Timeline for How much of the axiom of choice do you need in mathematics?
Current License: CC BY-SA 4.0
19 events
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| Aug 21, 2023 at 15:30 | comment | added | Timothy Chow | @MishaLavrov I would argue that in combinatorics, one is not really interested in $k$-colorings of $\mathbb{R}$. One is really interested in $k$-colorings of every finite subset of $\mathbb{R}$. For example, it's convenient to speak of "the chromatic number of the plane" but what a combinatorialist is really interested in is the maximum chromatic number of a finite subset of the plane. If one forces a choice between the two concepts by removing AC, then the combinatorialist will choose the finitary concept. | |
| Aug 14, 2023 at 2:54 | comment | added | Nik Weaver | @J.vanDobbendeBruyn I understand your concern. I feel confident in saying that I've never needed a pure state on a von Neumann algebra, but I recognize that my experience is very limited compared to the general community. | |
| Aug 13, 2023 at 23:18 | comment | added | J. van Dobben de Bruyn | @NikWeaver my worry is more that some very basic results which we take for granted might rely on compactness of the state space. One example would be the existence of pure states (or irreducible representations). But I don't oversee the textbook theory of $C^*$- or von Neumann algebras well enough to fully understand the importance of this in the overall theory. Perhaps you're right that we can do everything we want without it. In any case, there is no point in me further arguing down this line until I can give a concrete example of an non-trivial result that crucially relies on this. 🙂 | |
| Aug 13, 2023 at 21:19 | comment | added | Nik Weaver | Well, I'm sure there are people who care about state spaces of von Neumann algebras, but non-normal states tend to be pretty awful --- I think most of us are generally a lot more interested in normal states. | |
| Aug 13, 2023 at 21:04 | comment | added | J. van Dobben de Bruyn | @NikWeaver Fair enough. But are you saying the only interesting (state spaces of) $C^*$-algebras are the separable ones? I find that hard to believe, even though I don't have a concrete application at hand. I imagine that it would be impossible to have, say, a general theory of von Neumann algebras without this, but maybe I'm overestimating the importance of compactness of the state space. | |
| Aug 13, 2023 at 20:18 | comment | added | Nik Weaver | @J.vanDobbendeBruyn I care a lot about state spaces of separable C${}^*$-algebras, but the state space of $B(l^2)$? That's a pretty pathological object --- why do we care whether it's compact? | |
| Aug 13, 2023 at 19:21 | comment | added | J. van Dobben de Bruyn | “Nor have I ever used Alaoglu's theorem in the dual of a nonseparable space.” How about in proving that the state space of a unital $C^*$-algebra is compact? Even $B(\ell^2)$, arguably the simplest example of an infinite-dimensional $C^*$-algebra, is non-separable. I think this is an important application of Alaoglu's theorem in the dual of a non-separable Banach space. | |
| Aug 13, 2023 at 14:51 | comment | added | Mikhail Katz | @AndreasBlass, it would be interesting to see a more detailed version of your comment. | |
| Aug 12, 2023 at 18:44 | comment | added | Misha Lavrov | One use of uncountable Tychonoff in combinatorics (not sure how essential this is!) is to prove that the space of functions from $\mathbb R$ to $\{1,\dots,k\}$ (i.e. the space of $k$-colorings of $\mathbb R$) is compact, with the usual product topology. By a compactness argument, if we want to find a $k$-coloring of $\mathbb R$ with a certain property, and all obstructions have a fixed finite size, then it's enough to find a $k$-coloring of every finite subset of $\mathbb R$. For example, Theorem 5.2.2 in Alon & Spencer's Probabilistic Method uses this trick. | |
| Aug 12, 2023 at 15:48 | comment | added | Z. M | @TimothyChow I do not understand your trick. If I understand correctly, one point is Deligne's completeness theorem (equivalent to the ultrafilter lemma): the Zariski locale (or some valuative locale) has enough points, thus we can reduce to stalks, i.e. local rings (or valuation rings), somehow sort of "stalk-local global principle". I do not mean that AC or the ultrafilter lemma is un-eliminable (some partial progress was made by Lombardi–Quitté and so on), but that current mathematics build on them, and it is nontrivial to eliminate. | |
| Aug 11, 2023 at 23:37 | comment | added | Timothy Chow | @Z.M One standard "trick" for avoiding AC when you need a maximal ideal is to simply define a commutative ring to be a "commutative ring with a maximal ideal." Does this trick get increasingly cumbersome as you develop more theory? For example, maybe you need constructions that don't guarantee a maximal ideal even if the inputs to the construction have maximal ideals? | |
| Aug 11, 2023 at 18:42 | comment | added | Z. M | @TimothyChow There are many situations where AC (or at least, the ultrafilter lemma) is necessary for the current framework to work. For example, Krull's existence of maximal ideals needs the full generality of AC for non-Noetherian rings. We also need the abundance of valuation rings (and that they are stable under ultraproducts) to work with v-topology and arc-topology, where we need at least the ultrafilter lemma. Non-Noetherian rings appear naturally: even if you start with a variety, modern techniques such as perfectoid spaces will quickly lead you to non-Noetherians. | |
| Aug 11, 2023 at 18:25 | comment | added | new account | @AndreasBlass Thanks. I'm aware of that and I mentioned it in my answer. My understanding is it actually provides an algorithm to transform an ultrafilter proof to an ultrafilter-free proof? | |
| Aug 11, 2023 at 18:23 | comment | added | Andreas Blass | @newaccount Some uses of ultrafilters could "in principle" be eliminated by generically adjoining a suitable ultrafilter to the universe, proving the desired theorem in that forcing extension, and then returning to the original universe by some absoluteness result. Of course you might need a set theorist to design a notion of forcing to produce your "suitable" ultrafilter. | |
| Aug 11, 2023 at 17:03 | comment | added | Nik Weaver | @newaccount good question. Ultrafilters are certainly used in functional analysis. However, most of those uses are not needed, in the sense that they can be avoided in the applications of central interest. Whether that is always the case, for "ordinary" mathematics, will surely be a judgement call. I will stand on my position that all the necessary uses of ultrafilters that I know about are exotic, though I am sure others would disagree. | |
| Aug 11, 2023 at 16:11 | comment | added | new account | Do you ever use an ultrafilter? I thought they were at least occasionally useful in functional analysis | |
| Aug 11, 2023 at 11:12 | comment | added | Timothy Chow | @Z.M Can you elaborate on your comment? Just because one is studying non-Noetherian rings doesn't automatically mean that AC is needed. | |
| Aug 11, 2023 at 3:43 | comment | added | Z. M | We use much more AC than (countable) DC in commutative algebra and algebraic geometry. Many important rings, especially those have something to do with analytic geometry, such as valuation rings, are non-Noetherian. | |
| Aug 11, 2023 at 2:05 | history | answered | Nik Weaver | CC BY-SA 4.0 |