Talk:Countable Union of Finite Sets is Countable

merge request
No. ACC for finite sets is a weaker choice principle than ACC.

ACC for finite sets is equivalent to the countable union condition for finite sets.

ACC is not equivalent to the countable union condition. --Dfeuer (talk) 06:59, 9 May 2013 (UTC)


 * This information is not much use in the context of a talk page. The site might benefit from the distinction being made on the pages in question. --prime mover (talk) 07:53, 9 May 2013 (UTC)

Definition of $h$
Regarding the current Explain tag: I'm unsure whether the question is rhetorical, so for the benefit of other readers I'll treat it as if it weren't. The meaning of this concatenation is evaluation of the function $q \left( {g_1 \left( {n} \right)} \right)$, which is itself a bijection from $|f \left({g_1 \left( {n} \right)}\right)|$ to $f \left({g_1 \left( {n} \right)}\right)$, at the point $g_2 \left({n}\right)$. By construction of $g$, $g_1$ is the index of a member $F$ of $\mathcal F$, and $g_2$ is the index of an element of $F$.


 * It wasn't rhetorical, I genuinely did not know what it meant. If possible, we need to either explain it (perhaps via a link to a page which explains the notation), because it is not obvious, or rewrite it in the same mathematical notation of mappings that has been documented already on.

I would add this explanation to the text, but I'll leave this to a user more experienced with ProofWiki's style guides. Jumpythehat (talk) 21:54, 2 October 2020 (UTC)


 * I think this would be clearer if we could perhaps abbreviate $q \left( {g_1 \left( {n} \right)} \right)$ by something like $q_{g_1 \left( {n} \right)}$, but again, I know the style guides are strict here and I'm aware this may not be allowed. Jumpythehat (talk) 22:01, 2 October 2020 (UTC)


 * As long as the notation is explained, then it is all good. But $q \left( {g_1 \left( {n} \right)} \right)$ means, to the unenlightened undergraduate used to conventional notation, the composition of functions: $q$ of $g_1$ of $n$. If this is not what it means, then we need to go some way towards either evolving new notation or rewriting it in something more standard. This sort of style: $q_{g_1 \left( {n} \right)}$ is not recommended purely because putting complex expressions into subscripts makes them difficult to read. --prime mover (talk) 23:13, 2 October 2020 (UTC)

Regarding the hypothesis $ZF^-$
Since its first version this article states that some equivalence holds in $ZF^-$.

I incautiously added yesterday —on behalf on some recent readings of Set theory, an introduction to independence proofs (Kenneth Kunen)— that $ZF^-$ is ZF minus the axiom of foundation. Alas, I have since checked another textbook (Set Theory, by Thomas Jech); this author defines $ZF^-$ as… ZF minus the power set axiom!

There is obviously few consensus in the literature about what $ZF^-$ should be; and I've removed my careless allegation. Since only the creator of the article knows what (s)he meant by $ZF^-$, a brusque suggestion is to replace $ZF^-$ by $ZF$. --ヒナゲシさん (talk) 04:05, 12 October 2021 (UTC)


 * The original submitter of this page is no longer contributing. While a fine mathematician, his approach did not coincide with the philosophical position of (he did not feel he was obligated to adhere to house style; he did not feel the need to explain his notation; he did not think it was necessary to provide links; he did not believe it was worth his while to add source citations) so a lot of his contributions contains material which remains unexplained.


 * I do not endorse the approach "replace $ZF^-$ by $ZF$" -- as it is likely the original contributor was deliberately excluding some axiom from $\mathrm{ZF}$.


 * A close study of this proof (not me, I have neither the patience nor the aptitude) may provide the insight into which of the axioms is omitted from $\mathrm{ZF}^-$ during the course of the exposition. It may well turn out to be exactly the definition as provided by Jech. --prime mover (talk) 05:24, 12 October 2021 (UTC)