Basis for Topological Subspace

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Theorem

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

Let $\varnothing \subseteq H \subseteq A$ and so let $T_H = \left({H, \tau_H}\right)$ 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 $T$ if and only if:

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:

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

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:

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

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 $H$.

$\blacksquare$


Sources