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For a set $X$, the Lindenbaum number of $X$, $\aleph^*(X)$, is the least non-zero ordinal $\alpha$ such that there is no surjection $X\to\alpha$. It seems to be well-known that for infinite sets $X$ and $Y$,

$$\aleph^*(X)\times\aleph^*(Y)\leq\aleph^*(X\times Y)\leq\aleph^*(X)^+\times\aleph^*(Y)^+.$$

Is this a folklore result? If not, where in the literature can this be found?

It doesn't appear to be in Tarski-Lindenbaum Communication sur les recherches de la th'eorie des ensembles.

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  • $\begingroup$ I'd hazard a guess that this wasn't really known, or at least never explicitly stated. I certainly didn't know that before Elliot Glazer told me about it. $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 5 at 18:58
  • $\begingroup$ I think I saw the first of these inequalities as an (unattributed) exercise in a set theory class, but I'm not sure. I don't think I've seen the second inequality explicitly. I suspect these are folklore, sadly. $\endgroup$
    – Noah Schweber
    Commented Dec 5 at 19:18
  • $\begingroup$ @Noah: I mean, the first one is truly trivial. Picking two maps. We can easily lift them to maps from the product... $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 5 at 20:38
  • $\begingroup$ See my class slides: drive.google.com/file/d/1XGTzhPHYCYpmOlWzl7mUOpoD2OTwG37y/… $\endgroup$
    – Guozhen Shen
    Commented Dec 6 at 5:10
  • $\begingroup$ I'd call this fact "near-productivity of Lindenbaum numbers," while non-productivity refers to ZF models in which this inequality is sharp (see here for an example mathoverflow.net/a/456549/109573). $\endgroup$
    – Elliot Glazer
    Commented Dec 6 at 21:29

1 Answer 1

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Karl-Heinz Diener proved in On the transitive hull of a κ‐narrow relation that for all class relations $R$, if $R$ is $\kappa$-narrow in the sense that $\aleph^\ast(R^{-1}[\{x\}])\leqslant\kappa$ for all sets $x$, then the transitive closure $R^\ast$ of $R$ is $\kappa^+$-narrow.

The result you mentioned is a corollary of this result.

It can also be proved in $\mathsf{ZF}$ that $\aleph^\ast(\kappa\times x)=\kappa^+\cdot\aleph^\ast(x)$. See Lemma 3.6 of A generalized Cantor theorem in ZF for a proof of the case $\kappa=\omega$. The general case is proved by Peng and is not published yet.

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    $\begingroup$ This should also give the more general fact that $\aleph^*(X^{<\omega}) \le \aleph^*(X)^+.$ $\endgroup$
    – Elliot Glazer
    Commented Dec 6 at 21:33

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