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Nearly ten years ago, I explained in a blog post that, assuming only ZF, a cardinal number $\mathfrak{n}$ is finite if and only if it satisfies this monstrous inequality: $$2^{2^{2^{2^{\mathfrak{n}}}}} \lt \left(2^{2^{2^{2^{\mathfrak{n}}}}}\right)^2 = 4^{2^{2^{2^{\mathfrak{n}}}}}$$ Where "finite" is meant in the strictest Tarskian sense: in bijection with a finite ordinal.

My question is a bit vague since "simple" has a lot of interpretations but here it is:

Is there a simpler cardinal inequality that is equivalent to finiteness and uses only cardinal exponentiation?

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    $\begingroup$ Oh that's hideous, I love it! $\endgroup$
    – Noah Schweber
    Commented May 22 at 15:09
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    $\begingroup$ I suspect 4 could be replaced by 3 but even that requires some serious work. I expect the answer to be negative in some sense but I don't know what that "sense" is. $\endgroup$
    – François G. Dorais
    Commented May 22 at 15:32

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Läuchli proved in 1961 that $\mathfrak{n}$ is finite if and only if $2^{2^{\mathfrak{n}}}＜2^{2^{\mathfrak{n}}}\cdot2$. As a consequence, $\mathfrak{n}$ is finite if and only if $2^{2^{2^{\mathfrak{n}}}}＜(2^{2^{2^{\mathfrak{n}}}})^2$. It is still open (asked by Läuchli) whether $\mathfrak{n}$ is finite if and only if $2^{2^{\mathfrak{n}}}＜(2^{2^{\mathfrak{n}}})^2$.

Läuchli, H., Ein Beitrag zur Kardinalzahlarithmetik ohne Auswahlaxiom, Z. Math. Logik Grundlagen Math. 7, 141-145 (1961). ZBL0114.01005.

For an English translation of Läuchli's paper, see here.

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  • $\begingroup$ I'm not sure I follow the second iff here. Could you explain? $\endgroup$
    – François G. Dorais
    Commented May 22 at 18:19
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    $\begingroup$ @François The right hand side of the second iff implies the right hand side of the first iff. $\endgroup$
    – Guozhen Shen
    Commented May 22 at 18:33
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    $\begingroup$ I got lost in the stack of 2s... Thanks for pointing out Läuchli's question. $\endgroup$
    – François G. Dorais
    Commented May 22 at 23:42
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    $\begingroup$ On my homepage, there is an English translation of Läuchli's paper. $\endgroup$
    – Guozhen Shen
    Commented May 24 at 5:08
  • $\begingroup$ Thank you so much for this wonderful answer and your translation of Läuchli's paper! $\endgroup$
    – François G. Dorais
    Commented Jun 2 at 0:42

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