Sub-Basis for Topological Subspace

Theorem
Let $\left({X,\tau}\right)$ be a topological space.

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

Let $\left({S, \tau'}\right)$ be a subspace of $\left({X, \tau}\right)$.

Let $K' = \left\{{ U \cap S: U \in K }\right\}$. 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 subbasis, there is a finite subset $K_V$ of $K$ such that
 * $\displaystyle V = \bigcap K_V$

Let $K_V' = \left\{{ S \cap U: U \in K_V }\right\}$

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

so
 * $\displaystyle V' = \bigcap_{P \in K_V} \left({S \cap P}\right) = \bigcap K_V'$

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