Open Set in Open Subspace

Theorem

Let $T = \struct{X, \tau}$ be a topological space.

Let $T_U = \struct{U, \tau_U}$ be a subspace of $T$ where $U \subseteq X$ is open.

Let $V \subseteq U$ be a subset.

Then $V$ is open in $T_U$ if and only if $V$ is open in $T$.

Proof

Necessary Condition

Let $V$ be open in $T$.

By Intersection with Subset is Subset, $V\cap U = V$.

By definition of topological subspace, $V$ is open in $T_U$.

Sufficient Condition

Let $V$ be open in $T_U$.

By definition of topological subspace, there exists an open subset $W\subseteq X$ such that $V=U\cap W$.

Because $U$ and $W$ are open in $T$, $V=U\cap W$ is open in $T$.

$\blacksquare$

Also see

• Closed Set in Closed Subspace

Sources

• 2011: John M. Lee: Introduction to Topological Manifolds (2nd ed.) ... (previous) ... (next): $\S 3$: New Spaces From Old: Subspaces