Sub-Basis for Topological Subspace

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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 sub-basis, 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:

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

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

$\blacksquare$