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What is an example of a connected Riemannian manifold containing a non-compact closed bounded set?

By the Hopf-Rinow Theorem, I know that the closed bounded sets of a connected Riemannian manifold are compact if and only if the manifold is complete. So I have to pick a non complete Riemannian manifold. Let's say I choose the punctured sphere or an open hemisphere with the induced metric from Euclidean space. "Where is" that closed bounded subset of the sphere which is not compact? Or am I in the wrong track?

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Take the open unit disk in the plane. It is closed in itself but not compact.

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  • $\begingroup$ Or more intuitively (but not really) any diagonal that runs across it. It wont be compact as the endpoints are missing! This uses the fact that compactness is intrinsic and a segment missing endpoints is not compact either as a subset of plane or as a subset of part of plane. $\endgroup$ Jul 29, 2022 at 9:20
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In any metric space, if $(x_n)$ is a Cauchy sequence which does not converge, then the subset $\{x_n\}$ is closed, bounded, and not compact. Alternatively, if your entire manifold is bounded (as in the examples you name), you can just take the closed bounded set to be the entire manifold.

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  • $\begingroup$ Thank you! I was just trying to figure out why the subset $\{x_n\}$ of a non convergent Cauchy sequence has to be closed. Can't it have more than one accumulation point? $\endgroup$
    – Strider
    Dec 7, 2015 at 0:54
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    $\begingroup$ Nope, if a Cauchy sequence accumulates at a point, it must converge to that point. For if you can choose some very large $N$ such that $x_N$ is close to $p$, then $x_n$ will be close to $p$ for all $n>N$ as well since the sequence is Cauchy. $\endgroup$ Dec 7, 2015 at 1:03

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