# Continuous Image of Separable Space is Separable

## Definition

Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.

Let $f: T_1 \to T_2$ be a continuous mapping.

If $T_1$ is separable, then so is the image $f \sqbrk {T_1}$.

That is, separability is a continuous invariant.

## Proof

By definition, $T_1 = \struct {S_1, \tau_1}$ is separable if and only if there exists a countable subset $D \subset S_1$ which is (everywhere) dense in $T_1$.

We need to show that if there exists a continuous mapping $f: T_1 \to T_2$, then $f \sqbrk {T_1}$ is also separable.

That is, that there exists a countable subset of $f \sqbrk {S_1}$ which is dense in $T_2$.

Let $x_2$ be any point in the image $f \sqbrk {S_1}$ of $S_1$ under $f$.

Let $U \in \tau_2$ be an arbitrary open set such that $x_2 \in U$.

By definition of image set, there exists some $x_1 \in S_1$ with $\map f {x_1} = x_2$.

Since $f$ is continuous, $f^{-1} \sqbrk U$ is open in $T_1$.

By definition of preimage, $x_1$ is in this set.

We are given that $D$ is a countable subset of $S_1$ which is dense in $T_1$.

By definition of dense, $D \cap f^{-1} \sqbrk U \ne \O$

Thus there exists some $d \in D$ such that $d \in f^{-1} \sqbrk U$.

Therefore $\map f d \in U$.

Since $U$ was arbitrary, it follows that $f \sqbrk D$ is dense in $T_2$.

By Image of Countable Set under Mapping is Countable, $f \sqbrk D$ is countable.

Hence $T_2$ is separable.

$\blacksquare$