Weight of Discrete Topology equals Cardinality of Space

Theorem
Let $T = \left({S, \tau}\right)$ be a discrete topological space.

Then:
 * $w \left({T}\right) = \left\vert{S}\right\vert$

where
 * $w \left({T}\right)$ denotes the weight of $T$
 * $\left\vert{S}\right\vert$ denotes the cardinality of $S$.

Proof
By Basis for Discrete Topology the set $\mathcal B = \left\{{\left\{{x}\right\}: x \in S}\right\}$ is a basis of $T$.

By Set of Singletons is Smallest Basis of Discrete Space $\mathcal B$ is smallest basis of $T$:
 * for every basis $\mathcal C$ of $T$, $\mathcal B \subseteq \mathcal C$.

Then by Subset implies Cardinal Inequality:
 * for every basis $\mathcal C$ of $T$, $\left\vert{\mathcal B}\right\vert \leq \left\vert{\mathcal C}\right\vert$.

Hence $\left\vert{\mathcal B}\right\vert$ is minimal cardinalty of basis of $T$:
 * $w \left({T}\right) = \left\vert{\mathcal B}\right\vert$ by definition of weight.

Thus by Cardinality of Set of Singletons:
 * $w \left({T}\right) = \left\vert{S}\right\vert$.