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Nov 11 at 11:34 answer added Anonymous timeline score: 3
Nov 11 at 1:30 answer added Jakobian timeline score: 3
Nov 11 at 0:16 comment added Jakobian @Anonymous sorry, I do see how to refine this argument to see that it works now
Nov 11 at 0:04 comment added Jakobian @Anonymous I don't know, the same argument doesn't apply to it
Nov 11 at 0:02 history undeleted Jakobian
Nov 10 at 23:57 history deleted Jakobian via Vote
Nov 10 at 23:33 comment added Anonymous Can't we just let $X = \beta \omega \setminus (A \cup \{p\})$ where $A$ is a countably infinite discrete subset of $\beta \omega \setminus \omega$ and $p$ is an accumulation point of $A$?
Nov 10 at 17:23 history asked Jakobian CC BY-SA 4.0