Continuous Functions on Separable Metric Spaces

# Continuous Functions on Separable Metric Spaces

Recall from the Separable Metric Spaces page that a metric space $(M, d)$ is said to be separable if there exists a countable and dense subset [$S \subseteq M$, where we said $S$ is dense if for all $x \in M$ and for all $r > 0$, $B(x, r) \cap S \neq \emptyset$, i.e., every open ball in $(M, d)$ contains points of $S$.

We will now look at a nice theorem which tells us that if we have a continuous function $f$ from one metric space to another, and if the domain metric space is separable, then the range as a metric subspace of the codomain metric space is also separable.

 Theorem 1: Let $(S, d_S)$ and $(T, d_T)$ be metric spaces and let $f : S \to T$ be continuous. If $(S, d_S)$ is separable, then $(f(S), d_T)$ is separable.

Note that $(f(S), d_T)$ is a metric subspace of $(T, d_T)$.

• Proof: Let $(S, d_S)$ be a separable metric space. Then there exists a subset $A \subseteq S$ such that $A$ is countable and for all $x \in M$ and for all $r > 0$ we have that:
(1)
\begin{align} \quad B_S(x, r) \cap A \neq \emptyset \end{align}
• Now $f(A)$ is clearly countable and we claim that $f(A)$ is dense in $Y$. To show this, suppose instead that there exists a $y \in f(S)$ and an $r > 0$ such that:
(2)
\begin{align} \quad B_{f(S)} (y, r) \cap f(A) = \emptyset \end{align}
• Then taking the inverse image of both sides gives us that:
(3)
\begin{align} \quad f^{-1}(B_{f(S)}(y, r)) \cap A = \emptyset \end{align}
• Since $f$ is continuous and $B_{f(S)} (y, r)$ is open in $Y$ we have that $f^{-1}(B_{f(S)}(y, r))$ is open in $X$. Let $x \in f^{-1}(B_{f(S)}(y, r))$. Then since this set is open, there exists an $r_0 > 0$ such that $B(x, r_0) \subseteq f^{-1}(B_{f(S)}(y, r))$. But since $f^{-1}(B_{f(S)}(y, r)) \cap A = \emptyset$, this means that:
(4)
\begin{align} \quad B(x, r_0) \cap A = \emptyset \end{align}
• But this is a contradiction since $A$ is dense in $X$. So the assumption that there exists a $y \in f(S)$ and an $r > 0$ for which $B_{f(S)} (y, r) \cap f(A) = \emptyset$ is false, i.e., for all $y \in f(S)$ and for all $r > 0$ we have that $B_{f(S)} (y, r) \cap f(A) \neq \emptyset$. So $f(A)$ is a countable dense subset of $f(S)$ and so $(f(S), d_T)$ is a separable metric space. $\blacksquare$