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

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. Thus $$\bar{A}$$~$$B$$ and $$B$$ is countable since $$\bar{A}$$ is countable. Thus $$B$$ is a dense countable subset of $$\hat{X}$$ and hence $$(\hat{X},\hat{d})$$ is separable.

Is this the right way of approaching it?