Timeline for answer to Axiom of choice, Banach-Tarski and reality by Nik Weaver
Current License: CC BY-SA 3.0
Post Revisions
29 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 19, 2022 at 8:20 | comment | added | saolof | It may be worth mentioning that countable choice is implied by a large number of other axioms, some of which are mutually incompatible with choice, and that it is constructively valid. Also, countable choice is arguably necessary for finite sets to really be well behaved with an injection from the natural numbers to any infinite set. Without it, you can have an infinite set that cannot be written as the union of two nonempty infinite sets. | |
| Aug 5, 2017 at 15:12 | comment | added | user112009 | Questionable axioms? There is no doubt that the axiom of choice has been proven wrong. Its consequence, every set can be well-ordered, is as "real" as the result: there are ten different natural numbers less than 3. | |
| Jan 30, 2017 at 20:39 | comment | added | Delio Mugnolo | @NikWeaver Well, even without dealing with $B(H)$, Hahn-Banach is used to prove basic properties of operator semigroups, like the existence of duality sets needed in the Lumer-Phillips Theorem. If one studies so-called open quantum systems, operator semigroups acting on the space $\mathcal S_1$ of trace class operators must be introduced. Even at a much more elementary level, the whole theory of $H$-valued integration and $H$-valued holomorphic functions need HB. | |
| Jan 30, 2017 at 16:36 | comment | added | Nik Weaver | @DelioMugnolo: I've been working with $B(H)$ for twenty-five years and to the best of my recollection I've never needed to use Hahn-Banach on it. "Most operator theory would fail" --- can you give a specific example? | |
| Jan 30, 2017 at 10:27 | comment | added | Delio Mugnolo | > he Hahn-Banach theorem is a counterexample, but only if you want it for nonseparable spaces, and I can't think of any time I've ever needed this. The existence of Banach limits is an obvious case where HB is needed in its non-separable version. Worse even, the space of bounded linear operators on any infinite dimensional Hilbert space is not separable, so most operator theory would fail without non-separable HB. | |
| Jan 23, 2017 at 14:33 | comment | added | Stefan Rigger | @NikWeaver I wouldn't put it that way, but it's somewhere along the lines of what I always thought (I don't have expert knowledge of set theory, so that might very well be far from the truth). How do you judge if an axiom is true if not by its consequences? | |
| Jan 23, 2017 at 0:20 | comment | added | Nik Weaver | @StefanRIgger: Yes, I think the ZF axioms were formulated in a very ad hoc way, and the nonsensical "iterative conception" was only cooked up later. Do you dispute my assertion? You feel we can decide whether an axiom is true based on whether we like its consequences? | |
| Jan 22, 2017 at 17:32 | comment | added | Stefan Rigger | @NikWeaver "We can't decide whether an axiom is true based on whether we like its consequences". But isn't that how axioms where introduced? People didn't prove theorems from set-theoretic axioms before the 20th century, and when the axioms were invented, weren't they designed in such a way that they would express all that people thought was true mathematics as true theorems? | |
| Jan 22, 2017 at 16:25 | comment | added | Nik Weaver | It's hard to answer that in a comment ... can I refer you to my book Truth & Assertibility, Chapters 4 and 6? | |
| Jan 22, 2017 at 15:19 | comment | added | user21820 | @NikWeaver: In that post I was assuming that the natural numbers and power-set operation was granted, and that one could get the union of the levels labelled by any sequence. I don't see a serious conceptual problem if you interpret the power-set operation as obtaining just the definable subsets. But even then one can never get an uncountable number of levels, since there's no sequence of uncountable length to begin with. Anyway do you think intuitionistic logic is preferable to a 3-valued logic for impredicative collections, and why? | |
| Jan 22, 2017 at 1:09 | comment | added | Nik Weaver | That's hard to answer because I don't find the iterative conception cogent. | |
| Jan 21, 2017 at 15:43 | comment | added | user21820 | @NikWeaver: Yes I'm very interested. I've just looked at your pdf linked from mathoverflow.net/a/133598/50073 and much of it is similar to my personal viewpoint, except that I think it makes more sense to have a universal type and Kleene's 3-valued logic instead of an open domain supported by intuitionistic logic. And do you agree with this post that the iterative conception of the cumulative hierarchy does not explain Replacement? | |
| Jan 21, 2017 at 6:09 | comment | added | Nik Weaver | @user21820: indeed, I do reject the power set axiom. I've written extensively about this; there are a bunch of papers on the arXiv, if you're interested. | |
| Jan 21, 2017 at 5:05 | comment | added | Wojowu | @user21820 Ultrafilters can be used to give rather neat proofs of many combinatorial statements, like Ramsey or Hindman theorem. Whether you count it as a "real-world" application might be arguable, but they are definitely quite down-to-earth applications. | |
| Jan 21, 2017 at 2:58 | comment | added | user21820 | @DavidRoberts: Eating is an operation, and if you chew off half a rook then you would have changed $\{rook\}$ into $\{half(rook)\}$. =) | |
| Jan 21, 2017 at 2:06 | comment | added | user21820 | @Wojowu: But is there any theorem that over ZF requires the ultrafilter lemma to prove and has any real-world significance? | |
| Jan 21, 2017 at 2:03 | comment | added | user21820 | "because I don't know how to construct the universe." -- That's one possible justification, but then it rejects the power-set axiom (you can't construct almost all subsets of the natural numbers). My point is that I can readily accept the validity of various justifications for meaning of each ZF axiom schema, but none of them work for all the schemas, and are incompatible if we attempt to combine them, at least from my viewpoint. | |
| Jan 21, 2017 at 0:43 | comment | added | David Roberts♦ | @user21820 what are the elements of that set? What if I eat half of a rook? :-) | |
| Jan 20, 2017 at 20:34 | comment | added | Qfwfq | "We can't decide whether an axiom is true based on whether we like its consequences" - Well.... I would say whether we like or not its consequences is an important point for deciding if we want to pursue an axiom! | |
| Jan 20, 2017 at 16:27 | comment | added | Wojowu | Although not very relevant to the discussion at hand, I'd like to point out that a very important and widely applicable result, namely the ultrafilter lemma, depends on more than just the countable (or even dependent) choice. Indeed, there are models of ZF in which DC holds but there are no nonprincipal ultrafilters on any set. | |
| Jan 20, 2017 at 16:24 | comment | added | Nik Weaver | "your justification seems to apply equally to the universal complement" --- no, because I don't know how to construct the universe. If a set is given constructively this means I know how to construct it. For any decidable property, I can follow the same procedure but only accept the elements which have the property. | |
| Jan 20, 2017 at 15:57 | comment | added | user21820 | As for eating... We can eat a set of chocolate chess pieces. =) | |
| Jan 20, 2017 at 15:53 | comment | added | user21820 | @NikWeaver: Hmm do you mind explaining what kind of "meaning" you have in mind then? As for the comprehension axiom, your justification seems to apply equally to the universal complement, since if $S$ has been previously constructed we can 'construct its complement using it'. | |
| Jan 20, 2017 at 15:51 | comment | added | Nik Weaver | @user21820: I think you're using the word "meaning" in a different way than I am. I agree that comprehension seems somewhat ad hoc, but on the other hand it does have a clear constructive justification. (If we can decide, for each $x \in S$, whether $x$ has property $P$, then we can construct the desired subset by hand, at least on my interpretation of these terms.) | |
| Jan 20, 2017 at 15:44 | comment | added | user21820 | I'm curious about your last paragraph. Why is it that you feel that set-theoretic assertions are objectively meaningful, in the sense of having ontological status beyond the syntactical formal system? Specifically for example, does the axiom of comprehension have objective meaning in the way it does not allow constructing the universal complement of a previously constructed set $S$ even though membership in the complement is simply $\neg x \in S$? | |
| Jan 20, 2017 at 15:42 | comment | added | Nik Weaver | The uncountable rings that appear in analysis typically come with natural topologies with respect to which they are separable, and one can use this. If you're talking about untopologized uncountable rings, I wouldn't consider it "bread and butter" mathematics (just my opinion, of course). | |
| Jan 20, 2017 at 15:40 | comment | added | Nik Weaver | Good point, although for countable rings you don't need any choice for this result. | |
| Jan 20, 2017 at 14:58 | comment | added | Francesco Polizzi | Another very important consequence of AC is that, in a commutative ring with $1$, every proper ideal is contained in a maximal ideal. It seems to me that this is not a consequence of countable choice, at least for non-noetherian rings. | |
| Jan 20, 2017 at 3:36 | history | answered | Nik Weaver | CC BY-SA 3.0 |