Sub-Basis for Topological Subspace

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

Let $K$ be a sub-basis for $\tau$.

Let $\struct {S, \tau'}$ be a subspace of $\struct {X, \tau}$.

Let $K' = \set {U \cap S: U \in K}$.

That is, $K'$ consists of the $\tau'$-open sets in $S$ corresponding to elements of $K$.

Then $K'$ is a sub-basis for $\tau'$.

Proof
Let $B$ be the basis for $\tau$ generated by $K$.

By Basis for Topological Subspace, $B$ generates a basis, $B'$, for $\tau'$.

We will show that $K'$ generates $B'$.

Let $V' \in B'$.

Then for some $V \in B$:
 * $V' = S \cap V$

By the definition of a sub-basis, there is a finite subset $K_V$ of $K$ such that:
 * $\ds V = \bigcap K_V$

Let $K_V' = \set {S \cap U: U \in K_V}$.

Then:
 * $\ds V' = S \cap V = S \cap \bigcap K_V$

so:
 * $\ds V' = \bigcap_{P \mathop \in K_V} \paren {S \cap P} = \bigcap K_V'$

Thus $B'$ is generated by $K'$.