Timeline for F is ultrafilter over a Boolean algebra implies that for every b, either b or not-b is in F?
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 26, 2010 at 12:48 | vote | accept | Paul Crowley | ||
| Mar 22, 2010 at 1:34 | history | edited | Mikael Vejdemo-Johansson | CC BY-SA 2.5 |
Removed Zorn's lemma.
|
| Mar 22, 2010 at 1:26 | comment | added | Mikael Vejdemo-Johansson | @Paul: I'm including the above argument in my answer. | |
| Mar 22, 2010 at 1:25 | comment | added | Mikael Vejdemo-Johansson | @Paul: By stating that $F'$ is the smallest filter containing $F$ and $b$, I am closing it upwards. It's not necessarily given by $F\cup\{b\wedge f: f\in F\}$, as I could potentially include even more things by the upwards closure. However, if $0\in F'$, then $0$ will not be included by upwards closure, since $0$ is the bottom of the lattice. Hence, the only way $0$ could be in there is if it occurs as the meet of things in the filter. And if it's the meet of things included by upwards closure, I may as well pick the thing at the bottom to meet with - thus I can assume it to be a $b\wedge f$. | |
| Mar 22, 2010 at 1:22 | comment | added | Mikael Vejdemo-Johansson | Thank you Theo, I was throwing Zorn's lemma in without thinking it through completely. Somewhat Cargo Cult-ish. | |
| Mar 21, 2010 at 23:03 | comment | added | Paul Crowley | Damn, no comment preview and no comment editing? Sorry for the errors in the above markup. "If I understand properly" is the start of my text. Let's see if I can get it right this time: $F' = F \cup \{b \wedge f|f \in F\}$ | |
| Mar 21, 2010 at 22:59 | comment | added | Paul Crowley | > If $0 \in F$, then this means that there is some $f \in F$ such that $b \wedge f=0$ If I understand properly, this assumes that if $F$ is a filter, then $F' = F \cup {b \wedge f|f \in F}$ is a filter (which would clearly then be the smallest filter containing $F$ and $b$. I can see that it's closed under meets, but I can't work out how to prove that it's closed upwards. | |
| Mar 21, 2010 at 20:11 | comment | added | François G. Dorais | In fact, the proof is complete without the fourth paragraph: you have shown that $F$ is not maximal. | |
| Mar 21, 2010 at 18:08 | comment | added | Theo Johnson-Freyd | You shouldn't need Zorn's lemma to prove the statement. If you have a maximal filter $F$, then this argument shows that for each $b$, $F$ must include either $b$ or $\neg b$. But if a proper filter $F$ includes either $b$ or $\neg b$ for each $b$, then it must be maximal: there's no more room to add any elements and keep it proper. So therefore proper filters are maximal iff they include $b$ or $\neg b$ for each $b$. Now, without Zorn, you don't know if there are any nonprinciple ultrafilters, but that's a different question. | |
| Mar 21, 2010 at 17:37 | history | answered | Mikael Vejdemo-Johansson | CC BY-SA 2.5 |