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Oct 26, 2013 at 0:22 vote accept user8991
Oct 25, 2013 at 20:29 comment added user8991 @todd : Pietro Majer seems to reply indirectly to it. so the reply seems to be no.
Oct 25, 2013 at 12:44 history closed BS.
Andrey Rekalo
Carlo Beenakker
Ricardo Andrade
David White
Needs details or clarity
Oct 25, 2013 at 8:19 review Close votes
Oct 25, 2013 at 12:44
Oct 25, 2013 at 7:26 comment added Ben McKay @IgorBelegradek: What if $K$ is empty, or $K$ is a finite set of points? Are $K$ and $Y-K$ homeomorphic?
Oct 25, 2013 at 2:01 comment added Todd Trimble It seems to me Pietro Majer already addressed this in his answer here: mathoverflow.net/a/143737/2926 Right?
Oct 25, 2013 at 1:18 comment added Igor Belegradek Any separable Frechet (or Banach) space is homeomorphic to the countable product of lines, so my answer above applies.
Oct 25, 2013 at 1:09 history edited user8991 CC BY-SA 3.0
added 35 characters in body
Oct 25, 2013 at 1:07 comment added user8991 just modelled on a vector space. like Banach manifold, Frechet manifold.
Oct 25, 2013 at 1:04 answer added Tom Goodwillie timeline score: 9
Oct 25, 2013 at 1:04 comment added Igor Belegradek If this is your definition, then the answer is no, because if $K$ is a compact subset of a manifold $Y$ modelled on the countable product of lines, then $Y$ and $Y-K$ are homeomorphic.
Oct 25, 2013 at 0:58 review Low quality posts
Oct 25, 2013 at 1:13
Oct 25, 2013 at 0:56 comment added Igor Belegradek How do you define an infinite dimensional manifold? Is it modelled on the countable product of lines?
Oct 25, 2013 at 0:56 answer added johndoe timeline score: 1
Oct 25, 2013 at 0:42 history asked user8991 CC BY-SA 3.0