# Talk:Axiom of countable choice

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## Notation

Is this ACω or ACω. I'd use the latter, noting that it's related to certain subsets of Vω being non-empty. — Arthur Rubin | (talk) 23:45, 24 October 2007 (UTC)

## Reverting

My edit which spelled out the definition, added a reference and a couple of interesting facts, was reverted without comment. It would probably be more productive to discuss this first here. AxelBoldt (talk) 03:29, 14 November 2008 (UTC)

It would have been more productive, had the definition been even approximately correct, and the "facts" been interesting. — Arthur Rubin (talk) 03:34, 14 November 2008 (UTC)
Could you elaborate what you believe was wrong with the definition? Also, your scare quotes around "facts" seem to suggest that you doubt that the sequence characterization of accumulation points is equivalent to ACω. Is that so? Granted, whether this is an interesting fact or not is a matter of taste, but deleting that information doesn't help the readers in any way -- those who agree can ignore it easy enough. Cheers, AxelBoldt (talk) 16:12, 14 November 2008 (UTC)
I think I fixed the definition, if you want to be specific. I can see the explanations as possibly appropriate in another section, but I think the lead should probably be limited to a simple description and the definition. And I would ask you to add my late mother's book, Consequences of the Axiom of Choice (ISBN 978-0821809778 with updates at http://www.math.purdue.edu/~jer/cgi-bin/conseq.html and http://consequences.emich.edu/conseq.htm) to the references as a pointer to other equivalents of ACω, and possibly a pointer to the proof that ACω does not imply DC. I think I may have an actual conflict of interest, as the book still produces a small royalty stream, which I'm likely to inherit. — Arthur Rubin (talk) 17:02, 14 November 2008 (UTC)

Yes, the article needs a couple more references for sure. As to the explanation of the axiom, I think we can agree that under ZF, the two statements

if A is a function with domain N and A(n) is a non-empty set for every n∈N, then there exists a function f with domain N such that f(n) ∈ A(n) for every n∈N.

and

if B is some set and Bn is a non-empty subset of B for every n∈N, then there exists a function f : N → B with f(n) ∈ Bn for every n∈N.

are equivalent (with Bn=A(n) and B = ∪A(n)). While I like my version better (because this is how the axiom is used in practice, and the cases B = R and B = N yield interesting weaker axioms) I don't care enough to change it back. AxelBoldt (talk) 23:38, 14 November 2008 (UTC)

## Third sentence

Currently the third sentence of the article is:

Spelled out, this means that if A is a function with domain N (where N denotes the set of natural numbers) and A(n) is a non-empty set for every n ∈ N, then there exists a function f with domain N such that f(n) ∈ A(n) for every n ∈ N.

This is not a special case of the axiom of choice as usually formulated. For it to be that, it would have to be revised to:

Spelled out, this means that if A is a function with domain N (where N denotes the set of natural numbers) and A(n) is a non-empty set for every n ∈ N, then there exists a function g whose domain is the range of A such that g(A(n)) ∈ A(n) for every n ∈ N.

Right? JRSpriggs (talk) 21:46, 25 August 2011 (UTC)

It doesn't strike me as significantly different, and the f formulation is easier to state.
• To get from f to g, one can take
${\displaystyle g(X)=f((\mu n)(A(n)=X)),}$
where (μ n)φ(n) denotes the smallest n such that φ(n) holds.
• To get from g to f, one can take
${\displaystyle f(n)=g(A(n)).}$
Arthur Rubin (talk) 22:27, 25 August 2011 (UTC)
Yes. I hesitated to mention this since the formulations are obviously interconvertible. However, g satisfies our definition of a choice function while f does not. JRSpriggs (talk) 02:40, 26 August 2011 (UTC)

## Formal statement

It would be maybe helpful to some if the axiom were stated also formally:

Let ${\displaystyle X}$ be a non-empty set and ${\displaystyle A:\mathbb {N} \rightarrow {\mathcal {P}}(X)\setminus \{\emptyset \}}$ (where ${\displaystyle {\mathcal {P}}(X)}$ denotes the power set of ${\displaystyle X}$). Then there exists a function ${\displaystyle f:\mathbb {N} \rightarrow X}$ such that ${\displaystyle f(n)\in A(n)}$.

Howeworth (talk) 21:40, 6 July 2012 (UTC)

I'm afraid that notation is confusing, as ${\displaystyle X}$ is arbitrary, and it's not obvious that it doesn't affect the results.. The formal notation would be. Let ${\displaystyle A}$ be a function on ${\displaystyle \mathbb {N} }$ such that ${\displaystyle A(n)}$ is always a nonempty set. Then there is a function ${\displaystyle f}$ on ${\displaystyle \mathbb {N} }$ such that ${\displaystyle (\forall n\in \mathbb {N} )(f(n)\in A(n))}$.
This may be less formal, but the codomain of ${\displaystyle A}$ doesn't have to be exactly ${\displaystyle {\mathcal {P}}(X)\setminus \{\emptyset \}}$, even if ${\displaystyle X}$ is specified for some reason. — Arthur Rubin (talk) 22:17, 6 July 2012 (UTC)
What results are you referring to?
And what do you mean by "exactly" in your statement "the codomain of ${\displaystyle A}$ doesn't have to be exactly ${\displaystyle {\mathcal {P}}(X)\setminus \{\emptyset \}}$"? A set either is the codomain of a function or it isn't. There's no level of exactness. Howeworth (talk) 16:19, 27 August 2012 (UTC)
If A is injective, then X is infinite. If X is infinite, then A cannot be surjective, that is, the image of A is a proper subset of P(X)-{0}. JRSpriggs (talk) 00:40, 28 August 2012 (UTC)
You're assuming the "functions" here are the 3-component function, rather than the 1-component function. That does not go without saying.
Even if the function (A) had a well-defined codomain, we don't know that it is of the form ${\displaystyle {\mathcal {P}}(X)\setminus \{\emptyset \}}$. If the codomain of A is taken to be a fixed set Z (with ${\displaystyle \emptyset \notin Z}$) , we can take the codomain of f to be ${\displaystyle \cup Z}$, but there may be reasons to select a different codomain. — Arthur Rubin (talk) 01:26, 28 August 2012 (UTC)

## countable axiom of choice

The axiom discussed on this page is often referred to informally as the "countable axiom of choice". Should we make a note of that? I was thinking of creating a redirect at least. Tkuvho (talk) 10:34, 9 January 2014 (UTC)

I've created a redirect, since it's a plausible search term. Probably best not to add it to the article without a reference, however. --Zundark (talk) 12:50, 9 January 2014 (UTC)
Jech's article "About the axiom of choice" in the handbook here uses this term. Tkuvho (talk) 13:26, 9 January 2014 (UTC)