Relative closure theorem Theorem 6.4
Let $(X, \mathcal T )$ be a topological space and let $Y$ be a subset of
$X$. Let $A $ be a subset of $Y$ . The closure of $A$ in the relative topology $T_Y$ is
$Y \cap Cl(A)$, where $Cl(A)$ is the closure of $A$ in the topology T.
I am studying for an exam and would like to know how to prove the above theorem?

By definition, $O$ is open in $A$ iff there is an open $O'$ of $X$ such that $O = O '\cap A$.
It follows that $C$ is closed in $A$ iff there is a closed set $C'$ of $X$ such that $C = C' \cap A$.
Proof: if $C$ is closed in $A$, this means $A \setminus C$ is open in $A$ (relative complement) iff there is an open $O'$ of $X$ such that $A\setminus C = O' \cap A$. But then indeed $C = (X\setminus O') \cap A$, and we can use $C' = X\setminus O'$ as the closed set. The reverse is quite similar.
Now to closure of $Y \subseteq A$, which can be defined as the smallest closed set that contains $Y$. So $\operatorname{cl}_A(Y)$ is a closed set of $A$ that contains $Y$. By the above we can write $\operatorname{cl}_A(Y) = C' \cap A$ for some closed subset $C'$ of $X$. As $Y \subseteq \operatorname{cl}_A(Y)$, we see that $Y \subseteq C'$ and by minimality of closure (in $X$) $\operatorname{cl}_X(Y) \subseteq C'$, so that $$\operatorname{cl}_X(Y) \cap A \subseteq C' \cap A = \operatorname{cl}_A(Y)$$
which shows one inclusion. On the other hand, $\operatorname{cl}_X(Y)$ is a closed subset of $X$ so $\operatorname{cl}_X(Y) \cap A$ is closed in $A$ and as $Y \subseteq \operatorname{cl}_X(Y)$ and $Y \subseteq A$, we see by minimaility of closure (in $A$ now) that $$\operatorname{cl}_A(Y) \subseteq \operatorname{cl}_X(Y) \cap A$$
which then shows the equality $\operatorname{cl}_A(Y) = \operatorname{cl}_X(Y) \cap A$ for all $Y \subseteq A$.
