Continuous Image of Everywhere Dense Set is Everywhere Dense

Theorem

Let $A_T = \struct {A, \tau_A}$ and $B_T = \struct {B, \tau_B}$ be topological spaces.

Let $f : A \to B$ be an everywhere continuous surjection.

Let $S \subseteq A$ be everywhere dense in $A_T$.

Then, $f \sqbrk S$ is everywhere dense in $B_T$.

Proof

\(\ds \paren {f \sqbrk S}^-\)

\(\supseteq\)

\(\ds f \sqbrk {S^-}\)

Continuity Defined by Closure

\(\ds \)

\(=\)

\(\ds f \sqbrk A\)

Definition 1 of Everywhere Dense

\(\ds \)

\(=\)

\(\ds B\)

Definition 2 of Surjection

As $\paren {f \sqbrk S}^- \subseteq B$, by set equality:

$\paren {f \sqbrk S}^- = B$

Therefore, $f \sqbrk S$ is everywhere dense in $B$ by definition.

$\blacksquare$