Timeline for Is there such a thing as the sigma-completion of a Boolean algebra?
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| Dec 3, 2014 at 14:09 | history | edited | Goldstern | CC BY-SA 3.0 |
latex syntax: bar{...}
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| Nov 25, 2010 at 18:08 | comment | added | Phil Wild | Sorry if I wasn't sufficiently clear, but Todd's interpretation of my OP is correct - I have no guarantee that the f's in which I'm interested preserve any infinite joins that may exist in B. I appreciate all the comments nonetheless, they are very interesting. | |
| Nov 25, 2010 at 15:49 | comment | added | Andreas Blass | An example that may clarify this is the countable atomless Boolean algebra B (a.k.a. free on countably many generators). Its r.o. completion is the Cohen algebra C, but it also embeds naturally into the random algebra R (Borel sets in [0,1] mod measure 0 sets) and that embedding doesn't factor through C in a sup-preserving way. In fact, there are no complete homomorphisms from C to R (the random real model doesn't contain Cohen reals). A nice exercise is to see where the embedding of B in R fails to preserve sups. None of this contradicts what Joel and Todd said; I hope it concretizes it. | |
| Nov 25, 2010 at 4:49 | comment | added | Todd Trimble | Well, I know that's what you meant, but this is not a context stated in the original question, so "desired for this context" made me wonder whose desire and whose context. I think we can probably agree that each of us has given a correct answer but to different questions. (I will however point out again that Phil's question explicitly concerned countable sups, not arbitrary ones.) | |
| Nov 25, 2010 at 4:00 | comment | added | Joel David Hamkins | What I meant was that my interpretation of the question arises by asking not only for $g$ to commute with sups, but also $i$, when those sups exist in $\mathbb{B}$. | |
| Nov 25, 2010 at 3:45 | comment | added | Todd Trimble | "which would seem to be what is desired for this context" Not sure I follow, but this is getting to be a long discussion. If you like, you can email me at topological musings (one word) at gee-mail dot com. | |
| Nov 25, 2010 at 3:43 | comment | added | Todd Trimble | In all fairness, some of the confusion might stem from words like "smallest" in OP's question. The type of completion you are talking about is admittedly classical, but since OP explained what he meant by asking for a precise universal property, I followed that precise interpretation. And it's a fine question; it's not as if it's a weird question and he should have asked something else! | |
| Nov 25, 2010 at 3:43 | comment | added | Joel David Hamkins | What I said is correct not just for my completion, but for any completion that respects the suprema that already exist in $\mathbb{B}$, which would seem to be what is desired for this context. But I guess it is clear after all this discussion that there are variations on the question with different answers. Thanks for the helpful explanations! | |
| Nov 25, 2010 at 3:29 | comment | added | Todd Trimble | "But obviously the extension of f to a completion cannot respect infinitary joins if f doesn't even do so itself." No, you're wrong about that (unless by "a completion" you mean your completion). Consider for example the Dedekind completion of a poset P consisting of all down-sets of P. Given a poset map P --> Q to a complete poset (not necessarily preserving sups), there exists a unique sup-preserving map from the Dedekind completion to Q that preserves sups. The word "completion" has different senses, but I am sticking close to the words of OP here to take his meaning. | |
| Nov 25, 2010 at 3:15 | comment | added | Joel David Hamkins | But obviously the extension of $f$ to a completion cannot respect infinitary joins if $f$ doesn't even do so itself. So I didn't take the question to be asking for that. Meanwhile, the regular open completion seems to have all the universal properties that one would want for a completion. | |
| Nov 25, 2010 at 3:02 | comment | added | Todd Trimble | Let me just add that his question (as I interpret it) is a perfectly natural one from a categorical perspective. | |
| Nov 25, 2010 at 3:00 | comment | added | Todd Trimble | No! He said nothing about f preserving infinitary joins that exist in B. He said "homomorphism f from (the Boolean algebra) B to a countably complete Boolean algebra B". I interpreted "homomorphism" as Boolean algebra homomorphism, nothing more. If indeed OP intended that f also preserve all infinitary joins that exist in B, then he should have said so, and then we are in perfect agreement, but I see nothing in his words that warrants the meaning you are giving them. | |
| Nov 25, 2010 at 2:49 | comment | added | Joel David Hamkins | What I am saying is that if $f$ respects those infinitary joins that exist in $\mathbb{B}$, an obvious necessary requirement, then the extension of $f$ to the completion $\bar\mathbb{B}$ will respect all infinitary joins. And this seems to be exactly what the OP wants, right? | |
| Nov 25, 2010 at 2:47 | comment | added | Todd Trimble | closely at your statements, they do not express this universal property (between quotes). | |
| Nov 25, 2010 at 2:45 | comment | added | Todd Trimble | algebra maps to the category of Boolean algebras and Boolean algebra maps has a left adjoint; that is exactly what the universal property would be saying: "given a complete Boolean algebra C and a Boolean algebra map B --> C, there exists a unique sup-preserving Boolean algebra extension B' --> C." And I was saying that B' doesn't exist. If it did, then the underlying-set functor from complete Boolean algebras and complete Boolean algebra maps to sets would have a left adjoint. And it doesn't, by the Gaifman-Hales result. See Categories for the Working Mathematician, page 123. If you look ... | |
| Nov 25, 2010 at 2:36 | comment | added | Todd Trimble | Joel, look at OP's question again: "By "smallest" I mean that the inclusion i:B↪B′ has the obvious universal property, i.e. for every homomorphism f from B to a countably complete Boolean algebra C there exists a unique homomorphism g:B′→C such that g∘i=f (it would be nice if g turned out to commute with countable sups too)." The "would be nice" part is what I am explicitly addressing in my answer. What I am saying is that there is no analog of that "would be nice" in your case. You would be saying that the underlying functor from complete Boolean algebras and sup-preserving Boolean ... | |
| Nov 25, 2010 at 1:45 | comment | added | Joel David Hamkins | And I see that is exactly how Solovay does it in the paper to which you link. My problem is: I don't really see how that kind of freeness, which seems to concern you, has much to do with the OP's question. Could you explain? | |
| Nov 25, 2010 at 1:19 | comment | added | Joel David Hamkins | The Gaifman result is immediate from the forcing perspective (after all these years), since the regular open algebra of the forcing to collapse a cardinal $\kappa$ to $\omega$ is countably generated (since the generic filter is determined by a real), but has size $\kappa$. Gaifman told me once that originally he had a more complex construction. | |
| Nov 25, 2010 at 1:07 | comment | added | Joel David Hamkins | ...con't...To my mind, freeness has to do not with Boolean algebra homomorphisms at all, but with extensions of arbitrary set maps $X\to \mathbb{C}$. I think of the Gaifman result as the fact that there are arbitrarily large countably-generated Boolean algebras, and this is what prevents a free algebra, but that seems to have little to do with the question here. | |
| Nov 25, 2010 at 1:06 | comment | added | Joel David Hamkins | Todd, I don't really understand your objections here. The completion of a Boolean algebra that I mention interacts well with complete homomorphisms also: every complete homomorphism of $\mathbb{B}$ into a complete Boolean algebra $\mathbb{C}$ extends uniquely to a complete homomorphism (preserving arbirary meets and joins) of $\bar\mathbb{B}$ into $\mathbb{C}$. ...con't | |
| Nov 25, 2010 at 0:20 | comment | added | Todd Trimble | projecteuclid.org/DPubS/Repository/1.0/… | |
| Nov 25, 2010 at 0:17 | comment | added | Andrés E. Caicedo | @Todd : I never knew this was due to Gaifman (and Hales). Any chance you know a reference? | |
| Nov 25, 2010 at 0:03 | comment | added | Todd Trimble | We're both right. :-) You are answering a question where one doesn't demand that the extension preserve arbitrary joins, but be just a homomorphism of Boolean algebras, yes? I was referring to morphisms of complete Boolean algebras, where the condition is that the extension preserve arbitrary joins. This is where Gaifman-Hales comes in. | |
| Nov 24, 2010 at 23:55 | comment | added | Joel David Hamkins | Every homomorphism of $\mathbb{B}$ to a complete boolean algebra $\mathbb{C}$ extends uniquely to a homomorphism of $\bar\mathbb{B}$ into $\mathbb{C}$. Isn't this what we should want? | |
| Nov 24, 2010 at 23:46 | comment | added | Todd Trimble | But this is not a free construction that the OP was asking about (see his universal property). There is a theorem due to Gaifman and Hales that free complete Boolean algebras do not exist. | |
| Nov 24, 2010 at 23:43 | history | edited | Joel David Hamkins | CC BY-SA 2.5 |
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| Nov 24, 2010 at 23:38 | history | answered | Joel David Hamkins | CC BY-SA 2.5 |