Space is First-Countable iff Character not greater than Aleph 0
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Theorem
Let $T$ be a topological space.
$T$ is first-countable if and only if:
- $\map \chi T \le \aleph_0$
where $\map \chi T$ denotes the character of $T$.
Proof
Sufficient Condition
Let $T$ be first-countable.
By definition of first-countable:
- $\forall x \in T: \exists \BB \in \map {\mathbb B} x: \card \BB \le \aleph_0$
where $\map {\mathbb B} x$ denotes the set of all local bases at $x$.
Then by definition of character of a point:
- $\forall x \in T: \map \chi {x, T} \le \aleph_0$
Hence by definition of character of topogical space:
- $\map \chi T \le \aleph_0$
$\Box$
Necessary Condition
Let $\map \chi T \le \aleph_0$.
By definition of character of topogical space:
- $\forall x \in T: \map \chi {x, T} \le \aleph_0$
Then by definition of character of a point:
- $\forall x \in T: \exists \BB \in \map {\mathbb B} x: \card \BB \le \aleph_0$
Thus by definition of first-countable:
- $T$ is first-countable.
$\blacksquare$
Sources
- Mizar article TOPGEN_2:6