# A metric space is path connected and countable then it is complete

I have to show that if a metric space is path connected and countable then it is complete. I'm pretty lost where to start this at all. I have the basic definitions of complete, path-connected, compact and sequentially compact spaces.

Any help how to do this would be great (this is a past paper question-non assesed, just for practice so I think it should be reasonable simple)

Thanks very much for any help

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Hint: a connected countable metric space has at most one point. –  Chris Eagle May 4 '12 at 18:09
If it has at most one point then it is obviously sequentially compact and so is compact and so it is complete? –  hmmmm May 4 '12 at 18:56
That's a rather convoluted way to show it, but yes. –  Chris Eagle May 4 '12 at 18:57
If you haven't seen the result mentioned by Chris before, note the following: (1) a metric $d(\cdot,\cdot)$ is a continuous map into $\mathbb{R}$, so if $x$ is in the metric space, then $d(x,\cdot)$ is likewise a continuous map into $\mathbb{R}$; (2) the connected subspaces of $\mathbb{R}$ are $\emptyset$, $\mathbb{R}$, singletons, and intervals; (3) path-connected spaces are connected; (4) continuous images of connected spaces are connected. –  Cameron Buie May 4 '12 at 19:00
Much simpler: the only sequence in a one-point space is the constant sequence, and it converges. Since every sequence in $X$ converges, $X$ is complete. –  Brian M. Scott May 4 '12 at 19:02