Closed Set in Topological Subspace

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Theorem

Let $T$ be a topological space.

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


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


Corollary

Let subspace $T'$ be closed in $T$.

Then $V \subseteq T'$ is closed in $T'$ if and only if $V$ is closed in $T$.


Proof

Necessary Condition

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:

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

Thus $V$ is closed in $T$.

$\Box$


Sufficient Condition

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

Then $T' \setminus V = T' \setminus \paren {T' \cap W} = T' \cap \paren {T \setminus W}$ which is open in $T'$.

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

$\blacksquare$


Also see


Sources