Closed Sets in Topological Subspace

Theorem

Let $T$ be a topological space.

Let $T' \subseteq T$ be a subspace of $T$.

Then $V \subseteq T'$ is closed in $T'$ iff $V = T' \cap W$ for some $W$ closed in $T$.

Corollary

Suppose the above defined subspace $T'$ is closed in $T$.

Then $V \subseteq T'$ is closed in $T'$ iff $V$ is closed in $T$.

Proof

• Suppose $V \subseteq T'$ is closed in $T'$.

Then $T' \setminus V$ is open in $T'$ by definition.

So, by definition of subspace topology, $T' \setminus V = T' \cap U$ for some $U$ open in $T$.

Then:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle V\)	\(=\)	\(\displaystyle T' \setminus \left({T' \setminus V}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Relative Complement of Relative Complement
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle T' \setminus \left({T' \cap U}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	from above
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle T' \setminus U\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Set Difference with Intersection is Difference
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle T' \cap \left({T \setminus U}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Thus $T \setminus U$ is closed in $T$.

• Conversely, suppose $V = T' \cap W$ where $W$ closed in $T$.

Then $T' \setminus V = T' \setminus \left({T' \cap W}\right) = T' \cap \left({T \setminus W}\right)$ which is open in $T'$.

So $V$ is closed in $T'$.

$\blacksquare$

Proof of Corollary

If $V \subseteq T'$ is closed in $T'$ then $V = T' \cap V$ is closed in $T$.

If $V$ is closed in $T$ then $V = T' \cap W$ where $W$ is closed in $T$.

Since $T'$ is closed in $T$, it follows by Topology Defined by Closed Sets that $V$ is closed in $T$.

$\blacksquare$