Set of Singletons is Smallest Basis of Discrete Space

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

Let $\mathcal B = \left\{{\left\{{x}\right\} : x \in S}\right\}$.

Then $\mathcal B$ is the smallest basis of $T$.

That is:
 * $\mathcal B$ is a basis of $T$

and:
 * for every basis $\mathcal C$ of $T$, $\mathcal B \subseteq \mathcal C$.

Proof
By Basis for Discrete Topology $\mathcal B$ is a basis of $T$.

Let $\mathcal C$ be a basis of $T$.

Let $A \in \mathcal B$.

By definition of the set $\mathcal B$ there exists $x \in T$ such that
 * $A = \left\{x\right\}$.

By definition of singleton:
 * $x \in B$.

By definition of basis there exists $B \in \mathcal C$ such that
 * $x \in B \subseteq A$.

Then by Singleton of Element is Subset:
 * $\left\{x\right\} \subseteq B$.

Hence $B = A$ by definition of set equality.

Thus $A \in \mathcal C$.