Timeline for Incomplete subsets of the free boolean algebra on countably many generators
Current License: CC BY-SA 3.0
30 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 4, 2016 at 2:04 | comment | added | Joel David Hamkins | The Lindenbaum algebra and the term algebra are basically identical, so you can see how I argued with it. Since any term (formula) has only finitely many generators (variables) in it, there is some generator not appearing in it, and this is what causes problems, which you can see by considering the automorphisms I mentioned. | |
| Aug 4, 2016 at 1:17 | comment | added | dwymark | Thanks @Joel, I see why the continuum argument works now. As for whether it's "easy to see": you are right, there needs to be a proof, and I'm sure yours works. But on the Lindenbaum-Tarski algebra view- to me it seems obvious that (1) the tautology is an UB, and (2) no other element could possibly serve as an UB since this would require an infinitely long formula, which is disallowed by the syntax. (2) must hold because for a formula F to be an UB for a set of generators S, each generator $s\in S$ must show up at least once in F. Right? Or am I skirting over some details? | |
| Aug 4, 2016 at 0:26 | comment | added | Joel David Hamkins | @dwymark Concerning (infinite antichains) + (complete) implies (size at least continuum), this is easy. If A is an infinite antichain and every subset of A has a LUB, then the power set P(A) algebra embeds in the Boolean algebra, by mapping every subset of A to its LUB. This is injective, and in fact a Boolean embedding. Since P(A) has size at least continuum if A is infinite, the Boolean algebra is at least this large. | |
| Aug 4, 2016 at 0:24 | comment | added | Joel David Hamkins | @dwymark Is it "easy to see" that the LUB of an infinite set of generators in the Lindenbuam algebra is the tautology? I didn't see this without the arguments mentioned in my answer. Of course, no candidate LUB comes easily to mind, but to prove that there isn't one, it seems one must argue as in my answer. | |
| Aug 4, 2016 at 0:08 | comment | added | dwymark | Can you briefly clarify the following? (infinite antichains) + (complete) => (size continuum). This is an argument I have seen a few times but I am still grappling with it. I do not see why the fact that distinct subsets have distinct lubs leads to size continuum. What about the fact that (every element in the antichain) <= 1? | |
| Aug 4, 2016 at 0:03 | comment | added | dwymark | Just a small comment: For the special case of the free boolean algebra on countably many generators, if you view it as the set of equivalence classes for propositional logic (= the Lindenbaum-Tarski algebra for propositional logic), the fact that every subset of the generators has a lub is easy to see- it is just the disjunction if the set is finite, and the tautology if the set is infinite. | |
| Aug 2, 2016 at 18:40 | comment | added | Joel David Hamkins | Yes, that suffices, since if those equations are true, then the $a_i$ will be atoms in the generated algebra. But in fact there is no need to express this equationally; in the theorem at the end, the situation is that we have a Boolean algebra that happens to have some atoms, and those atoms happen to generate the whole Boolean algebra. | |
| Aug 2, 2016 at 18:37 | comment | added | მამუკა ჯიბლაძე | In fact I do not quite see how to express equationally that the generators are atoms. Does $a_i\land a_j=\bot$ suffice for that? | |
| Aug 2, 2016 at 12:07 | comment | added | Joel David Hamkins | I edited to include your nice argument, Andreas, as an alternative. | |
| Aug 2, 2016 at 11:45 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
deleted 152 characters in body
|
| Aug 2, 2016 at 11:37 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
added 472 characters in body
|
| Aug 2, 2016 at 11:17 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
Improved exposition for clarity
|
| Aug 1, 2016 at 5:19 | comment | added | მამუკა ჯიბლაძე | @AndreasBlass Your argument is certainly "more" constructive than mine - I need ultrafilters and you only (some form of) excluded middle. I think in fact excluded middle can be avoided too but don't know how to do it... | |
| Aug 1, 2016 at 4:43 | comment | added | Andreas Blass | Here's an alternative way to see that $1$ is the only upper bound for an infinite set $X$ of generators. Let $u$ be any upper bound for $X$ and let $p$ be an element of $X$ that doesn't occur in the term $u$. The algebra has an automorphism that sends $p$ to $\neg p$ and fixes all the other generators. So it fixes $u$, and since $p\leq u$, we also have $(\neg p)\leq u$. Therefore $1=p\lor(\neg p)\leq u$. | |
| Aug 1, 2016 at 2:25 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
added 256 characters in body
|
| Aug 1, 2016 at 0:39 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
added 10 characters in body
|
| Aug 1, 2016 at 0:07 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
added 20 characters in body
|
| Jul 31, 2016 at 23:57 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
added 44 characters in body
|
| Jul 31, 2016 at 23:04 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
deleted 10 characters in body
|
| Jul 31, 2016 at 22:54 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
deleted 10 characters in body
|
| Jul 31, 2016 at 22:50 | comment | added | Joel David Hamkins | I have now edited with the revised argument. Every subset has a least upper bound! This is very surprising. | |
| Jul 31, 2016 at 22:45 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
Every subset has a LUB.
|
| Jul 31, 2016 at 21:02 | comment | added | Joel David Hamkins | I'll edit and post a revised argument later, unless you want to do it. | |
| Jul 31, 2016 at 20:59 | comment | added | Joel David Hamkins | Yes, I came to the same conclusion myself while I was doing some woodwork this afternoon. So every subset of the generators has a least upper bound! | |
| Jul 31, 2016 at 19:02 | comment | added | მამუკა ჯიბლაძე | I believe any infinite set $X$ of generators has least upper bound $1$: take any $b\in B$. There is a generator $p\in X$ such that $b$ does not depend on $p$. Then if $b\ne1$, there is a homomorphism $B\to2$ sending $b$ to $0$ and $p$ to $1$, so one cannot have $p\leqslant b$. | |
| Jul 31, 2016 at 15:56 | comment | added | Joel David Hamkins | This point seems to break the co-finite part of the argument also. I am now confused about it... | |
| Jul 31, 2016 at 15:40 | comment | added | Joel David Hamkins | Ah, you are right! I'll think about this and edit. | |
| Jul 31, 2016 at 15:38 | comment | added | Andreas Blass | This was the first example that I thought of and I even started to write an answer, but I don't think it works. The trouble is in the next to last sentence, where you say that one of the generators, $q$, is disjoint from some other generators, those in $X$. In the free Boolean algebra on a set of generators, distinct generators are not disjoint. I think the argument can be repaired by using some disjoint things in place of the generators, for example $p_0\land p_1\land\dots\land p_{n-1}\land\neg p_n$ for all $n\in\mathbb N$. | |
| Jul 31, 2016 at 14:39 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
added 14 characters in body
|
| Jul 31, 2016 at 14:02 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |