Relative closure theorem

Question

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?

Answer 1

For $x\in\overline{A}^{Y}$, then find a net $(x_{\delta})\subseteq A$ such that $x_{\delta}\rightarrow x$ in $Y$, but then $x_{\delta}\rightarrow x$ in $X$, so $x\in\overline{A}$. On the ohther hand, $x\in Y$ since the closure of $A$ is taken in $Y$, this shows $x\in Y\cap\overline{A}$.

Now let $x\in Y\cap\overline{A}$, then $x\in\overline{A}$ and there is some net $(x_{\delta})\subseteq A$ such that $x_{\delta}\rightarrow x$ in $X$. Since $x_{\delta},x\in Y$, we have $x_{\delta}\rightarrow x$ in $Y$, so $x\in\overline{A}^{Y}$.

Another way: For $x\in\overline{A}^{Y}$, then surely $x\in Y$. For open set $G$ that containing $x$, $G\cap Y$ is an open set in $Y$ that containing $x$, so $\in G\cap Y\cap A\ne\emptyset$, so $y\in G\cap Y\cap A$ for some $y$, and hence $y\in G\cap A$, so $G\cap A\ne\emptyset$, so $x\in\overline{A}$, this shows $\overline{A}^{Y}\subseteq Y\cap\overline{A}$.

For $x\in Y\cap\overline{A}$, for any open set $G$ in $Y$ that contaning $x$, there is some open set $H$ in $X$ such that $G=Y\cap H$, so $x\in H$, and hence $H\cap A\ne\emptyset$, say, $y\in H\cap A$ for some $y$, so $y\in A\subseteq Y$, and hence $y\in G$, so $x\in\overline{A}^{Y}$, this shows $Y\cap\overline{A}\subseteq\overline{A}^{Y}$.

Answer 2

Let $F$ be a closed set of $(X, T)$ containing $A$. Then $Y - F = (X - F) \cap Y$ is an open set of $T_Y$, so that $Y \cap F$ is a closed set of $(Y, T_Y)$.

Conversely, if $F'$ is a closed set of $(Y, T_Y)$ containing $A$, then there exists $U$ open in $(X, T)$ such that $Y \cap U = Y - F'$; then $F' = Y \cap (X - U)$.

From the above, we deduce that the family of closed sets of $Y$ containing $A$ is the intersection of $Y$ with the family of closed sets of $X$ containing $A$. This must be true also for the smallest such set for inclusion, so $Y \cap \bar{A}$ is the closure of $A$ in $Y$.

Answer 3

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 $Y \subseteq A$.