Density not greater than Weight

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

Then
 * $d \left({T}\right) \leq w \left({T}\right)$

where
 * $d \left({T}\right)$ denotes the density of $T$,
 * $w \left({T}\right)$ denotes the weight of $T$.

Proof
By definition of weight there exists a basis $\mathcal B$ of $T$:
 * $w \left({T}\right) = \left\vert{\mathcal B}\right\vert$

where $\left\vert{\mathcal B}\right\vert$ denotes the cardinality of $\mathcal B$.

By Axiom of Choice define a mapping $f: \left\{{U \in \mathcal B: U \ne \varnothing}\right\} \to S$:
 * $\forall U \in \mathcal B: U \ne \varnothing \implies f \left({U}\right) \in U$

We will prove that
 * $\forall U \in \tau: U \ne \varnothing \implies U \cap \operatorname{Im} \left({f}\right) \ne \varnothing$

where $\operatorname{Im} \left({f}\right)$ denotes the image of $f$.

Let $U \in \tau$ such that
 * $U \ne \varnothing$

By definition of empty set:
 * $\exists x: x \in U$

Then by definition of basis:
 * $\exists V \in \mathcal B: x \in V \subseteq U$

By definition of image:
 * $f \left({V}\right) \in \operatorname{Im} \left({f}\right)$

By definition of $f$:
 * $f \left({V}\right) \in V$

By definition of subset:
 * $f \left({V}\right) \in U$

Then by definition of intersection:
 * $f \left({V}\right) \in U \cap \operatorname{Im} \left({f}\right)$

Thus by definition of empty set:
 * $U \cap \operatorname{Im} \left({f}\right) \ne \varnothing$

Then:
 * $\forall x \in S: \forall U \in \tau: x \in U \implies U \cap \operatorname{Im} \left({f}\right) \ne \varnothing$

Hence by Condition for Point being in Closure:
 * $\forall x \in S: x \in \left( {\operatorname{Im} \left({f}\right)}\right)^-$

where $^-$ denotes the topological closure.

Then by definition of subset:
 * $S \subseteq \left( {\operatorname{Im} \left({f}\right)}\right)^- \subseteq S$

Thus by definition of set equality:
 * $S = \left( {\operatorname{Im} \left({f}\right)}\right)^-$

Thus by definition:
 * $\operatorname{Im} \left({f}\right)$ is dense

By definition of density:
 * $d \left({T}\right) \leq \left\vert{ \operatorname{Im} \left({f}\right) }\right\vert$

$f$ as $\left\{{U \in \mathcal B: U \ne \varnothing} \right\} \to \operatorname{Im} \left({f}\right)$ by definition is surjection.

Therefore by Surjection iff Cardinal Inequality:
 * $\left\vert{\operatorname{Im} \left({f}\right)}\right\vert \leq \left\vert{\left\{{U \in \mathcal B: U \ne \varnothing} \right\}}\right\vert$

By definition of subset:
 * $\left\{{U \in \mathcal B: U \ne \varnothing} \right\} \subseteq \mathcal B$

Then by Subset implies Cardinal Inequality:
 * $\left\vert{\left\{{U \in \mathcal B: U \ne \varnothing} \right\}}\right\vert \leq \left\vert{\mathcal B}\right\vert$

Thus:
 * $d \left({T}\right) \leq w \left({T}\right)$