Talk:Axiom of choice

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I like the new one, but it has the same technical problem as the old one. Take the colored item. No choice axiom needed for that. -- YohanN7 (talk · contribs) 10:11, 5 April 2017‎ (UTC)

Axiom of non-choice

"Redirected from Axiom of non-choice"... however, "non-choice" does not appear anywhere in the article. DAVilla (talk) 21:17, 22 June 2019 (UTC)

No one uses "axiom of non-choice" as far as I know. So what should it redirect to?

This article has a section on Axiom of choice#Stronger forms of the negation of AC which is probably as close as one can get to that. JRSpriggs (talk) 08:21, 23 June 2019 (UTC)

Should the redirect Axiom of non-choice be deleted? Or perhaps be turned back into an article (Old revision of Axiom of non-choice)? – Tea2min (talk) 09:14, 23 June 2019 (UTC)

If I understood JRSpriggs right, the text of the Old revision should be included in Axiom of choice. I agree with that; the redirect should then go to the new subsection. However, shouldn't that be part of section Axiom of choice#Weaker forms, since AC implies ANC, but not vice versa? - Jochen Burghardt (talk) 20:22, 23 June 2019 (UTC)

Until I read the later comments, I did not realize that this redirect had previously been an article. After reading that article, I changed the redirect back into the article and added a line about this being a theorem in classical ZF. Please excuse my ignorance of constructive set theory.

I would not merge that article into this one because this one is about classical set theory, not constructive set theory. But we could add a link to it. JRSpriggs (talk) 01:21, 24 June 2019 (UTC)

Actually, I think that "axiom of non-choice" is a misleading name for that theorem. (1) It is not an axiom in classical set theory. (2) It is not denying choice, but merely avoiding using it. JRSpriggs (talk) 01:32, 24 June 2019 (UTC)

Ok for me. I'd also appreciate if you'd add a link (e.g. a sentence in section Axiom of choice#In constructive mathematics, including your above remarks about the name). - If you have a source available confirming the provability in ZF, could you add it at Axiom of non-choice? I see that the replacement axiom easily matches the "non-choice" axiom, but I don't see immediately the latter's proof from that. - Jochen Burghardt (talk) 10:24, 24 June 2019 (UTC)

I think the simple path here is just to delete the redirect, unless this is a term that's actually used in the wild. Such redirects are "mostly harmless", but not completely; it has some potential to mislead people into thinking the "axiom of non-choice" is some particular thing (especially if someone links to it), and it's spawning noise in this talk page. --Trovatore (talk) 18:28, 24 June 2019 (UTC)

Oh, I didn't see that this is now apparently an article. Is that term really used? I have some familiarity with that milieu but not an awful lot. --Trovatore (talk) 18:29, 24 June 2019 (UTC)

set builder

Why isn't $\forall X\left[\varnothing \notin X\implies \exists f\colon X\rightarrow \bigcup X\quad \forall A\in X\,(f(A)\in A)\right]\,$ written as $\forall X\left[X\not =\varnothing \implies \exists f\colon X\rightarrow \bigcup X\quad \forall A\in X\,(f(A)\in A)\right]\,?$ Nikolaih^☎️📖 03:12, 6 June 2020 (UTC)

The first,

$\forall X\left[\varnothing \notin X\implies \exists f\colon X\rightarrow \bigcup X\quad \forall A\in X\,(f(A)\in A)\right]\,,$

is a statement about (possibly empty) sets that do not contain the empty set as an element (that is, sets of nonempty sets). The second,

$\forall X\left[X\not =\varnothing \implies \exists f\colon X\rightarrow \bigcup X\quad \forall A\in X\,(f(A)\in A)\right]\,,$

is a statement about nonempty sets.

Consider the nonempty set consisting of the empty set alone, $X=\{\varnothing \}$. Now, $\bigcup X=\varnothing$. The second expression you gave now postulates the existence of a function $f\colon X\rightarrow \varnothing$, and there is no such function for a nonempty $X$.  –Tea2min (talk) 06:33, 6 June 2020 (UTC)

Thank you very much for the clarification. Nikolaih^☎️📖 22:47, 6 June 2020 (UTC)

X as a collection or set?

I'm still learning but shouldn't the boxed statement of the axiom refer to X as a collection, not a set?

In ZFC set theory, every entity is a set. Every collection is a set of sets. See pure set. JRSpriggs (talk) 21:02, 25 July 2021 (UTC)

Essay on reasons to use choice

I reverted this paragraph:

One may notice that many of the above statements (and also some of the statements in the next sections using weaker forms of choice) come in one of two following types: First, there are those statements which assert that a given property pertaining to a type of mathematical structure is closed under arbitrary Cartesian products (e.g. the Cartesian product of nonempty sets is nonempty, or the Cartesian product of connected topological spaces is connected). Second, there are those statements that assert the existence of a maximal element of some special collection of subsets in a given mathematical structure, and moreover assert that any such subset can be extended to a maximal one (e.g., the Hausdorff Maximal Principle, Krull's Theorem, and the assertion that every vector space has a basis). This pattern is not a coincidence, as can be seen by the following reasoning: To prove any of these statements when the Cartesian product (in the case of the first type) or the special collection of subsets (in the case of the second type) is finite, one simply can use finite induction on the size of these collections or indexing sets. However, to extend these statements to infinite cardinalities, one needs to go beyond finite induction to transfinite induction, and as mentioned earlier in the article, this in general requires the Axiom of Choice.

So the first objection is that, helpful or not, this seems to be a collection of personal reflections, not sourced to the literature.

Beyond that, there's a confusion here about transfinite induction. Transfinite induction per se does not rely on the axiom of choice. Not in any way whatsoever.

What does rely on the axiom of choice is inductive arguments that involving "picking" a particular element at each induction step. But the choice is used for the picking, not for the induction. We really have to be clear on this, because this is a common misconception. I haven't scoured the article to see if there's anywhere else this needs to be clarified. --Trovatore (talk) 19:31, 19 April 2022 (UTC)

Just to mention, I have seen the term "principle of transfinite induction" used in the literature to actually refer to forms of the Axiom of Choice such as Zorn's Lemma. For example, there is Terence Tao's classic text Analysis I. In Section 8.4 in this book, he uses the phrase "principle of transfinite induction" to refer to Zorn's Lemma. Also, the classic abstract algebra textbook by Dummit and Foote concedes in Appendix A that Zorn's Lemma can viewed as a sort of "infinite induction", i.e. a generalization of finite induction to infinite partially ordered sets. The point is that the principle of finite induction can be reframed in the context of partial orders, to essentially assert the existence of a maximal chain (or element) in any finite partially ordered set; Zorn's Lemma naturally extends this fact to infinite partially ordered sets. One must recognize that this is a different usage of the term "principle of transfinite induction" than is typically used in the context of ordinals. However, there are some merits in the way this term is used as above. Often, novices have a hard time understanding the intuition behind Zorn's Lemma, but hopefully presenting it as a sort of "infinite induction" can help aid understanding. Logic314 (talk) 22:22, 23 April 2022 (UTC)

Controversy

This part is misleading: “Although originally controversial, the axiom of choice is now used without reservation by most mathematicians,[3] and it is included in the standard form of axiomatic set theory, Zermelo–Fraenkel set theory with the axiom of choice (ZFC).”

That’s not what the reference says. The reference acknowledges that AC is the most contested axiom, and that it is less controversial in recent years. However it still remains very controversial, with many mathematicians avoiding its use. The other sentence is also misleading because AC is not automatically included in standard set theory. It is independent. That’s why there is ZF, ZFC, and ZF with negated C. 184.169.45.4 (talk) 19:24, 8 October 2022 (UTC)

Your fourth sentence in the second paragraph is not correct. AC is part of default set theory. --Trovatore (talk) 19:27, 8 October 2022 (UTC)

The text in the Wiki article is still not reflecting the citation, because it makes the claim that AC is not controversial anymore, when in fact it remains controversial. The cited source just says it is less controversial. The text should be edited to remove bias and acknowledge the ongoing controversy. 107.194.133.235 (talk) 02:54, 11 October 2022 (UTC)