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Feb 5, 2020 at 17:09 history edited Martin Sleziak CC BY-SA 4.0
http -> https (the question has been bumped anyway)
Jan 26, 2020 at 18:17 history edited Pietro Majer CC BY-SA 4.0
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Jan 26, 2020 at 10:07 answer added Pietro Majer timeline score: 4
Jan 25, 2020 at 20:01 answer added Aryeh Kontorovich timeline score: 2
Jan 25, 2020 at 16:24 answer added jlewk timeline score: 6
May 11, 2012 at 10:52 vote accept gondolier
May 11, 2012 at 10:52 vote accept gondolier
May 11, 2012 at 10:52
May 10, 2011 at 23:06 comment added Anton Petrunin If the dimension is finite, there is no need to use Hausdorff's maximal principle. Choose a dense countable set in U and extend the map to set of all rational points. The obtained map can be extended to a Lipschitz one on whole space.
May 10, 2011 at 18:31 comment added Bill Johnson If the space is separable, you can choose a countable dense set of the complement of the domain of the function and recursively define the extension to include that countable dense set and then extend to the whole space by continuity. The entire proof is then explicit once you have the countable dense set and an ordering on it.
May 10, 2011 at 18:29 comment added Bill Johnson The key step in the proof of Kirszbraun's theorem involves extending the function to one more point. You write down the conditions on an extension which make the extension have the same Lipschitz constant and show that it is possible to satisfy the conditions. It is easy to make the extension explicit. TBC
May 10, 2011 at 17:30 answer added Sergei Ivanov timeline score: 10
May 10, 2011 at 16:11 comment added Theo Buehler Offhand I don't know how constructive this is, but are you aware of the paper of Lang-Pavlovic-Schroeder's springerlink.com/content/5pd0u4yr5frrvbyk ?
May 10, 2011 at 15:49 history asked gondolier CC BY-SA 3.0