Prove that if a metric space $(X,d)$ is separable, then its completion $(\hat{X}, \hat{d})$ is separable.
So we want to show there exists a countable dense subset in $\hat{X}$.
My attempt:
Suppose a metric space $(X,d)$ is separable.
Then there exists a dense countable subset $A \subseteq X$.
Since $A$ is dense in $X$, then $\bar{A}=X$.
Let $(\hat{X},\hat{d})$ be the completion of $(X,d)$. Then there exists a dense subset $B\subseteq \hat{X}$ so that $(X,d)$ is isometric to $(B,{\hat{d}|}_{AxA})$.
Thus $f:X\rightarrow B$ is a bijection
and so $f:\bar{A} \rightarrow B$ is also bijective.
How would I go about showing $B$ is countable or is it countable because a bijection exists?
Best Answer
$B$ may not be countable (just like $X$ may not be countable!), but $f[A]$ is countable and dense in $B$, and thus dense in $\overline B=\widehat X$.