Closed Set 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 $T' - V$ is open in $T'$ by definition.

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

Then:

Thus $T - U$ is closed in $T$.


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

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

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

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 $T$.