Local Basis of Open Subspace iff Local Basis

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Theorem

Let 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 $\BB \subseteq \powerset U$.

Then:

$\BB$ is a local basis of $s$ in $\struct{U, \tau_U}$ if and only if $\BB$ is a local basis of $s$ in $\struct{S, \tau}$

Proof

Let $\map \BB s$ denote the set of open sets containing $s$ in $\struct{S, \tau}$

Let $\map \CC s$ denote the set of open sets containing $s$ in $\struct{U, \tau_U}$

From Open Set in Open Subspace:

$\BB \subseteq \map \BB s$ if and only if $\BB \subseteq \map \CC s$

Necessary Condition

Let $\BB$ be a local basis of $s$ in $\struct{U, \tau_U}$.

Let $W \in \map \BB s$.

By definition of subspace topology:

$W' = W \cap U \in \tau_U$

By definition of local basis:

$\exists V \in \BB : x \in V \subseteq W'$

From Subset Relation is Transitive:

$V \subseteq W$

We have:

$\exists V \in \BB : x \in V \subseteq W$

It follows that:

$\BB$ is a local basis of $s$ in $\struct{S, \tau}$

$\Box$

Sufficient Condition

Let $\BB$ be a local basis of $s$ in $\struct{S, \tau}$.

Let $W \in \map \CC s$.

From Open Set in Open Subspace:

$W \in \map \BB s$.

Hence:

$\exists V \in \BB : x \in V \subseteq W$

It follows that:

$\BB$ is a local basis of $s$ in $\struct{U, \tau_U}$

$\blacksquare$