1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

$B$ may not be countable (just like $X$ may not be countable!), but $f[A]$ is countable and dense in $B$, and thus dense in $\overline B=\widehat X$.

$\endgroup$
  • $\begingroup$ is $f[A]$ the image of $f:X\rightarrow B$ where A is acting as the input? $\endgroup$ – s_healy May 8 at 0:40
  • 1
    $\begingroup$ $f[A]=\{f(x)\,:\, x\in A\}$ and $f$ is your isometry $f:X\to B$ $\endgroup$ – Saucy O'Path May 8 at 0:40
  • $\begingroup$ 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 $\endgroup$ – s_healy May 8 at 0:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.