Sub-Basis for Topological Subspace
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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 $K'$.
$\blacksquare$