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Feb 24, 2014 at 14:22 comment added Gerald Edgar All separable, infinite-dimensional Banach spaces are homeomorphic to each other (and to $\mathbb R^\infty$), so if you like you can use one of those other spaces.
Jan 13, 2014 at 21:21 answer added Sam Nead timeline score: 2
Sep 15, 2010 at 8:09 comment added Andrew Stacey Meta discussion on whether to close as duplicate or not: tea.mathoverflow.net/discussion/672/are-these-duplicates
Sep 15, 2010 at 7:50 comment added Andrew Stacey (Looking again at my answer to the contractiblity of the sphere, I realised one could weaken the conditions so the "stable" in the above is stronger than needed.)
Sep 15, 2010 at 7:18 comment added Andrew Stacey Technically, this is a duplicate of: mathoverflow.net/questions/198/… , my answer there works for any space that is "stable" in the sense that $X \oplus \mathbb{R} \cong X$.
Sep 15, 2010 at 5:54 answer added Dick Palais timeline score: 8
Sep 15, 2010 at 4:03 vote accept Nikita
Sep 15, 2010 at 3:36 answer added fedja timeline score: 23
Sep 15, 2010 at 3:16 answer added Richard Borcherds timeline score: 7
Sep 15, 2010 at 2:30 answer added Bill Johnson timeline score: 14
Sep 15, 2010 at 2:15 history asked Nikita CC BY-SA 2.5