Separable Topological Space remains Separable in Coarser Topology

Theorem

Let $\struct {X, \tau_2}$ be a separable topological space.

Let $\tau_1$ be a topology on $X$ such that:

$\tau_1 \subseteq \tau_2$

That is, such that $\tau_1$ is coarser than $\tau_2$.

Then $\struct {X, \tau_1}$ is separable.

Proof

Let $\cl_1$ and $\cl_2$ denote the topological closure in $\struct {X, \tau_1}$ and $\struct {X, \tau_2}$ respectively.

Let $D$ be a countable everywhere dense subset of $\struct {X, \tau_2}$.

From Topological Closure in Coarser Topology is Larger, we have:

$\map {\cl_2} D \subseteq \map {\cl_1} D$

Since $D$ is everywhere dense in $\struct {X, \tau_2}$, we have $\map {\cl_2} D = X$.

So:

$\map {\cl_1} D = X$

So $D$ is everywhere dense in $\struct {X, \tau_1}$.

$\blacksquare$