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On The Existence Of Shortest Paths

Differential Geometry
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Tsakanikas Nickos
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On The Existence Of Shortest Paths

#1

Post by Tsakanikas Nickos » Tue Jul 19, 2016 9:37 pm

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.
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Grigorios Kostakos
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Re: On The Existence Of Shortest Paths

#2

Post by Grigorios Kostakos » Wed Jul 20, 2016 9:11 am

$X$ is compact, so is complete. The rest is a consequent of Hopf–Rinow theorem.
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Re: On The Existence Of Shortest Paths

#3

Post by Tsakanikas Nickos » Wed Jul 20, 2016 10:38 am

$X$ 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 Arzela-Ascoli's Theorem and the "semicontinuity of the lengths of curves".
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Re: On The Existence Of Shortest Paths

#4

Post by Grigorios Kostakos » Wed Jul 20, 2016 10:53 am

Tsakanikas Nickos wrote:$X$ is assumed to be boundedly compact. Can you justify the reduction to the case that $X$ is compact?
No, I don't(*). Because it is not true that a compact metric space is boundedly compact. (Example?)
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 Bolzano-Weierstrass theorem.
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Re: On The Existence Of Shortest Paths

#5

Post by Tsakanikas Nickos » Wed Jul 20, 2016 6:47 pm

Let's prove some of the assertions then!
  • 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 Hopf-Rinow theorem)?
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