Talk:Axiom of dependent choice

WikiProject class rating
This article was automatically assessed because at least one WikiProject had rated the article as stub, and the rating on other projects was brought up to Stub class. BetacommandBot 03:48, 10 November 2007 (UTC)

Why?
Surely, there are a gazillion axioms of set theory that assert the existance of what one wants and not anything else as opposed to what one wants and unfortunately gets but can't avoid some extra benifits.

This is one of them. It's plain it'll never be accepted anyway, why the heck shuld AC be abbrupted just when it was waking up, at just that particular moment? Nah!

"Sorry old choice-function, youre TOO OLD..." Sure, that's really convincing. Just about as convincing it is to see arguments against AC and for Countable Choice.

What I'm objecting to is that the silly articles are getting the same attention level as the accepted ones. By that I mean simply that "objections to ZFC" is lowlevel. Even though it happen to be highlevel to these working with it¨¨. —Preceding unsigned comment added by YohanN7 (talk • contribs) 14:57, 21 July 2009 (UTC)


 * The axiom of dependent choices is not some newly invented and unused thing. On the contrary, it has a long history of being studied by mathematicians. JRSpriggs (talk) 17:34, 21 July 2009 (UTC)


 * Right &mdash; actually DC is a fairly important fragment of choice, one that allows a lot of arguments to go through in real analysis, but doesn't let you wellorder the reals and come up with perceived "pathological" sets of reals. The way to think of it is, DC is what you need to do a transfinite induction of countable length, in which you're allowed to make one choice at each step.
 * In the very important model L(R), DC is actually true, but AC is not. --Trovatore (talk) 18:31, 21 July 2009 (UTC)


 * I stand corrected. Are there versions of DC that allow and disallaw Zorns Lemma as well? What I mean is that to me personnally, AC is ok, wellorder is murky, and Zorn is somewhere in between, probably something I want. (Yes, I do know that they are all equivalent under ZF + AC.) By the way, when it comes to paradoxes, there must surely be a few if one introduces NOT AC in the same way that NOT special relativity would yield a host of strange results. Does DC cure that? YohanN7 (talk) 02:02, 22 July 2009 (UTC)
 * I think you're taking an overly "foundational" approach to DC. I don't know that there are too many folks who are arguing that DC is true but AC is not.  The usual view among those who think the question makes sense, is that full AC is true.
 * But even if you think full AC is true, it's still important to know about DC. Because (oversimplifying here) that's the version of choice you can have in determinacy models.  So it's good to know what it can do, what it can't do, and how to reason informally in a context where you can use DC but not AC. --Trovatore (talk) 03:45, 22 July 2009 (UTC)
 * Yup, your'e right. Besides, I'm here to learn, not to criticize. Iv'e been complaining too much lately;) YohanN7 (talk) 08:37, 22 July 2009 (UTC)

Couldn't find a ref, so gave a proof
I recently added footnotes 2 and 3 to take care of verifying two statements. However, I was unable to find a reference source for the equivalence between $$\mathsf{DC}$$ and every pruned tree with ω levels having a branch. Since the proof is straightforward, I have included it. This proof should also help readers understand how $$\mathsf{DC}$$ can be used. The proof comes up collapsed so readers can easily skip it. (By the way, I left out "nonempty" when describing a tree with ω levels since such a tree is clearly nonempty.) --RJGray (talk) 20:45, 17 September 2017 (UTC)

Unnecessary TeX?
Curiously, this article uses the math environment to denote the axioms mentioned, that is, $$\mathsf{DC}$$ ( $$\mathsf{DC}$$ ) instead of just DC and $$ \mathsf{AC} $$ ( $$ \mathsf{AC} $$ ) instead of just AC. Is there a specific reason for this? Why would you use inline images instead of just letters? – Tea2min (talk) 07:10, 20 July 2020 (UTC)

Non-Lebesgue measurable set?
It is currently stated that DC is insufficient to show the existence of a non-measurable subset of the real numbers. However, DC implies countable AC implies Ultrafilter Lemma implies Hahn-Banach theorem implies non-measurable set. And this is done using only ZF. (Check the Wikipedia entry for Ultrafilter Lemma and the things it implies.)

So why are these two statements contradictory? Have we finally found that ZF is inconsistent? 134.250.190.90 (talk) 23:16, 22 May 2023 (UTC)


 * The title of the cited paper:
 * Foreman, M.; Wehrung, F. (1991). "The Hahn–Banach theorem implies the existence of a non-Lebesgue measurable set" (PDF). Fundamenta Mathematicae. 138: 13–19. doi:10.4064/fm-138-1-13-19.
 * is somewhat misleading. The measure used is defined in the same way as Lebesgue measure, but without assuming AC is only finitely additive rather than countably additive. So there is no contradiction. 134.250.202.110 (talk) 22:12, 25 May 2023 (UTC)
 * Indeed. So, if I read this correctly, the statement seems to be "there is no non-measurable subset of the reals, when you define a measure to be finitely additive, which is how you must define a measure, when working with DC". I can't fix this article based on these statements, so I will tag it instead. 67.198.37.16 (talk) 03:46, 7 December 2023 (UTC).
 * I haven't followed all of the above, but DC is consistent with AD, which implies "every set of reals is Lebesgue measurable", and Lebesgue measure is still countably additive, not just finitely additive, in such a model. --Trovatore (talk) 04:30, 7 December 2023 (UTC)
 * Oh, this might be the key step? DC implies countable choice for reals, not in general.  --Trovatore (talk) 04:31, 7 December 2023 (UTC)
 * Strike the last comment &mdash; I was thinking that DC by default meant $$\textrm{DC}_\mathbf{R}$$, which is the bit that's most often applicable in descriptive set theory, but I looked it up in Moschovakis, and you're right, full DC is stronger than countable choice. OK then &mdash; why do you think countable choice implies the ultrafilter lemma?  That seems unlikely to me.  The natural proof of the ultrafilter lemma by transfinite recursion involve wellordering the subsets of the overall containing set, iterating through them, and deciding which ones go into the ultrafilter.  That means making many more than countably many choices. --Trovatore (talk) 06:00, 7 December 2023 (UTC)