Basis for Topological Subspace

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Theorem

Let $T = \struct {A, \tau}$ be a topological space.

Let $\O \subseteq H \subseteq A$ and so let $T_H = \struct {H, \tau_H}$ be a subspace of $T$.

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

Let $\BB_H$ be defined as:

$B_H = \set {U \cap H: U \in \BB}$


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


Proof

$\BB \subseteq \powerset A$ is a synthetic basis for $T$ if and only if:

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


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

Then:

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

for some $\mathbb S \subseteq \BB$.

Hence:

\(\displaystyle H\) \(=\) \(\displaystyle A \cap H\) Intersection with Subset is Subset‎
\(\displaystyle \) \(=\) \(\displaystyle \paren {\bigcup_{S \mathop \in \mathbb S} S} \cap H\)
\(\displaystyle \) \(=\) \(\displaystyle \bigcup_{S \mathop \in \mathbb S} \paren {S \cap H}\) Intersection Distributes over Union

But if $S \in \BB$, then $S \cap H \in \BB_H$

So $H$ is a union of sets from $\BB_H$.


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

Let $U_1, U_2 \in \BB_H$.

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

Then $U_1 \cap U_2 = B_1 \cap H \cap B_2 \cap H = \paren {B_1 \cap B_2} \cap H$.


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

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

for some $\mathbb S \subseteq \BB$.

Hence:

\(\displaystyle U_1 \cap U_2\) \(=\) \(\displaystyle \paren {B_1 \cap B_2} \cap H\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {\bigcup_{S \mathop \in \mathbb S} S} \cap H\)
\(\displaystyle \) \(=\) \(\displaystyle \bigcup_{S \mathop \in \mathbb S} \paren {S \cap H}\) Intersection Distributes over Union

But if $S \in \BB$, then $S \cap H \in \BB_H$.

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

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

$\blacksquare$


Sources