Question on Closure of Relatively Closed Set. If $A \subset \mathbb{R}^{n}$ is closed, then
$\overline{A} \cap \mathbb{R}^{n} = A$,
where $ \overline{ \vphantom{A}\cdot\vphantom{A} }$ denotes the closure operator in $\mathbb{R}^{n}$.
My question is:
If $A$ is relatively closed in $X \subset \mathbb{R}^{n}$, does it then hold that
$\overline{A} \cap X = A$?
Edit: From this entry in the Encyclopedia of Mathematics, it seems that the second equation is in fact equivalent to $A$ being relatively closed in $X$.
 A: Yes it is, $A$ is relatively closed in $X$ so it exists $C$ closed subset in $E$ such as $A = C \cap X$.
$\overline{C \cap X} \subset \bar C\cap \bar X$.
Consequently : $\bar A \cap X = \overline{C \cap X} \cap X \subset \bar C\cap \bar X \cap X = C\cap X = A$
Since $A \subset X$ and $A \subset \bar A$ the other inclusion is obvious.
Finally $$\bar A \cap X = A$$
A: Yes. The closure  of a set $S$ is the $\subseteq$-minimum closed set that has $S$ as a subset, so it is the intersection of all the closed sets that have $S$ as a subset. So the complement of the closure of $S$ is the union of all open sets that are disjoint from $S.$
And so if $S$ is closed then $S$ is its own closure.
Let $T_Y$ be a topology on a set $Y,$ and let $A\subseteq X\subseteq Y.$ The subspace topology on $X$ is $T_X=\{t\cap X: t\in T_Y\}.$
We have  $$Cl_X(A)= X \setminus \cup \{t'\in T_X: t'\cap A=\phi\}=$$ $$= X \setminus \cup \{t\cap X:t\in T_Y\land (t\cap X)\cap A=\phi\}=$$ $$=X \setminus \cup \{t\cap X: t\cap A=\phi\}=$$ $$=X \setminus (Y  \setminus Cl_Y(A)\,)=$$ $$= X\cap Cl_Y(A).$$   So any $A\subseteq X$ is closed in the space $X$ iff $A=Cl_X(A)=X \cap Cl_Y(A).$
