Basis for Topological Subspace

Theorem
Let $T = \left({A, \vartheta}\right)$ be a topological space.

Let $\varnothing \subseteq H \subseteq A$ and so let $T_H = \left({H, \vartheta_H}\right)$ be a subspace of $T$.

Let $\mathcal B$ be a (synthetic) basis for $T$.

Let $\mathcal B_H$ be defined as:
 * $B_H = \left\{{U \cap H: U \in \mathcal B}\right\}$

Then $\mathcal B_H$ is a (synthetic) basis for $H$.

Proof
$\mathcal B \subseteq \mathcal P \left({A}\right)$ is a synthetic basis for $T$ iff:


 * B1: $A$ is a union of sets from $\mathcal B$
 * B2: If $B_1, B_2 \in B$, then $B_1 \cap B_2$ is a union of sets from $\mathcal B$.

Let $A = \mathbb S$ is a union of sets from $\mathcal B$.

Then:


 * $\displaystyle A = \bigcup_{S \mathop \in \mathbb S} S$

for some $\mathbb S \subseteq \mathcal B$.

Hence:

But if $S \in \mathcal B$, then $S \cap H \in \mathcal B_H$

So $H$ is a union of sets from $\mathcal B_H$.

In the same way we investigate $U_1$ and $U_2$.

Let $U_1, U_2 \in \mathcal B_H$.

Then $U_1 = B_1 \cap H, U_2 = B_2 \cap H$ for some $B_1, B_2 \in \mathcal B$.

Then $U_1 \cap U_2 = B_1 \cap H \cap B_2 \cap H = \left({B_1 \cap B_2}\right) \cap H$.

As $\mathcal B$ is a (synthetic) basis for $A$, we have that:


 * $\displaystyle B_1 \cap B_2 = \bigcup_{S \mathop \in \mathbb S} S$

for some $\mathbb S \subseteq \mathcal B$.

Hence:

But if $S \in \mathcal B$, then $S \cap H \in \mathcal B_H$.

So $U_1 \cap U_2$ is a union of sets from $\mathcal B_H$.

So $\mathcal B_H$ fulfils the conditions to be a synthetic basis for $H$.