Talk:Infinite Set has Countably Infinite Subset/Proof 2

The "intuitive proof" cannot be refined into a formal proof. There should be an explicit reference to needing at least the Axiom of Countable Choice, even if actual proof using it is not provided.


 * Feel free to set yourself up a user name and start contributing properly. You'd be more than welcome. --prime mover 14:12, 20 June 2011 (CDT)

ok Done, but see note on my user page. Arthur 07:15, 22 June 2011 (CDT)

The "formal proof" currently added to the "intuitive proof" is clearly wrong. It could (and should, very usefully) be used in ProofWiki to (classically) illustrate the very common fallacy explained in the Wikipedia entry on Axiom of Countable Choice.

That entry also states the axiom required, explains that this axiom can be proved from ZFC but not from ZF and gives a correct formal proof that infinite set has countable subset using that axiom.

I recommend that author of the current entry adapt it to use for a separate entry illustrating the common fallacy at the same time as rewriting the correct proof here. Nobody would have a better insight into explaining the fallacy than a previous "believer" ;-) Arthur 08:17, 22 June 2011 (CDT)


 * Hi. I might be wrong, but I think the proof can be made correct by using explicitly the Recursion Theorem. There is not a proof here on proofwiki of the recursion theorem (I shall create it before correcting this one article), but the theorem says: If $X$ is any set, $\varphi:X\to X$ is a function, and $a\in X$, then there exists a function $f:\N\to X$ such that $f\left(1\right)=a$ and $f\left(n+1\right)=\varphi\left(f\left(n\right)\right)$. I don't think the theorem "Infinite Set has Countable Subset" requires the Axiom of Countable Choice, but this is just pure intuition. If anyone has a reference that says that it does, please share it :) --Pedro Angelo 00:22, 24 June 2011 (CDT)


 * The link provided in previous paragraph does provide the reference you want. As stated above it explicitly mentions that you cannot prove from just ZF, hence not from the recursion theorem of ZF. A .pdf file linked from that page mentions that an even weaker axiom than countable choice could be used, namely that finite and Dedekind finite are equivalent (with a citation to Jensch as footnote 19). The fact that it is weaker means there is provably no way to prove this theorem without at least the Axiom of Countable Choice. --Arthur 01:34, 24 June 2011 (CDT)


 * There are already some posts on here abut recursion. Can't see them at the moment, but you might want to do a quick search. --prime mover 00:24, 24 June 2011 (CDT)
 * ... Aha - here it is: Principle of Recursive Definition, but it's in the context of a Naturally Ordered Semigroup, an abstract-algebraic construct used during the process of defining the natural numbers and proving their properties from the ZF axioms. So while it can be proved for natural numbers, we can not then use that proof in the course of proving results about the natural numbers.


 * However, as it is later demonstrated that the Natural Numbers are a Naturally Ordered Semigroup, this ends up not actually being a problem, and the Principle of Recursive Definition can indeed be applied to the Natural Numbers. --prime mover 00:36, 24 June 2011 (CDT)


 * Hmmmm, good :) I will study these articles, and my books, because I have actually never attempted to understand the Recursion Theorem's proof. Then, I will see what I can do. If, in the meantime, anyone wants to rewrite he article, please go on. --Pedro Angelo 01:04, 24 June 2011 (CDT)


 * Please read the Wikipedia link for Axiom of countable choice provided at top of this page very carefully. Recursion will only prove an "Axiom of Finite Choice". One can choose from a set of size 1, if one can choose from a set of size n then one can choose from a set of size n + 1. Therefore one can choose from a set of size any finite size n no matter how big. Your handwaving fallacy is to assume that you can choose from a countably infinite set. As explained at the link above, for that you need an Axiom of Countable Choice and a different proof (which is provided at that link).--Arthur 01:34, 24 June 2011 (CDT)

Jumping in here ... I would be sceptical about quoting proofs about infinite sets etc. from a book on real analysis. They often don't treat this deep set-theoretical stuff in a rigorous wey because their focus is on the structure of the real (or complex etc.) numbers. Therefore the set-theory results are often taken as by-and-large "axiomatic" - their application to measure theory and complete spaces are what is important. For a proper discussion on axiomatics of infinite sets, go to a book on set theory - but be prepared for some hard work before you are able to understand it. --213.123.213.185 03:17, 24 June 2011 (CDT)

Hi. I took my time to study at least the basics of set theory from books that focus on se theory itself, and now I understand the point you all were making. To be honest, I'm still not veeery sure that Infinite Set has Countably Infinite Subset is undecidable in ZF alone, but I'm pretty convinced that it is far from being trivial in ZF. Anyway, I'm not removing the Handwaving and Questionable templates myself, but I would like someone to proofread it, and if it is the case, take them off. Thanks --Pedro Angelo 04:56, 15 August 2011 (CDT)


 * I've tidied up a bit (i.e. added line breaks between thoughts - please familiarise yourself with the house style) but I still think that it still only copes with finite sets. It's subtle. Just because you've proved it for all finite $n$ you can't then just say that it applies to the smallest infinite cardinal. (I know enough about set theory to have learned this, but I need to take it a lot further before I know what I'm talking about.) I plan on coming back this way again before the end of the universe, I'll give it a going over then. --prime mover 13:45, 15 August 2011 (CDT)


 * Hmmm, but I think that everything I want to prove is indeed about finite naturals. The domain of the function $f$ is $\N$, all of whose elements are finite naturals. When I use the expression $f\left(\N\right)$ in the text, I mean the set $\left\{f\left(x\right) : x\in\N \right\}$, that is, I'm regarding $\N$ as a (improper) subset of the domain of $f$, and not as an element of the ordinal $\N\cup\left\{\N\right\} = \omega + 1$. --Pedro Angelo 20:14, 15 August 2011 (CDT)


 * Just to make it clear, I do know the difference between proving something for all finite $n$, that is, for all $n\in\N$, and proving it for all $n$ less than a fixed $k$. All I want is to show that one can define the function $f$ (and for that, I need the axiom of choice), and then that $f$ is injective. In order to prove these two assertions, everything I need to show concerns just finite naturals, because the domain of $f$ is a collection of (all) finite naturals. --Pedro Angelo 20:14, 15 August 2011 (CDT)

Every time I come across this page I feel that the argument used is actually rigorous. Therefore I plead for removing the handwavery and non-validity templates. The proofread can then be dealt with later. --Lord_Farin 17:05, 3 December 2011 (CST)


 * I'm coming round to agreeing. Feel free to do so. Okay, so if someone then decides it's inadequate in any way, I would encourage them to fix it themselves rather than just saying "it's wrong, go away and learn how to fix it" as there's a finite number of microseconds in a millennium here and (speaking for myself) I have other stuff to do. --prime mover 17:18, 3 December 2011 (CST)