Neighborhood in Open Subspace

Theorem

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

Let $U \subseteq S$ be an open subset.

Let $\tau_U$ denote the subspace topology on $U$.

Let $s \in U$

Let $N \subseteq U$ be a subset.

Then:

$N$ is a neighborhood of $s$ in $\struct{U, \tau_U}$

if and only if

$N$ is a neighborhood of $s$ in $\struct{S, \tau}$

Proof

Necessary Condition

Let $N$ be a neighborhood of $s$ in $\struct{U, \tau_U}$.

By definition of neighborhood:

$\exists V \in \tau_U : x \in V \subseteq N$

From Open Set in Open Subspace:

$V \in \tau$

Hence:

$\exists V \in \tau : x \in V \subseteq N$

It follows that $N$ is a neighborhood of $s$ in $\struct{S, \tau}$ by definition.

$\Box$

Sufficient Condition

Let $N$ be a neighborhood of $s$ in $\struct{S, \tau}$.

By definition of neighborhood:

$\exists W \in \tau : x \in W \subseteq N$

Let $V = W \cap U$.

From Subset Relation is Transitive:

$V \subseteq N$

By definition of topological space:

$V \in \tau$.

From Open Set in Open Subspace:

$V \in \tau_U$

Hence:

$\exists V \in \tau_U : x \in V \subseteq N$

It follows that $N$ is a neighborhood of $s$ in $\struct{U, \tau_U}$ by definition.

$\blacksquare$

Also see

• Open Set in Open Subspace
• Closed Set in Closed Subspace