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 $X$. ∎

Title	closed set in a subspace
Canonical name	ClosedSetInASubspace
Date of creation	2013-03-22 15:33:32
Last modified on	2013-03-22 15:33:32
Owner	yark (2760)
Last modified by	yark (2760)
Numerical id	9
Author	yark (2760)
Entry type	Theorem
Classification	msc 54B05