Timeline for Is there a known construction for heavy topologies of all sizes?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 16, 2017 at 15:07 | vote | accept | Gorka | ||
| Feb 16, 2017 at 0:40 | comment | added | Danielle Ulrich | Shouldn't "normal" be "uniform" here? I have always seen "normal" mean "closed under diagonal interesections" | |
| Feb 15, 2017 at 14:44 | comment | added | Will Brian | @AndreasBlass: Thanks for clearing that up. It is indeed a (pretty straightforward) diagonal argument, and now I'm not really sure what I was thinking when I wrote my comment. | |
| Feb 14, 2017 at 22:01 | comment | added | Andreas Blass | @WillBrian To show that no family of $|X|$ sets in $U$ can be a base, enumerate the family as $(X_\xi)_{\xi<|X|}$ and then, by induction on $\xi$, pick $x_\xi$ and $y_\xi$ in $X_\xi$, distinct from each other and from earlier choices. Then each $X_\xi$ meets both the set of $x_\xi$'s and its complement, so the $X_\xi$'s can't be a base for $U$. Then it follows that they can't be a base for the topology either. | |
| Feb 14, 2017 at 20:39 | comment | added | Will Brian | It's not clear to me how to use a "diagonal argument" to prove that every uniform ultrafilter on $X$ has weight bigger than $|X|$. (That being said, I can prove that it's true for some uniform ultrafilters using independent sets, and this is good enough for an affirmative answer to the original question.) Am I missing something? | |
| Feb 14, 2017 at 20:16 | history | answered | Tomasz Kania | CC BY-SA 3.0 |