I think the infinite dimensional sphere satisfies the following criteria. However, I was hoping that someone could come up with a more elementary example. Thanks.
Find an example of a complete bounded metric space which is not compact.
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4An infinite discrete space. – David Mitra Jul 04 '13 at 00:10
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1@David: That's a great example. – Asaf Karagila Jul 04 '13 at 00:11
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does a set {0,1 ,1/2,1/3....} with usual metric count? – James Wang May 14 '19 at 05:29
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@JamesWang. It's compact. – DanielWainfleet May 14 '19 at 10:27
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As you can see by the A by Ittay Weiss, "bounded" is not a topological property: We cannot declare a subset of a metric space to be unbounded by only looking at the topology. – DanielWainfleet May 14 '19 at 10:30
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Other than discrete spaces, you can take the following general approach. Take any non-compact metric space which is complete. To turn it into a bounded metric space without changing its non-compactness nor its completeness, just change the metric to $\min\{d(-,-),1\}$. So, for instance, $\mathbb R$ is complete, not compact, nor bounded (with the usual metric). After truncating the metric as above you get $\mathbb R$ with a metric such that it is bounded, is complete, but not compact. This little trick shows why in the context of a metric the concept of total boundedness is more useful than boundedness.
Ittay Weiss
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A variant is $e(x,y)=d(x,y)(1+d(x,y),)^{-1},$ which induces the same topology. as $d$. If $d$ is complete then so is $e,$ and if $d$ is unbounded then $e$ does not attain its $lub$ (which is $1$). – DanielWainfleet May 14 '19 at 10:23
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Hint: In a Banach space the closed unit ball is compact if and only if the dimension is finite; but a closed subset of a complete metric space is complete.
Asaf Karagila
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If you want a "simple" example, then I can't think of anything different than a subset of a Banach space. Being complete means being close (since the whole space is complete), and for a closed bounded subset of a vector space to be non-compact, you have to be in an infinite dimensional vector space.
Daniel Robert-Nicoud
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