# Closure of subspace topology

Let $X$ be a topological space and $A\subseteq X$ a subset of $X$. Let $Y$ be a subspace of $X$ so that $A\subseteq Y$, and let $A_X$ and $A_Y$ be the closure of $A$ in $X$ and $Y$ respectively. I need to show that $A_Y=A_X\cap Y$.

So far (I think) I've shown that $A_X\cap Y \subseteq A_Y$. Let $x \in A_X\cap Y$, then in particular $x\in A_X$. As the closure of $A$ is the intersection of all closed sets containing $A$, $x$ must be in each of those. Let $K$ be any one of those closed sets. Then $X\setminus K$ is open, and by the definition of the subspace topology $Y\cap (X\setminus K)=Y\setminus K$ is open in the subspace topology on $Y$. Hence $K$ is closed in $Y$ as well. I think I've shown that any set which is closed in $X$ is closed in $Y$, hence if $x$ is in $A_X$, it must be in $A_Y$.

I have absolutely no idea how to show the opposite inclusion. Any help would be greatly appreciated!

$A_{Y}$ is closed in $Y$ so $A_{Y}=Y\cap F$ for some $F$ closed in $X$. Then $A\subseteq A_{Y}\subseteq F$ and consequently $A_{X}\subseteq F$.

This leads to $Y\cap A_{X}\subseteq Y\cap F=A_{Y}$.

converse:

$A\subseteq Y\cap A_{X}$ and $Y\cap A_{X}$ is closed in $Y$ so $A_{Y}\subseteq Y\cap A_{X}$ .

To show the opposite inclusion, let $x \in A_Y$. Then clearly $x \in Y$. It remains to show $x \in A_X$, which is equivalent to showing that $x$ is in every closed set $K \subseteq X$ which contains $A$. But $K \cap Y$ is a closed set in the subspace topology on $Y$, and $A \subseteq K \cap Y$. Thus by assumption, $x \in K \cap Y$. In particular, $x \in K$.

$A\subset A_X\cap Y$, as $A_X\cap Y$ is a closed subset in $Y$, then $A_Y\subset A_X\cap Y$.

If order to show $A_Y \subset A_X\cap Y$, remark that $A_Y$ is the smallest closed set in $Y$ that contains $A$, and $A_X\cap Y$ is one closed set in $Y$ that contains $A$