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On The Existence Of Shortest Paths

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On The Existence Of Shortest Paths
Let $(X,d)$ be a boundedly compact metric space. Show that whenever two points are joined by a rectifiable curve (i.e. a curve of finite length), then there exists a shortest path (minimal geodesic) between them.
 Grigorios Kostakos
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Re: On The Existence Of Shortest Paths
$X$ is compact, so is complete. The rest is a consequent of Hopf–Rinow theorem.
Grigorios Kostakos

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Re: On The Existence Of Shortest Paths
$X$ is assumed to be boundedly compact. Can you justify the reduction to the case that $X$ is compact?
Furthermore, there is also a proof using ArzelaAscoli's Theorem and the "semicontinuity of the lengths of curves".
Furthermore, there is also a proof using ArzelaAscoli's Theorem and the "semicontinuity of the lengths of curves".
 Grigorios Kostakos
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Re: On The Existence Of Shortest Paths
No, I don't(*). Because it is not true that a compact metric space is boundedly compact. (Example?)Tsakanikas Nickos wrote:$X$ is assumed to be boundedly compact. Can you justify the reduction to the case that $X$ is compact?
But, every boundedly compact metric space is complete(**). So we can now aply the Hopf–Rinow theorem.
(*) I was careless!
(**) This is a generalization of the BolzanoWeierstrass theorem.
Grigorios Kostakos

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Re: On The Existence Of Shortest Paths
Let's prove some of the assertions then!
Can you now fill in the details of the initial proof (by using my hints and avoiding HopfRinow theorem)?
 Every compact metric space is boundedly compact (i.e. every closed and bounded subset is compact) : It is a fact (prove it if you want to) that every closed subset of a compact space is compact. In particular, any closed and bounded subset of a compact space is compact. But this precisely means that the space is boundedly compact.
 Reduction to the case that $X$ is compact : Pick any two points $x,y$ that are joined by a rectifiable curve, say $\gamma$. Assume that $ \mathcal{l} $ is the length of this curve. Consider the ball $ B_{x}(\mathcal{l}) $ centered at $x$ of radius $ \mathcal{l} $. Then its closure is a closed and bounded subset of $X$ which contains $\gamma$ (and, of course, any other shorter curve joining $x$ and $y$). By hypothesis, $B_{x}(\mathcal{l}) $ is compact. Hence it suffices to show existence of a shortest path inside this ball.
 Every boundedly compact space is complete : Sketch of proof : Consider a Cauchy sequence of elements of $X$. Then it is bounded, so all its terms are contained in some ball. The closure of this ball is compact, by hypothesis. Therefore, the sequence in question has a convergent subsequence, and so the sequence itself converges, as was assumed to be Cauchy.
Can you now fill in the details of the initial proof (by using my hints and avoiding HopfRinow theorem)?