Cauchy sequences are not preserved between homeomorphic metric spaces, e.g. the sequence $x_n=1/(n+1)$ in $(0,1)$ is cauchy but not when passing to $\mathbb{R}$ by $1/x$. A natural follow up is if every metric space has a metric under it is complete. I would think that $\mathbb{Q}$ fails this, but it's not clear how to show this.

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    $\begingroup$ You could use the Baire category theorem $\endgroup$ Sep 21 at 11:19
  • $\begingroup$ Metrisable spaces for which there exists a compatible complete metric are called "completely metrisable" it seems, if you'd like to look into this more. $\endgroup$
    – Bruno B
    Sep 21 at 11:30
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    $\begingroup$ There is a well known result in general topology (will try to find a reference or to write a proof later when I have the time) that states that a metrisable space $Y$ is completely metrizable if and only if its image through any embedding inside a complete metric space $X$ is a $G_\delta$. $\endgroup$
    – Pelota
    Sep 21 at 11:31
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    $\begingroup$ The following are equivalent for a metric space $X$: 1) $X$ is completely metrizable; 2) $X$ is a $G_\delta$ whenever embedded in a complete metric space; 3) $X$ is a $G_\delta$ is its Stone-Čech compactification; 4) $X$ is a $G_\delta$ whenever densely embedded in a Tychonoff space. $\endgroup$
    – Ningxin
    Sep 21 at 12:49
  • $\begingroup$ Presumably you mean that the new metric should induce the same topology as the original one. Otherwise the answer is trivially yes, because every set can be equipped with the discrete metric, which is complete. $\endgroup$ Sep 22 at 0:12

1 Answer 1


As you guessed, $\mathbb Q$ is a counter example:

  1. points are not open

  2. in metric spaces points are closed

  3. thus $\mathbb Q$ with any metric has no isolated points

  4. $\mathbb Q$ is countable union of points, that are closed sets with empty interiors

  5. Complete metric spaces withouht isolated points are not countable union of closed sets with empty interior (this is the Baire Theorem, you can easily prove it by proving -- via a recursive argument --- the contrapositive: countable intersection of open dense sets is dense)

4)+5) prove that $\mathbb Q$ cannot have a complete metric.


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