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