Space is First-Countable iff Character not greater than Aleph 0

Theorem
Let $T$ be a topological space.

$T$ is first-countable :
 * $\chi \left({T}\right) \leq \aleph_0$

where $\chi \left({T}\right)$ denotes the character of $T$.

Sufficient Condition
Let $T$ be first-countable.

By definition of first-countable:
 * $\forall x \in T: \exists \mathcal B \in \mathbb B \left({x}\right): \left\vert{\mathcal B}\right\vert \leq \aleph_0$

where $\mathbb B \left({x}\right)$ denotes the set of all local bases at $x$.

Then by definition of character of a point:
 * $\forall x \in T: \chi \left({x, T}\right) \leq \aleph_0$

Hence by definition of character of topogical space:
 * $\chi \left({T}\right) \leq \aleph_0$

Necessary Condition
Let $\chi \left({T}\right) \leq \aleph_0$.

By definition of character of topogical space:
 * $\forall x \in T: \chi \left({x, T}\right) \leq \aleph_0$

Then by definition of character of a point:
 * $\forall x \in T: \exists \mathcal B \in \mathbb B \left({x}\right): \left\vert{\mathcal B}\right\vert \leq \aleph_0$

Thus by definition of first-countable:
 * $T$ is first-countable.