Exercise 1.3.1

Proof. First note that for $Y\subseteq X$ we have $(X\setminus V)\cap Y=Y\setminus (V\cap Y)$. Here is a proof for those that are interested: If $x\in (X\setminus V)\cap Y$, then $x\in X,x\notin V,x\in Y$. Thus $x\in Y$, and $x\notin V\cap Y$, hence $x\in Y\setminus (V\cap Y)$. If $x\in Y\setminus (V\cap Y)$, then $x\in Y,x\notin V$. But $Y\subseteq X$, thus $x\in X$, hence $x\in (X\setminus V)\cap Y$.

$\text{(⇒)}$ Suppose $E$ is relatively closed with respect to $Y$. Taking the complement in $Y$ we get that $Y\setminus E$ is relatively open with respect to $Y$. Now using the results of part (a) we know that there is some $V\subseteq X,V$ open, such that $Y\setminus E=V\cap Y$. Now take $K=X\setminus V$ which we know is closed in $X$, and clearly $K\subseteq X$. But $K\cap Y=(X\setminus V)\cap Y=Y\setminus (V\cap Y)=Y\setminus (Y\setminus E)=E$

$\text{(⇐)}$ Suppose $E=K\cap Y$ for some set $K\subseteq X$ which is closed in $X$. Thus $X\setminus K$ is open. But, from what we noted above, we know $(X\setminus K)\cap Y=Y\setminus (K\cap Y)$, and since $X\setminus K$ is open, we know that $Y\setminus (K\cap Y)$ is relatively open with respect to $Y$. But $Y\setminus (K\cap Y)=Y\setminus E$, thus $E$ is relatively closed with respect to $Y$. □

Proof. $\text{(⇒)}$ Suppose $E$ is relatively closed in $Y$, thus every convergent sequence $\left\{x_n\right\}\subseteq E$ converges in $E$. That is, $x_n\to x\in E$.

Let $K=$ closure of $E$ in $(X,d)=\{x\in X:x$ is an adherent point of $E$ w.r.t. $(X,d)\}$. So $K$ is closed in $(X,d)$

Now we need to show that $E=K\cap Y$.

If $x\in E$, then clearly $x\in K$ and $x\in Y$. Hence $E\subseteq K\cap Y$. If $x\in K\cap Y$ then $x\in K$ and $x\in Y$. Thus there is $\left\{x_n\right\}\subseteq E$ such that $x_n\to x\in K$ with respect to $(X,d)$. But $\left\{x_n\right\}\subseteq Y$ and $x\in Y$, then $x_n\to x$ with respect to $\left(Y,{d|}_{Y\times Y}\right)$. But since $E$ is relatively closed with respect to $\left(Y,{d|}_{Y\times Y}\right)$, we get $x_n\to x\in E$. Hence $K\cap Y\subseteq E.$

Thus we have $E=K\cap Y$ as desired. □