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Jun 11, 2022 at 22:14 comment added Saúl RM Yeah this proof was written a bit too fast as I was in the middle of an exam week. I have tried to write it in more detail in MSE
Jun 11, 2022 at 18:59 comment added Tereza Tizkova Thank you! I am still quite unsure, after consulting that with colleagues, so I tried to rephrase your answer to the initial question on stackexchange and gave you credits. I am not sure if you have account there, but I highlighted what is still unclear and will appreaciate if you take a look. math.stackexchange.com/questions/4468793/…
Jun 10, 2022 at 21:42 comment added Saúl RM The procedure always gives $U$ as a union of clopens, the problem is that only finitely many of them may be nonempty (for example this happens if $A_1=U$). However this cannot happen if $U$ is not contained in any finite union of the $A_n$, because any finite union of the clopens we obtain is contained in a finite union of the sets $A_n$, so it can't be the whole $U$
Jun 10, 2022 at 16:45 comment added Tereza Tizkova Thank you, this is wonderful! The only part I dont understand is how the "This implies that $U\equiv\mathbb{Q}^\infty$..." follows from that we get infinitely many clopens if we pick $A_n$ so that no finite union covers $U$? How do we "get infinite many non empty clopens"? Thank you
Jun 10, 2022 at 9:30 vote accept Tereza Tizkova
Jun 9, 2022 at 17:36 history edited Saúl RM CC BY-SA 4.0
added 465 characters in body
Jun 9, 2022 at 17:30 history answered Saúl RM CC BY-SA 4.0