Timeline for Are there examples of compact infinite dimensional manifolds? [closed]
Current License: CC BY-SA 3.0
15 events
when toggle format | what | by | license | comment | |
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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
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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 |