Density not greater than Weight

Theorem
Let $T = \struct {S, \tau}$ be a topological space.

Then
 * $\map d T \le \map w T$

where
 * $\map d T$ denotes the density of $T$,
 * $\map w T$ denotes the weight of $T$.

Proof
By definition of weight there exists a basis $\BB$ of $T$:
 * $\map w T = \card \BB$

where $\card \BB$ denotes the cardinality of $\BB$.

By Axiom of Choice define a mapping $f: \paren {\BB \setminus \O} \to S$:
 * $\forall U \in \BB: U \ne \O \implies f \sqbrk U \in U$

We will prove that
 * $\forall U \in \tau: U \ne \O \implies U \cap \Img f \ne \O$

where $\Img f$ denotes the image of $f$.

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

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

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

By definition of image:
 * $f \sqbrk V \in \Img f$

By definition of $f$:
 * $f \sqbrk V \in V$

By definition of subset:
 * $f \sqbrk V \in U$

Then by definition of intersection:
 * $f \sqbrk V \in U \cap \Img f$

Thus by definition of empty set:
 * $U \cap \Img f \ne \O$

Then:
 * $\forall x \in S: \forall U \in \tau: x \in U \implies U \cap \Img f \ne \O$

Hence by Condition for Point being in Closure:
 * $\forall x \in S: x \in \paren {\Img f}^-$

where $^-$ denotes the topological closure.

Then by definition of subset:
 * $S \subseteq \paren {\Img f}^- \subseteq S$

Thus by definition of set equality:
 * $S = \paren {\Img f}^-$

Thus by definition:
 * $\Img f$ is dense

By definition of density:
 * $\map d T \le \card {\Img f}$

$f$ as $\paren {\BB \setminus \O} \to \Img f$ by definition is surjection.

Therefore by Surjection iff Cardinal Inequality:
 * $\card {\Img f} \le \card {\BB \setminus \O}$

From Set Difference is Subset:
 * $\BB \setminus \O \subseteq \BB$

Then by Subset implies Cardinal Inequality:
 * $\card {\BB \setminus \O} \le \card \BB$

Thus:
 * $\map d T \le \map w T$