Is it true that a subset that is closed in a closed subspace of a topological space is closed in the whole space? I have a non homework related question from a text and require a nice clear proof/disproof
please
Is it true that a subset that is closed in a closed subspace of a topological
space is closed in the whole space?
my ideas:
if $H$ is the subset of the topological space $X$
if the subset is closed in the closed subspace, the complement is open in the subspace, which means the complement is of form $U\cap H$ for some $U$ open in $X$
if the subspace is closed the complement is open which means complement of $H=U$ for some open $U$ in $X$
kind thanks
 A: Let $X'$ be a closed subset of $X$, and say that $H$ is closed in $X'$ in the subspace topology.  Then $X'\setminus H$ is open in $X'$, and is therefore of the form $G\cap X'$ for some set $G$ that is open in $X$.  Then $X\setminus G$ is closed in $X$, and $(X\setminus G)\cap X' = H$ is an intersection of sets closed in $X$ and is thus closed in $X$.
A: Suppose that $H$ is a closed subspace of $X$, and $F$ is closed in the subspace $H$. By definition of the relative topology there is a closed set $C$ in $X$ such that $F=C\cap H$. But then $F$ is the intersection of two closed subsets of $X$, so $F$ is closed in $X$.
If you don’t already know this characterization of closed sets in the relative topology, it’s worth proving as a separate 

Proposition. Let $Y$ be a subspace of a space $X$. Then a set $H\subseteq Y$ is closed in $Y$ if and only if $H=F\cap Y$ for some closed set $F$ in $X$.

Of course for this you do need to look at complements, but it’s very easy. $Y\setminus H$ is open in $Y$, so there is an open $U\subseteq X$ such that $Y\setminus H=U\cap Y$. Let $F=X\setminus U$; then $F$ is closed in $X$, and $F\cap Y=(X\setminus U)\cap Y=Y\setminus U=H$.
