It is currently Fri Apr 20, 2018 8:08 am


All times are UTC [ DST ]




Post new topic Reply to topic  [ 5 posts ] 
Author Message
PostPosted: Tue Jul 19, 2016 9:37 pm 
Team Member

Joined: Tue Nov 10, 2015 8:25 pm
Posts: 313
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.


Top
Offline Profile  
Reply with quote  

PostPosted: Wed Jul 20, 2016 9:11 am 
Team Member
User avatar

Joined: Mon Nov 09, 2015 1:36 am
Posts: 444
Location: Ioannina, Greece
$X$ is compact, so is complete. The rest is a consequent of Hopf–Rinow theorem.

_________________
Grigorios Kostakos


Top
Offline Profile  
Reply with quote  

PostPosted: Wed Jul 20, 2016 10:38 am 
Team Member

Joined: Tue Nov 10, 2015 8:25 pm
Posts: 313
$X$ 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".


Top
Offline Profile  
Reply with quote  

PostPosted: Wed Jul 20, 2016 10:53 am 
Team Member
User avatar

Joined: Mon Nov 09, 2015 1:36 am
Posts: 444
Location: Ioannina, Greece
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.

_________________
Grigorios Kostakos


Top
Offline Profile  
Reply with quote  

PostPosted: Wed Jul 20, 2016 6:47 pm 
Team Member

Joined: Tue Nov 10, 2015 8:25 pm
Posts: 313
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)?


Top
Offline Profile  
Reply with quote  

Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 5 posts ] 

All times are UTC [ DST ]


Mathimatikoi Online

Users browsing this forum: No registered users and 1 guest


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Search for:
Jump to:  
cron
Powered by phpBB® Forum Software © phpBB Group Color scheme created with Colorize It.
Theme created StylerBB.net