closed set in a subspace

In the following, let $X$ be a topological space.

Theorem 1.

Suppose $Y\mathrm{\subseteq }X$ is equipped with the subspace topology,and $A\mathrm{\subseteq }Y$.Then $A$ is closed (http://planetmath.org/ClosedSet) in $Y$ if and only if$A\mathrm{=}Y\mathrm{\cap }J$ for some closed set $J\mathrm{\subseteq }X$.

Proof.

If $A$ is closed in $Y$,then $Y\setminus A$ is open (http://planetmath.org/OpenSet) in $Y$,and by the definition of the subspace topology,$Y\setminus A=Y\cap U$ for some open $U\subseteq X$.Using properties of the set difference (http://planetmath.org/SetDifference),we obtain

$A$$=$$Y\setminus (Y\setminus A)$

$=$$Y\setminus (Y\cap U)$

$=$$Y\setminus U$

$=$$Y\cap U^{\mathrm{\complement }}.$

On the other hand, if $A=Y\cap J$ for some closed $J\subseteq X$,then $Y\setminus A=Y\setminus (Y\cap J)=Y\cap J^{\mathrm{\complement }}$,and so $Y\setminus A$ is open in $Y$,and therefore $A$ is closed in $Y$.∎

Theorem 2.

Suppose $X$ is a topological space, $C\mathrm{\subseteq }X$is a closed set equipped with the subspace topology,and $A\mathrm{\subseteq }C$ is closed in $C$.Then $A$ is closed in $X$.

Proof.

This follows from the previous theorem:since $A$ is closed in $C$,we have $A=C\cap J$ for some closed set $J\subseteq X$,and $A$ is closed in $X$.∎