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Aug 30, 2021 at 11:19 history made wiki Post Made Community Wiki by Stefan Kohl
Jan 24, 2011 at 22:18 comment added Lennart Meier Both Sperner's lemma and the non-draw in Hex are of course easier than Brouwer fixed point theorem since both posess elementary and short arguments.
Jan 24, 2011 at 16:09 comment added Timothy Chow @Joel: This is one area where reverse mathematics as it is currently set up does not quite capture the informal sense of "equivalent." Many people feel intuitively that Sperner's lemma and Brouwer's fixed-point theorem are "equivalent," in that the "tricky part" is the same and you can pass from one to the other via "straightforward" reasoning. However, Sperner's lemma is provable in $RCA_0$ while Brouwer's fixed-point theorem requires $WKL_0$. Very roughly speaking, you need greater logical strength to take a limit, but of course logical strength isn't the same as psychological difficulty.
Mar 25, 2010 at 22:07 comment added Sridhar Ramesh "determinacy" is not the right word to use here; the relevant fact is that Hex cannot end in a draw (a "topological" fact; any way to assign half the cells to each player gives at least one player a winning path).
Mar 25, 2010 at 19:11 comment added Mariano Suárez-Álvarez It is a much less technical sense of 'equivalent', I guess... If you know the truth of one of the statements, the truth of the other follows `easily'.
Mar 25, 2010 at 19:07 history edited Mariano Suárez-Álvarez CC BY-SA 2.5
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Mar 25, 2010 at 18:20 comment added Joel David Hamkins Mariano, do you really mean equivalent here? I would interpret that to mean that over a very weak base theory, as in Reverse Mathematics, you can prove Brouwer's theorem just from the assumption that Hex is determined. This seems unlikely for any finite instantiation of Hex, since the determinacy of finite games amounts to De Morgan's rules of logic. Perhaps you are referring to some kind of continuous analogue of Hex? Could you explain?
Mar 25, 2010 at 4:30 history answered Mariano Suárez-Álvarez CC BY-SA 2.5