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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.2012-05-04
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    If it has at most one point then it is obviously sequentially compact and so is compact and so it is complete?2012-05-04
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    That's a rather convoluted way to show it, but yes.2012-05-04
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    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.2012-05-04
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    *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.2012-05-04
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    @CameronBuie and as the metric space is countable so is the image so it is either the singleton or the empty set right?2012-05-04
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    That works. Or to show Chris's result, suppose the space has more than one point, so if $x,y$ distinct points, then $d(x,x)=0$ and $d(x,y)>0$, so the image of the space has more than one point, so is either a non-degenerate interval or all of $\mathbb{R}$, and in either case is uncountable, so cannot be mapped onto by a countable set, and so the space is uncountable.2012-05-04
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    Actually, your version is simpler. The image will necessarily be a singleton for a non-empty countable space, meaning $d(x,y)=0$ for all $y$ in the space, and so there is only one point in the space by metric properties.2012-05-04

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