Prove that if a metric space (X,d) is separable, then its completion ($\hat{X}, \hat{d}$) is separable.

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?

$$B$$ may not be countable (just like $$X$$ may not be countable!), but $$f[A]$$ is countable and dense in $$B$$, and thus dense in $$\overline B=\widehat X$$.
• is $f[A]$ the image of $f:X\rightarrow B$ where A is acting as the input? – s_healy May 8 '19 at 0:40
• $f[A]=\{f(x)\,:\, x\in A\}$ and $f$ is your isometry $f:X\to B$ – Saucy O'Path May 8 '19 at 0:40
• so because $f[A]$ is countable and dense in $B$, $f[A]$ is dense AND countable in $\overline B = \widehat X$ since $B \subseteq \overline B$? hence a dense countable subset exists in $\widehat X$ and thus ($\hat{X}, \hat{d}$) is separable – s_healy May 8 '19 at 0:48