Timeline for Spaces that never separate the Hilbert cube
Current License: CC BY-SA 3.0
9 events
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| Jul 2, 2013 at 19:47 | comment | added | Tom Goodwillie | Yes, Igor, there was an erroneous "not". I've deleted the comment. Here's what I meant: I cannot think, off the top of my head, of any subset $X\subset Q$ of the Hilbert cube that separates $Q$ and that does not have a subset homeomorphic to $Q$. But I have little experience with these things and have not thought very hard. | |
| Jul 2, 2013 at 14:53 | history | edited | Igor Belegradek | CC BY-SA 3.0 |
added 28 characters in body
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| Jul 2, 2013 at 14:46 | history | edited | Igor Belegradek | CC BY-SA 3.0 |
update
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| Jul 2, 2013 at 14:38 | history | edited | Igor Belegradek | CC BY-SA 3.0 |
update
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| Jul 1, 2013 at 19:30 | comment | added | Igor Belegradek | Also any separable metrizable space embeds into the Hilbert cube (by Urysohn's embedding). | |
| Jul 1, 2013 at 18:25 | comment | added | Igor Belegradek | @Tom, you surely meant to ask something else. Any finite dimensional compact subset of the Hilbert cube satisfies your condition; it does not separate by the Alexander duality, and it cannot contain an infinite dimensional subspace, such as the Hilbert cube. | |
| Jul 1, 2013 at 3:00 | comment | added | Igor Belegradek | Ryan, the argument in the linked paper seems to fail right before one invokes duality. Did you have in mind some modification? | |
| Jul 1, 2013 at 2:41 | comment | added | Ryan Budney | By the same argument $\sqcup_n \mathbb R^n$ also has this property. If you want the space to be connected you could take a wedge instead of a disjoint union. | |
| Jul 1, 2013 at 2:23 | history | asked | Igor Belegradek | CC BY-SA 3.0 |