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Jun 27, 2014 at 0:10 history edited David White CC BY-SA 3.0
Texified because it was on the front page anyway.
Oct 11, 2009 at 23:20 comment added Eric Wofsey Indeed. This follows from the fact that every compact subset is contained in some finite S^n, which then can be contracted in S^{n+1}. This sort of argument is more generally useful to show the contractibility of this sort of infinite-dimensional object when it may be nonobvious how to write down a contraction explicitly.
Oct 11, 2009 at 22:54 history answered Ilya Nikokoshev CC BY-SA 2.5