closed set in a subspace

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

Theorem 1.

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

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

 $\displaystyle A$ $\displaystyle=$ $\displaystyle Y\setminus(Y\setminus A)$ $\displaystyle=$ $\displaystyle Y\setminus(Y\cap U)$ $\displaystyle=$ $\displaystyle Y\setminus U$ $\displaystyle=$ $\displaystyle Y\cap U^{\complement}.$

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

Theorem 2.

Suppose $X$ is a topological space, $C\subseteq X$ is a closed set equipped with the subspace topology, and $A\subseteq C$ 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 ClosedSetInASubspace 2013-03-22 15:33:32 2013-03-22 15:33:32 yark (2760) yark (2760) 9 yark (2760) Theorem msc 54B05