Is it true that "Axiom of Choice is true or Axiom of Choice is false"?

Question

Is it true that "Axiom of Choice is true or Axiom of Choice is false" in ZF with classical logic?

It looks like LEM would suffice to prove it.

Maybe I should write it as: Is it true "'Axiom of Choice' OR NOT('Axiom of Choice')"?

And for 'Axiom of Choice' I mean an expression that express the Axiom of Choice.

Edit: sorry people, I should have provide more hints about my concerns. I chose Axiom Of Choice because I know it is undecidable under ZF. I am not so interested in AC by itself but I would like to use it as an example of an undecidabe statement. AC has been a bad choice from my side, sorry. Anyway all links looks promising but I am not sure I want to go into that direction now.

To come back to my original question (that I didn't write until now, sorry), is it ok to be in a framework like ZF, being able to express "AC or ¬AC", proving that is true and at the same time knowing that AC is undecidable under ZF?

Edit: Not sure how to improve the question. The short answer to the question is "Yes.". It is just LEM when P is AC. So it is provable by LEM axiom and there is no problem.

Please don't rely on the title of your MSE questions for important information. The body of the question should be self-contained and the title should just give a useful summary. There are several reasons for this, not least the fact that some interfaces to MSE such as the Android interface do not render the title very well. — Rob Arthan, Jun 08 '22 at 20:58
https://math.stackexchange.com/questions/3955048/does-the-law-of-excluded-middle-imply-the-axiom-of-choice — amWhy, Jun 08 '22 at 21:00
https://math.stackexchange.com/questions/1998009/axiom-of-choice-implies-law-of-excluded-middle — amWhy, Jun 08 '22 at 21:01
https://math.stackexchange.com/questions/1017361/what-is-the-opposite-of-the-axiom-of-choice — amWhy, Jun 08 '22 at 21:06
You might be interested in Counter-intuitive consequences of $\lnot AC — amWhy, Jun 08 '22 at 21:11
Also not that LEM is consistent with classical logic, but not so with constructive logic. No conflict, necessarily, between LEM and $\lnot AC$. — amWhy, Jun 08 '22 at 21:20

score 4 · Accepted Answer · answered Jun 09 '22 at 18:59

4

Yes, we can prove (from $\mathsf{ZF}$, in classical logic) the sentence "$\mathrm{AC}\lor \lnot \mathrm{AC}$", where $\mathrm{AC}$ is the axiom of choice. This is just an instance of the tautology $P\lor \lnot P$, where $P$ is any sentence (the law of excluded middle).

There is no conflict with the fact that $\mathrm{AC}$ is independent of $\mathsf{ZF}$ (assuming $\mathsf{ZF}$ is consistent). Writing out the situation in words:

There are some models of $\mathsf{ZF}$ in which $\mathrm{AC}$ is false ($\mathsf{ZF}$ does not prove $\mathrm{AC}$) and some models of $\mathsf{ZF}$ in which $\mathrm{AC}$ is true ($\mathsf{ZF}$ does not prove $\lnot \mathrm{AC}$).
In every model of $\mathsf{ZF}$, $\mathrm{AC}$ is either true or false ($\mathsf{ZF}$ proves $\mathrm{AC}\lor\lnot \mathrm{AC}$).

An analogy with the theory of groups might help. The sentence $\varphi: \forall x\forall y\,(xy = yx)$ is true in a group $G$ if and only if $G$ is abelian. Now some groups are abelian and some are not. So $\varphi$ is independent from the theory of groups: we cannot prove $\varphi$ from the group axioms, and we cannot prove $\lnot \varphi$ from the group axioms. But we can prove $\varphi\lor \lnot \varphi$. This just means that for every group $G$, either $G$ is abelian, or $G$ is not abelian.

answered Jun 09 '22 at 18:59

Alex Kruckman

76,357

Thanks a lot. It is exactly what I was looking for. Can you elaborate a little the second point? A model should be defined in such a way it must decide any "translated" statement? Maybe I would need a precise definition on "interpretation" to further discuss this.... – Eduard Jun 10 '22 at 05:58
@Eduard Yes, if this is not clear, you should learn more about models. The point is that the definition of $M\models \lnot \varphi$ ("$M$ satisfies $\lnot\varphi$") is $M\not\models\varphi$ ("$M$ does not satisfy $\varphi$"). So for any sentence $\varphi$, $M$ satisfies either $\varphi$ or $\lnot\varphi$. – Alex Kruckman Jun 10 '22 at 11:36
Building a model is defining a set and a mapping between the syntax side and the semantic side. But the semantic side as a set would contain undecidable sentences too, isn't it? – Eduard Jun 11 '22 at 15:21
In order to proof that something is a model of ZF, what should I do? I need to proof that P or Not P is true for any P I would be able to construct within the language? – Eduard Jun 11 '22 at 15:25
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@Eduard Those questions take us far enough afield that you should ask another question - or read up on models of set theory. – Alex Kruckman Jun 11 '22 at 16:02

Is it true that "Axiom of Choice is true or Axiom of Choice is false"?

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