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 Intersection and Union of Closed Sets that $$V$$ is closed in $$T$$.