Talk:Axiom of countable choice

Latest comment: 10 years ago by Tkuvho in topic countable axiom of choice

Notation

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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)Reply

Reverting

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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)Reply

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)Reply
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)Reply
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)Reply

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)Reply

Third sentence

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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)Reply

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
     
    where (μ n)φ(n) denotes the smallest n such that φ(n) holds.
  • To get from g to f, one can take
     
Arthur Rubin (talk) 22:27, 25 August 2011 (UTC)Reply
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)Reply

Formal statement

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

Let   be a non-empty set and   (where   denotes the power set of  ). Then there exists a function   such that  .

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

I'm afraid that notation is confusing, as   is arbitrary, and it's not obvious that it doesn't affect the results.. The formal notation would be. Let   be a function on   such that   is always a nonempty set. Then there is a function   on   such that  .
This may be less formal, but the codomain of   doesn't have to be exactly  , even if   is specified for some reason. — Arthur Rubin (talk) 22:17, 6 July 2012 (UTC)Reply
What results are you referring to?
And what do you mean by "exactly" in your statement "the codomain of   doesn't have to be exactly  "? 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)Reply
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)Reply
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  . If the codomain of A is taken to be a fixed set Z (with  ) , we can take the codomain of f to be  , but there may be reasons to select a different codomain. — Arthur Rubin (talk) 01:26, 28 August 2012 (UTC)Reply

countable axiom of choice

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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)Reply

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)Reply
Jech's article "About the axiom of choice" in the handbook here uses this term. Tkuvho (talk) 13:26, 9 January 2014 (UTC)Reply