Existence of Subfamily of Cardinality not greater than Weight of Space and Unions Equal
Theorem
Let $T$ be a topological space.
Let $\FF$ be a set of open sets of $T$.
There exists a subset $\GG \subseteq \FF$ such that:
- $\ds \bigcup \GG = \bigcup \FF$
and:
- $\card \GG \le \map w T$
where:
- $\map w T$ denotes the weight of $T$
- $\card \GG$ denotes the cardinality of $\GG$.
Proof
By definition of weight of $T$ there exists a basis $\BB$ of $T$ such that:
- $(1): \quad \card \BB = \map w T$
Let:
- $\BB_1 = \set {W \in \BB: \exists U \in \FF: W \subseteq U}$
By definition of subset:
- $\BB_1 \subseteq \BB$
Then by Subset implies Cardinal Inequality:
- $(2): \quad \card {\BB_1} \le \card \BB$
By definition of set $\BB_1$:
- $\forall W \in \BB_1: \exists U \in \FF: W \subseteq U$
Then by the Axiom of Choice there exists a mapping $f$ from $\BB_1$ into $\FF$ such that
- $(3): \quad \forall U \in \BB_1: U \subseteq f \sqbrk U$
Let:
- $\GG = \Img f$
where $\Img f$ denotes the image of $f$.
- $\GG \subseteq \FF$
Then by Union of Subset of Family is Subset of Union of Family:
- $\ds \bigcup \GG \subseteq \bigcup \FF$
By definition of set equality, to prove $\ds \bigcup \GG = \bigcup \FF$ it is sufficient to show:
- $\ds \bigcup \FF \subseteq \bigcup \GG$
Let $\ds x \in \bigcup \FF$.
By definition of union there exists a set $A$ such that:
- $A \in \FF$ and $x \in A$
Because $A$ is open, then by definition of basis there exists $U \in \BB$ such that:
- $x \in U \subseteq A$
By definition of the set $\BB_1$:
- $U \in \BB_1$
Because $\GG = \Img f$:
- $f \sqbrk U \in \GG$
By $(3)$ and Set is Subset of Union:
- $\ds U \subseteq f \sqbrk U \subseteq \bigcup \GG$
Thus by definition of subset:
- $\ds x \in \bigcup \GG$
This ends the proof of inclusion.
By Cardinality of Image of Mapping not greater than Cardinality of Domain:
- $\card {\Img f} \le \card {\BB_1}$
Thus by $(1)$ and $(2)$:
- $\card \GG \le \map w T$
$\blacksquare$
Sources
- Mizar article TOPGEN_2:11