Talk:Axiom of choice/Archive 2

message: Correct me if I'm wrong, but Baire's theorem requires only inductive choice. That's quite different than the full axiom of choice. I suggest that it be removed from the list of dependencies. (the Baire category theorem page requires checking too). Gady 22:10, 12 Nov 2004 (UTC)
replies: text: The page says "Several central theorems in various branches of mathematics require the axiom of choice (or one of its weaker versions, such as the Boolean prime ideal theorem, the axiom of countable choice, or the axiom of dependent choice).", though, so while it doesn't explicitely list inductive choice as a weaker version, it does state that full AC may not be necessary to prove the heorems listed. -- Schnee 02:05, 14 Nov 2004 (UTC)

text: Feel free to update it so it's more clear. :) -- Schnee 16:39, 14 Nov 2004 (UTC)
thread_title:  Baire's theorem 

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thread_title:  [[User:LinkBot/suggestions/Axiom_of_choice|Link suggestions]] 

message: I once heard another definition of the AoC which I quite liked.  I haven't researched it or checked it, but it seems to be equiv.:
: The cross product of non-empty sets is a non-empty set
Or abusing notation:
: Given sets S_1, S_2, \ldots, the set S_1 \times S_2 \times \ldots contains an element.
I guess that by using integer subscripts it looks like there are countably many sets.  But you get the idea.
--Sean Kelly 01:09, 21 Dec 2004 (UTC)
thread_title:  Another dfn. 

message: Is the statement that the proof of the Banach Tarski paradox is nonconstructive relly true? I saw a proof (albeit a long time ago), but I had a definite idea that the proof actually constructed the a finite partition of the sphere... Schnolle 12:36, 16 Feb 2005 (UTC)
replies: text: The nonconstructivity is in step 3 of Banach-Tarski paradox#A sketch of the proof, where you use the axiom of choice to pick one element from every orbit of a particular action.  dbenbenn | talk 13:19, 16 Feb 2005 (UTC)
thread_title:  Banach-Tarski paradox nonconstructive? 

message: Why does X have to consist of non-empty sets? Or is this only in the first formulation? It doesnt seem necessary in the second def. -MarSch 14:58, 20 Apr 2005 (UTC)
replies: text: Save that one for April 1? Charles Matthews 15:31, 20 Apr 2005 (UTC)
replies: text: No this is serious. There exists a (unique) function from the empty set to itself, right? -MarSch 15:54, 20 Apr 2005 (UTC)

text: Let X be a set of non-empty sets. There exists a choice function f defined on X such that for each set S in X, f(S) is an element of S. So, if S is the empty set, how can we have f(S) an element of S? A choice function f here presumably means a function from X to Y, where Y is the union of all the elements Z of X. If the empty set were an element of X, no such function could exist. Charles Matthews 17:53, 20 Apr 2005 (UTC)
replies: text: It's one of those things that's too obvious to come up with on your own... we all make mistakes.  =) —Sean κ. ⇔ 21:46, 20 Apr 2005 (UTC)
replies: text: right, my bad :) thanks for clearing up my confusion. MarSch 12:50, 21 Apr 2005 (UTC)
thread_title:  non-empty 

message: All this talk about AC and ~AC has got me to thinking about the statement (P or ~P) which is taken to be true regardless of what P is. But suppose we are using ZF (w/o AC) and P = AC. Then is it really fair to say that AC or ~AC is true. Obviously, I'm not the first person to wonder about this and that's why there are people who reject th law of excluded middle, but it's the first time I've ever thought, of it and it is interesting. Any thoughts?--Heathcliff 03:40, 9 May 2005 (UTC)
replies: text: Yep, in standard logic "AC or ~AC" is a tautology, and always provable.  I don't find this troublesome.  Either there exists a choice function for every set of nonempty sets, or there doesn't. Josh Cherry 04:12, 9 May 2005 (UTC)

text: Which statement is more interesting, "For any P, P or ~P" or "There is P st. P or ~P, but neither is provable."  I find the first intuitive, since the law of the excluded middle is logically equivalent to proof by contradiction, which I couldn't imagine living without.  Does your comment concern the second or first statement with P=AC? —Sean κ. ⇔ 05:18, 9 May 2005 (UTC)
replies: text: Yep, in standard logic "AC or ~AC" is a tautology, and always provable.  I don't find this troublesome.  Either there exists a choice function for every set of nonempty sets, or there doesn't. I'm not sure I agree with this last statement in ZF. Since we cannot prove either statement to be true or false. I also believe that the statement you have made (P is True) or (~P is True) is stonger that (P or ~P) is True. As for P or ~P being a tautology in standard logic, well obviously. I was merely noting that it is not as obviously correct to me to accept it as a tautology as it once was. In other words perhaps standard logic is "wrong". As for Sean's question, I'm afraid I'm not sure what you're asking could you ask it a different way.--Heathcliff 13:28, 9 May 2005 (UTC)
replies: text: Feel free to become an intuitionist and reject the law of excluded middle and much of the type of reasoning that most of us take for granted. Josh Cherry 13:21, 9 May 2005 (UTC)

text: I think you answered my question by showing us you're an intuitionist ;).  Which is fine, just remember you're giving up proof by contradiction.  As far as how this discussion applies to the AoC, I would claim (without evidence) that ZF always assumes the axioms of standard propositional logic, which includes the law of the excluded middle.  (So if you want to talk about ZF without the LoEM, you should probably make note of it, ZF\LoEM er... maybe not). —Sean κ. ⇔ 17:44, 9 May 2005 (UTC)
replies: text:  I can't tell if this is directed at me or not. If it is. I am not an intuitionist and I'm not sure why you assume that I am. I hope that intuitionist are not the only people who ponder the validity of their own basic assumptions. If you were addressing Josh, I'm sorry to have butted in.--Heathcliff 23:08, 11 May 2005 (UTC)

text: It's funny this discussion came up. At the AMS convention in Atlanta, Paul Cohen and some other were giving a panel on the CH, and I think someone in the audience asked, "What do you think of the following proof: case 1: CH holds, blah, blah, blah; case 2: ~CH holds, blah, blah, blah; in each case, you prove X. Is X true?" I'm not sure what the response was, but the law of excluded middle seems to generate some emotional reactions. The way I understand it, those who work in logic and set theory often take an attitude akin to how we view the group axioms or the parallel postulate. Is the parallel postulate "true"? The question has no meaning. Sometimes, the mathematics that comes about from assuming it is more useful, other times from assuming its negation. Sometimes, you just want to see what happens in neutral geometry. Same thing here. MOST of the time, mathematics will need AC, or some weaker version of it to get by. You're discarding a lot by not availing yourself of it. On the other hand, it's possible to see how much is provable from ZF alone. And it's also possible to see what happens by assuming ~AC or ~(weaker versions of AC). We can do all this without getting caught up in trying to figure out "whether it's really true or not". This is one thing I didn't really "get" at that panel discussion in January...they really seemed to figure that CH or ~CH was some empirical question that we could "figure out" if we just thought about it long enough or something. Or maybe it was all just way over my head. In any case, set theorists and logicians aren't the ONLY people who think about AC and ~AC and various weakened forms of it...people interested in the foundations of analysis (real number system, TVS, functional analysis, descriptive set theory, construction of certain types of real functions) have to deal with these issues directly. Revolver 18:37, 11 May 2005 (UTC)
replies: text: And of course when we say AC implies such and such we really means ZFA implies such and such. Could someone tell me how much work has been done in significantly altering the other axioms of set theory?--Heathcliff 23:12, 11 May 2005 (UTC)
replies: text: Take a gander at non-well-founded set theory for one example. Revolver 02:33, 12 May 2005 (UTC)
replies: text: Thanks. That is interesting. Years ago I ask a couple of math professors why it was that a set could not contain itself. They seemed to think it was obvious and didn't really seem to understand why anyone would ask such a question. Neither ever mentioned non-well-founded set theory.--Heathcliff 12:48, 12 May 2005 (UTC)

text: Perhaps they were trying to prove "CH or ~CH" without the law of the excluded middle? —Sean κ. ⇔ 05:43, 12 May 2005 (UTC)
thread_title: P or ~P

message: I think I see why I was wrong in saying "in all models of ZF¬C"...e.g. there may be a model where countable choice holds, but not arbitrary choice, so in this model, statements which only require countable choice WILL be true, although choice doesn't hold in general. Is this the right reasoning? Can this be explained by someone who maybe understands it better? Revolver 04:43, 14 May 2005 (UTC)
replies: text: To elaborate, it may be useful to point out which, if any, (I don't know for certain) statements DO hold in all models. E.g. the statement about sequences involving continuous functions seems only to involve countable choice, so it probably doesn't hold in all models, but the statement about vector spaces would seem to hold in all models (isn't it equivalent to choice??) Revolver 04:44, 14 May 2005 (UTC)
replies: text: It's just that "in all models of ZF¬C" is just a more complicated way of saying "¬AC implies".  It's a pretty strong statement.  —Sean κ. ⇔ 04:53, 14 May 2005 (UTC)
replies: text: I understand my mistake. Thinking aloud again, I really don't like the wording "there exists a model", even if it is now corrected from what I put which was incorrect. The reason as I think was mentioned above, is that it makes it sound like this statement requires ZF~C. We don't say, "there exists a model of ZFC in which every vector space has a basis" is a consequence of ZFC, we simply say "every vector space has a basis" is a consequence of ZFC. If something is really equivalent to choice, then it seems it would hold in any model, and we should say this. On the other hand, the statement about sequences involving continuous functions only holds in certain models (namely, if I think right, those in which countable choice does not hold); it is possible to imagine a model where countable choice holds but not arbitrary choice, in this case, the statement is certainly NOT a "consequence of ZF~C"!, to my mind, a "consequence of ZF~C" SHOULD be a statement that holds in ALL models. Whereas AC has weaker versions, ~AC has _stronger_ versions (the negations of the weaker versions of AC), and it's kind of misleading the way it's worded, it makes it sounds like the statements follow from ZF~C, when really there just a model where it doesn't hold. Revolver 05:01, 14 May 2005 (UTC)

text: Okay, it doesn't say "consequences" it say "results requiring", this is technically true, I suppose (the results imply ~AC), but not in the spirit of the first part "results requiring AC". I say this because the first part on results requiring AC allows weaker versions, (not all results there require AC). Moreover, all the results there ARE consequences of AC, since the weaker versions are implied by arbitrary choice. This doesn't happen with ~AC...not all the results are consequences of ~AC...the result about sequences is one example. But the way the two sections seem to mirror each other, the reader is left with this impression. Revolver 05:23, 14 May 2005 (UTC)

text: I think I finally hit what bothers me most. Not all the statements under "results requiring AC" actually require AC...some require weaker versions, so the heading here is misleading. It is true that all statements in the ~AC part, without the qualifier "there exists a model", DO require ~AC, in the sense that in every model where the statement holds, ~AC is true. Again, there seems to be a lack of consistency between the two sections. Revolver 05:23, 14 May 2005 (UTC)

text: I see I was the one who originally titled the section "results requiring AC", so my bad! Revolver 05:23, 14 May 2005 (UTC)
replies: text: Yeah, the logical status of various propositions is unclear from the article.  There are three cases for propositions P of interest: 1. ZFC \vdash P and ZF \lnot C leaves P undecided. 2. ZFC \vdash P and ZF \lnot C \vdash \lnot P  3. ZFC leaves P undecided and ZF \lnot C \vdash \lnot P (three additional cases result from substituting \lnot P for P, but that doesn't really give us anything new).  How about a two-column chart showing the status (provably true, provably false, or undecided) of each proposition under ZFC and ZF \lnot C? Josh Cherry 01:56, 18 May 2005 (UTC)
thread_title:  models 

message: There is a quote in this article from A. K. Dewdney's "famous" April Fool's Day article in the Scientific American of April 1989, but after following up all the links and trying google, I could find no reference to the substance of the article. I may have missed something, but could someone who knows what this article said give a brief explanation of it either in this or the Dewdney article? Thanks --RMoloney 20:27, 17 May 2005 (UTC)
thread_title:  Dewdney link 

message: "Any union of countably many countable sets is itself countable" requires AoC. So I suppose there must be a flaw in this totally constructive proof. Where is it?
Let A0, A1, A2 etc. be countably infinitely many countably infinite sets, and let B be their union.
Let x0;0, x0;1, x0;2 etc. be the elements of A0; let x1;0, x1;1, x1;2 etc. be the elements of A1, and so far.
:This is the flaw right here, no need to look at the rest of the proof.  The fact that A0 is countable does not by itself give you a particular enumeration x0i of A0, it just tells you there is one.  So how do you pick such an enumeration for each An?  If AC fails, each An has an enumeration, but there may not be a sequence of such enumerations, one for each An. --Trovatore 17:52, 16 August 2005 (UTC)
replies: text: Oh, BTW -- you don't need full AC for this.  Countable AC, written ACω, is enough; it says that, given countably many nonempty sets, you can choose one element from each.  Therefore AD actually implies that the union of any countable collection of countable sets of reals is countable. --Trovatore 17:58, 16 August 2005 (UTC)
thread_title: Where's the flaw?

message: I haven't read the whole section yet, and perhaps I won't understand it all when I do, but it looks the problem with the axiom of choice may be summarized in the idea of 'interpenetration'.
The problem as I see it is, you can pick 1 object from infinitely many bins, but if some bins overlap (as some definitions or sets of defined numbers may sometimes overlap) you end up with many bins, not empty, but sharing '1' object - therefore non-empty, but, between them, they should only have a 'fractional' portion to give, if they were to each give equally to choosing function.  So, if you have an 'infinite' set of bins, but a finite set of objects, for instance, you run into a problem, unless the 'portion' of the 'give' of each object is run down to a limit or something.
Other ways to solve the problem might involve 'recursive' or fractal type phenomenon, things that take into account holism and self-similarity, so that reproduction of the same thing again and again doesn't occur - reuse of functions with 'function libraries' and function calls in programs might be an example of this - I mean, something existing only once in a 'direct' form and having other instances occuring as 'links' or 'reflections' of the original. (Note Indra's Net for further info on self-similarity, and Fritjof Capra's "Tao of Physics" for other metaphors created to show similarity between intuitive and rational views of the universe.)
Another example might be the idea of 'templates', where other instances are in fact 'transformations' of this template, the transformations not being inherent in the object, but for instance, stored separately, using the 'symmetries' of/between a given template and an object in order to 'compress information'. (something most useful with a complex 'image' type scenario, where only minor changes occur, so mapping the change as a function symmetrical to the original saves on 'raw data' at the expense of additional 'processing').
An example showing this effect might be the idea of decreasing probability of primes occuring as one progresses towards ever larger numbers.  ie (2*3*5*7*11*13*17*19...*n)^-1 = probability of there being primes greater than n ... if you instead added 1/2 (probability of number being even) + 1/3 +1/5 +1/7 etc, you would end up with the overlapping probabilities added in, ie even numbers divisible by 3 included in 1/2, again when you consider 1/3 of numbers being divisible by 3 .
I hope I'm making sense, I'm not a math wiz, just 'math interested' -
I sort of decided to randomly drop a post, since I got caught up in trying to learn some math stuff in wikipedia, and thought there might be something I might actually have a relevant idea about.  I'll cross link this post with my blog. Might remove later, depending.
m_G, http://www.livejournal.com/users/onfeynyuan_shen/78070.html
:I don't think this has too much bearing on the axiom of choice (AoC). It is simply an axiom of set theory that you could assume, just like all the other ones. If you have two one-element-sets and you want to make a set containing both elements you will have to do that using some axiom. If there is no axiom (or sequence of axioms) which allows it, then it is impossible. In the same way the AoC allows the construction of some set, only more complicated. --MarSch 09:53, 23 October 2005 (UTC)
thread_title:  interpenetration 

message: The following is listed as a result requiring AC:
:Every infinite game G(T,X) where X is either open or closed is determined.
The claim is true, but there are a few problems with it:
#I believe T here is supposed to be a tree, and X a set of branches of T. That's not clearly stated.
#No one has yet bothered to give a definition of such games G(T,X); there's a to-be-written section Determinacy#Games played on trees that someone might fill in.
#The claim is maybe a little misleading; if T is a tree on the naturals (or other wellordered set), no choice is required. On the other hand, in the completely general situation, you don't need infinite games at all— AC is equivalent to the determinacy of all length-2 games (which of course are trivially both open and closed). --Trovatore 20:30, 1 December 2005 (UTC)
thread_title: determinacy requiring AC

message: I don't believe I've ever encountered this abbreviation in the literature; in my experience it's always called ACω. Lethe, do you have a source? --Trovatore 01:22, 9 March 2006 (UTC)
replies: text: Sure.  The first book I picked up, Schechter's HAF uses CC.  -lethe talk [ +] 01:24, 9 March 2006 (UTC)

text: I've also got LM, HB and UF now.  Do you feel that they shouldn't go in?  It doesn't really matter much whether they go in or not, but it's hard to write things like ZF + DC + BP without some abbreviations, right? -lethe talk [ +] 01:51, 9 March 2006 (UTC)
replies: text: Now I'm reading a paper where the property "every set of reals has the Baire property" is represented with exactly that sentence, and not with an abbreviation.  Yuck!  But I guess BP is not standard. -lethe talk [ +] 01:58, 9 March 2006 (UTC)
replies: text: I guess I don't have any problem with using the abbreviation to simplify the exposition; I rewrote the sentence to try to avoid the implication that everyone in the field would recognize it. --Trovatore 02:01, 9 March 2006 (UTC)
replies: text:  BP = "Boolean Prime" ideal theorem?  Very few of these are standard.  In fact, CC may be ACω or ACω.  — Arthur Rubin |  (talk) 19:26, 8 May 2006 (UTC)
thread_title: Abbreviation "CC"

message: OK, first, again I'm not familiar with the abbreviation "BP" for the claim "every set of reals has the property of Baire" and I don't think this is standard; the text should avoid implying that it's a standard abbreviation.
More importantly, the text I removed claimed that from ZF+DC+"every set has the property of Baire" you can prove that every set is measurable. That's false. It's true that in the Solovay model, ZF+DC holds, and every set of reals is measurable and has the p.o.B., but it's not true that you can recover measurability from the p.o.B. In fact, ZF+DC+"every set has the p.o.B" is equiconsistent with ZFC, but ZF+DC+"every set is Lebesgue measurable" is stronger (I think the same consistency strength as ZFC+"there exists an inaccessible"). --Trovatore 01:29, 9 March 2006 (UTC)
replies: text: As far as the initialism goes, BP is used in HAF, and now I'm reading the paper of Shelah 1984 where he proves that ZF + DC + BP is consistent to figure out where my mistake was (I'm a little confused at the moment).  Obviously he uses the initialism BP as well.  -lethe talk [ +] 01:46, 9 March 2006 (UTC)
replies: text: What is HAF? --Trovatore 01:49, 9 March 2006 (UTC)
replies: text: Handbook of Analysis and Foundations.  http://math.vanderbilt.edu/~schectex/ccc/index.html.  I guess it's not very standard? -lethe talk [ +] 01:52, 9 March 2006 (UTC)
thread_title: Solovay model

message: We should put those in order (in ZF), right?  Something like: AC → DC → CC and AD → LM  → ¬AC, would be nice.  But maybe they don't make a total order...  -lethe talk [ +] 02:16, 9 March 2006 (UTC)
replies: text: I don't think they are linearly ordered, no. For example, I've never heard that either of what you're calling LM and BP implies the other, and since it's a natural question, presumably someone has models of ZF+DC+LM+¬BP and ZF+DC+BP+¬LM. (LM is stronger than BP consistency-wise, but I don't think it outright implies it.) --Trovatore 02:20, 9 March 2006 (UTC)
thread_title:  Weakened AC and strengthened ¬AC 

message: Intuitively speaking, AC says that given a collection of bins, each containing at least one object, then exactly one object from each bin can be picked and gathered in another bin - even if there are infinitely many bins, and there is no "rule" for which object to pick from each.
This does not make sense to me, as far as picking and gathering. Is there supposed to be a postcondition met as well?
Full Decent 00:26, 29 March 2006 (UTC)
replies: text: No postcondition.  That's the whole bit.  Maybe you could mention what postcondition you think is required, and we could figure out how to word it to make it clear.  Right now, I'm not sure what you're looking for. -lethe talk [ +] 06:55, 29 March 2006 (UTC)
thread_title:  Intuititively 

message: I don't get it.
RandomP 01:26, 1 May 2006 (UTC)
replies: text: Yeah, I'm curious about that too. I remember I was pretty annoyed at a dab line at the top of the mathematics article pointing to mathematics (producer), but really, people looking for that article should have some way to find it. I would have liked to use  at mathematics to help them out, but unfortunately no one seems to have a third article called "mathematics", which is what you're supposed to have before writing mathematics (disambiguation). --Trovatore 01:37, 1 May 2006 (UTC)

text: Oh, I see now that Michael hasn't removed the hat note, just reworded and reformatted. Can't say I care much either way. --Trovatore 01:42, 1 May 2006 (UTC)

text: But it didn't say "for other uses of the axiom of choice"; it just said "For other uses, see Axiom of choice (disambiguation)". Takes a little interpolation on the reader's part, but I don't think that's so terrible. Also the capital letter helps make the use-mention distinction. But as I say, it's not that I care much either way. --Trovatore 02:00, 1 May 2006 (UTC)
thread_title:  How was otheruses template offensive? 

message: (Yes, the Jean Rubin mentioned there is my late mother.)  My copies of the book are at work, but many of the questions about specific statements related to the axiom of choice are there.  Perhaps I'll look them up and comment later. — Arthur Rubin |  (talk) 09:01, 7 May 2006 (UTC)
replies: text: Oh?  I saw a Rubin referenced in one of my textbooks a couple months ago, and briefly entertained the idea that it might be you.  But it wasn't an A. Rubin.  Maybe it was your mother? -lethe talk [ +] 09:35, 7 May 2006 (UTC)
thread_title:  Consequences of the Axiom of Choice 

message: Note:  I'm going to refer to my mother's survey books as Equivalents, Equivalents II, and Consequences, to avoid retyping "of the Axiom of Choice" each time.  If there are any other survey books on forms of the Axiom of Choice or related forms, I'd be happy to use those, as well.
I was wondering if there would be any objection to noting (in an HTML comment) how these forms are denoted in those books, so we (or, at least those of us who have those books) can "easily" see whether we've stated the forms correctly. — Arthur Rubin |  (talk) 16:15, 26 May 2006 (UTC)
thread_title:  Naming issues