Denseness Preserved in Coarser Topology

Theorem

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

Let $T' = \struct {S, \tau'}$ be another topological space on $S$ such that $\tau'$ is coarser than $\tau$.

Let $H \subseteq S$ be everywhere dense in $T$.

Then $H$ is also everywhere dense in $T'$.

Proof

Let $H \subseteq S$ be everywhere dense in $T$.

As $\tau'$ is coarser than $\tau$, by definition $\tau' \subseteq \tau$.

Let $x \in S$.

As $H \subseteq S$ is everywhere dense in $T$, either $x \in H$ or $x$ is a limit point of $H$ in $T$.

Suppose $x \notin H$.

Then $x$ is a limit point of $H$ in $T$.

Then by definition of limit point of $H$:

$\forall U \in \tau, x \in U: \exists y \in H: y \in U, y \ne x$

As $\tau' \subseteq \tau$ we have that $U \in \tau' \implies U \in \tau$.

So:

$\forall U \in \tau', x \in U: \exists y \in H: y \in U, y \ne x$

That is, $x$ is a limit point of $H$ in $T\ '$.

As $x$ is arbitrary, this follows for all $x \in S, x \notin H$.

So, for all $x \in S$, either $x \in H$ or $x$ is a limit point of $H$ in $T'$.

Thus, by definition, $H \subseteq S$ is everywhere dense in $T'$.

$\blacksquare$