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Jun 15, 2020 at 7:27 history edited CommunityBot
Commonmark migration
Nov 8, 2017 at 20:37 comment added Joel David Hamkins Well, further consideration shows that this is not a path after all, and I now expect that this space is not path-connected.
Nov 8, 2017 at 19:15 comment added Dominic van der Zypen That would be great @JoelDavidHamkins, this has been open for some time. Thanks in advance for your effort!
Nov 8, 2017 at 18:45 comment added Joel David Hamkins It seems to me that it is; I'll write up something later when I have a chance.
Nov 8, 2017 at 17:02 comment added Joel David Hamkins If we use $J=\mathbb{Q}\cap[0,1]$ in place of $\omega$, is the map from the unit interval $x\in[0,1]$ to the rational cut $\{q\in J\mid q<x\}$ a path from $\emptyset$ to $J$ in $P(J)/\text{Fin}$? If so, then we could translate this all around and path-connect any point to the top, deducing path-connectedness of $P(\omega)/\text{Fin}$. That is, we basically map reals to their cut in the rationals. Is it a path?
Nov 8, 2017 at 10:45 history edited Dominic van der Zypen CC BY-SA 3.0
included definition of interval topology
Apr 13, 2017 at 12:57 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Mar 30, 2015 at 17:57 comment added Nik Weaver Oh, I misread the definition, sorry.
Mar 30, 2015 at 15:36 comment added Dominic van der Zypen No, otherwise $\mathcal{P}(\omega)/fin$ would be disconnected, but it is connected (see mathoverflow.net/questions/200784/… ). My question is if it is even path-connected
Mar 30, 2015 at 14:36 comment added Nik Weaver According to your definition of "interval topology", it looks as though every subbasic open set is also closed, in any poset ...
Mar 30, 2015 at 13:28 history edited Dominic van der Zypen CC BY-SA 3.0
edited title
Mar 30, 2015 at 13:27 history undeleted Dominic van der Zypen
Mar 25, 2015 at 9:46 history deleted Dominic van der Zypen via Vote
Mar 25, 2015 at 9:44 history asked Dominic van der Zypen CC BY-SA 3.0