show that $X-A$ had to be closed by trying to show the complementary had to be open (trying to express it as union or intersection of known opens), but I couldn't do it: $(X-A)$ has to be open, and that equals $(X-Y)\cup (Y-A)$, I can't prove $Y-A$ is open, thoguh.

I googled some solution and found this:

If $Y$ is closed then we have: $A$ is closed in $Y$ iff $A=Y\cap B$ and $B$ is closed in $X$ iff $A\subseteq Y$ and $A$ is closed in $X$.

I don't know what is $B$, and I don't understand the argument in general.

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$Y - A$ is open because $A$ is closed in $Y$. – Michael Albanese Aug 17 '13 at 14:56

If $Y-A$ is open, then it's equal to the intersection of an open set in $X$, let's call it $B$, with $Y$: $Y-A=Y\cap B$, and so $B$ contains $Y-A$, so $(X-Y)\cup B$ is open because it's the union of two open sets, and that is equal to $X-A$ because $B$ contains all $Y$ but $A$, and $X-Y$ contains all $X$ but $Y$.
Yes, you are right, just $(X \setminus Y) ∪ B$ is of course equal to $X \setminus A$ not $A$. Also note that for any $A, Y$ $\overline{A}^Y$ = $\overline{A}^X ∩ Y$. – user87690 Aug 17 '13 at 16:35
Your proof looks good. Note that what you have shown here is that if $A\subset Y$ is closed in $Y$, then there is a closed set $C$ in $X$ such that $A=Y\cap C$, just take $C=X-B$. – Stefan Hamcke Aug 17 '13 at 17:53