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Famously, there are the integers, and the real numbers, and the first is countably infinite, and the latter is uncountably infinite. The continuum hypothesis says that there's no infinity with cardinality in between the integers and reals. My question is: what about an infinity with cardinality below the integers? Can that exist? Or is it provable that such a thing doesn't exist (i.e. anything with cardinality below the integers is provably finite).

chausies
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  • I think this is an eminently reasonable, honest question, which is not directly addressed by standard sources. But I myself, today, do not want to write about various notions of "infinite"... so I'll leave it to someone who has more enthusiasm... :) – paul garrett Mar 11 '25 at 19:45
  • Within ZFC it is provable that no infinite cardinality is smaller that $\aleph_0,$ but ZF without C may be another matter. – Michael Hardy Mar 11 '25 at 19:58
  • @lulu : The alleged duplicate to which you link is hardly an exact duplicate, since it contemplates the possibility of an infinite set whose cardinality is not comparable with $\aleph_0. \qquad $ – Michael Hardy Mar 11 '25 at 20:00
  • Without C, cardinals aren't necessarily comparable, as far as I know. –  Mar 11 '25 at 20:02
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    Without choice you might have non finite sets which are incomparable with $\aleph_0$ however, the statement that any set $S$ which injects in $\omega$ is either finite or $|S|=\aleph_0$ is very much provable without choice and holds even in quite weak fragments of $ZF$. – Giorgio Genovesi Mar 11 '25 at 20:25
  • @MoisheKohan I don’t think it’s the same as the referenced theorem requires choice. – Porky Mar 11 '25 at 21:58
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    In the linked answer the axiom of countable choice is used only to ensure existence of a sequence of $n$-element subsets $A_n\subset X$ since the author did not assume that the set $X$ is a subset of $\mathbb N$. – Moishe Kohan Mar 11 '25 at 22:08
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    Showing every infinite set contains a countably infnitie subset requires choice. Showing that a set that is smaller than the set of naturals must be finite does not. – Porky Mar 11 '25 at 22:39

1 Answers1

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Here's a proof that doesnt use the axiom of choice, but it's rather long winded. Is there a shorter one?

Note: I am assuming a set is finite if it is equinumerous with a natural, and infinite otherwise.

Suppose $A \prec \omega$. Suppose $A$ is infinite. Then there is an injection $f$ such that $f:A \rightarrow \omega$ and so $f$ is a bijection between $A$ and $f[A]$ and $A \approx f[A]$.

Define $g$ on $\mathcal{P}(f[A])$ by:

\begin{equation} g: X \mapsto \left \{ \begin{aligned} &X\backslash\{\underline X\}, && \text{if}\ X \text{ is not a singleton set and not the empty set} \\ &X &&\text{ if X is a singleton set or the empty set} \end{aligned} \right. \end{equation}

Where $\underline X$ is the least element of $X$ (which exists as it is a set of naturals).

Then by the recursion theorem there is a $h$ on $\omega$ such that $h(\emptyset)=f[A]$ and $h(n+1)=g(h(n))$.

Then it's an easy induction to show $h(n)$ is infinite for all $n \in \omega$ (we are making use of the result that an infinte set less a finite set is infinite, which does not require choice).

Now let $r$ be a function on $\omega$ that maps $n$ the least element of $h(n)$, which exists as $h(n)$ is a non-empty set of naturals. Then $r$ is increasing and so is 1-1. And as $r:\omega \rightarrow f[A]$, $\omega \preceq f[A]$. And so $\omega \prec \omega$, a contradiction. Hence $A$ is finite.

Porky
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