Question about relatively closed set I know that $G$ is relatively open set in $U$ if and only if there exists open set $G_1$in $R$ such that $G=U \cap G_1$.
By using this, I want to show that

F is relatively closed set if and only if there exists closed set $F_1$ in $R$ such that $F = U \cap F_1$.

I have tried if $F$ is relatively closed in $R$ then $F^c \equiv U\setminus F $ is relatively open in $U$ so that there exists open set $G_1$ in R such that $U\setminus F = U \cap G_1$.
I don't know how to show this theorem in this way. Help me to prove it.
 A: Set theoretic relation we need


From the figure observe that $\forall V\subset \mathbb{R}$, $$\boxed {U\setminus (U\cap V)=U\cap (\mathbb{R}\setminus V)}.$$
Now let us see the proof of the statement:

F is relatively closed set if and only if there exists closed set $F_1$ in $R$ such that $F = U \cap F_1$.

Proof:

Assume $F$ is relatively closed in $U$.
Then $U \setminus F$ is relatively open in $U$,i.e.,$\exists$ openset $G_1$ in $\mathbb{R}$ such that $U\setminus F=U\cap G_1$.
$$\begin{align}\therefore U\setminus(U\setminus F)&=U\setminus(U\cap G_1)\\\implies F&=U\setminus(U\cap G_1)\\\implies F&=U\cap (\mathbb{R}\setminus G_1)\\\implies F&=U\cap F_1 \end{align}$$ where $F_1=\mathbb{R}\setminus G_1$ is a closed set.

Now assume there exists closed set $F_1$ in $\mathbb{R}$ such that $F=U\cap F_1$.

To prove $F$ is relatively closed in $U$.
Now $$\begin{align}U\setminus F&= U\setminus (U\cap F_1)\\\implies U\setminus F&=  U\cap (R\setminus F_1)\\\implies U\setminus F&=U \cap G_1\end{align}$$ where $G_1=R\setminus F_1$ is openset in $R$.
$\therefore U\setminus F$ is relatively open$\implies$ $U\setminus (U\setminus F)=F$ is relatively closed in $R$.

Hence,

$F$ is relatively closed in $U$ $\iff \exists \bar F_1=F_1\subset R$ such that $F=U\cap F_1$.

A: If $F$ is relatively closed (i.e. $U$ \ $F$ is relatively open in $U$) then $F= U \cap G_1^c $. Note that $G_1^c$ is closed in $R$. Conversely, if $F = U \cap F_1$ for some closed $F_1$ in $R$ then $U$ \ $F$ $ = U \cap F_1^c$. Note that $F_1^c$ is open in $R$ and so $U$ \ $F$ is relatively open in $U$.
