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Is completeness an intrinsic property of a space that is independent of metric? For example, since $\mathbb{R}^n$ is complete with the Euclidean metric, is it complete with any other metric?

If completeness is an intrinsic property, why is it intrinsic?

Thanks :)

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    Nope. $\mathbb R^n$ is homeomorphic to an open ball, and with the Euclidean metric on the ball, it is not complete.2012-02-29
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    You should really specify if you mean "any other metric" or "any other metric that generates the same topology."2012-02-29
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    Sorry, but actually that's something else I didn't know. I had always assumed different metrics would generate different topologies, but I see what you mean, e.g. if we take $d(x,y) = |x-y|$ and $d(x,y) = 2|x-y|$ they generate the same topology.2012-02-29

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The completness depend of the metric, for instance $\mathbb{Q}$ with the standard metric is not complete, but if you consider $\mathbb{Q}$ with the distance : $d(x,y) = 0$ if $x=y$ and $d(x,y) =1$ else, $\mathbb{Q}$ is complete. You can find the same type of example in $\mathbb{R}^n$.

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    But these do not give the same topology.2012-02-29
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    @Jim true, and I like your example better, but the OP did not specify that "space" means "topological space", nor that the metrics should generate the same topology.2012-02-29
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    Thanks, both examples were helpful.2012-02-29
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    @WillieWong: to me "space" means "topological space." Otherwise the term "set" would be used. Of course I'm probably just betraying my bias toward topology.2012-02-29
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There is a property called "completely metrizable" and a space has the property if it will admit a complete metric that generates the same topology. So in that sense completeness might be regarded as an intrinsic property of the space. For example, the irrationals with the usual topology are completely metrizable. One builds the new metric by essentially killing off any Cauchy sequences that don't converge.