# Existence of Subfamily of Cardinality not greater than Weight of Space and Unions Equal

## Theorem

Let $T$ be a topological space.

Let $\mathcal F$ be a set of open sets of $T$.

There exists a subset $\mathcal G \subseteq \mathcal F$ such that:

$\displaystyle \bigcup \mathcal G = \bigcup \mathcal F$

and:

$\left\vert{\mathcal G}\right\vert \leq w \left({T}\right)$

where:

$w \left({T}\right)$ denotes the weight of $T$
$\left\vert{\mathcal G}\right\vert$ denotes the cardinality of $\mathcal G$.

## Proof

By definition of weight of $T$ there exists a basis $\mathcal B$ of $T$ such that:

$(1): \quad \left\vert{\mathcal B}\right\vert = w \left({T}\right)$

Let:

$\mathcal B_1 = \left\{{W \in \mathcal B: \exists U \in \mathcal F: W \subseteq U}\right\}$

By definition of subset:

$\mathcal B_1 \subseteq \mathcal B$
$(2): \quad \left\vert{\mathcal B_1}\right\vert \leq \left\vert{\mathcal B}\right\vert$

By definition of set $\mathcal B_1$:

$\forall W \in \mathcal B_1: \exists U \in \mathcal F: W \subseteq U$

Then by the Axiom of Choice there exists a mapping $f$ from $\mathcal B_1$ into $\mathcal F$ such that

$(3): \quad \forall U \in \mathcal B_1: U \subseteq f \left({U}\right)$

Let:

$\mathcal G = \operatorname{Im} \left({f}\right)$

where $\operatorname{Im} \left({f}\right)$ denotes the image of $f$.

$\mathcal G \subseteq \mathcal F$
$\displaystyle \bigcup \mathcal G \subseteq \bigcup \mathcal F$

By definition of set equality, to prove $\displaystyle \bigcup \mathcal G = \bigcup \mathcal F$ it is sufficient to show:

$\displaystyle \bigcup \mathcal F \subseteq \bigcup \mathcal G$

Let $\displaystyle x \in \bigcup \mathcal F$.

By definition of union there exists a set $A$ such that:

$A \in \mathcal F$ and $x \in A$

Because $A$ is open, then by definition of basis there exists $U \in \mathcal B$ such that:

$x \in U \subseteq A$

By definition of the set $\mathcal B_1$:

$U \in \mathcal B_1$

Because $\mathcal G = \operatorname{Im} \left({f}\right)$:

$f \left({U}\right) \in \mathcal G$

By $(3)$ and Set is Subset of Union:

$\displaystyle U \subseteq f \left({U}\right) \subseteq \bigcup \mathcal G$

Thus by definition of subset:

$\displaystyle x \in \bigcup \mathcal G$

This ends the proof of inclusion.

$\left\vert{\operatorname{Im} \left({f}\right)}\right\vert \leq \left\vert{\mathcal B_1}\right\vert$

Thus by $(1)$ and $(2)$:

$\left\vert{\mathcal G}\right\vert \leq w \left({T}\right)$

$\blacksquare$