-1
$\begingroup$

For sets $A,B$ we write $A\approx B$ if there is a bijection between $S$ sand $B$.

If $\kappa$ is a cardinal, let $\kappa^{<\kappa}$ denote the collection of subsets of $\kappa$ having cardinality $<\kappa$.

If $\kappa,\lambda$ are cardinals, does $\kappa^{<\kappa} \approx \lambda^{<\lambda}$ imply $\kappa=\lambda$?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
7
$\begingroup$

No.

Consider when $2^{\aleph_0}=2^{\aleph_1}=2^{\aleph_2}=\aleph_3$, with $\kappa=\aleph_1$ and $\lambda=\aleph_2$.

If you allow for one of these to be singular, then consider $\kappa=\beth_\omega$ and $\lambda=\kappa^+$. Then $\lambda^{<\lambda}=\lambda^\kappa=2^\kappa\cdot\lambda$, and on the other hand since $\kappa$ is a strong limit cardinal, $\kappa^{<\kappa}=2^\kappa$ as well.

If, on the other hand, you require that both are regular, then there is no provable counterexample, since $\sf GCH$ implies that $\kappa^{<\kappa}=\kappa$ for all regular cardinals.

Improve this answer
$\endgroup$
3
  • 2
    $\begingroup$ Are you assuming $\lnot$CH? $\endgroup$
    – tj_
    Commented Dec 29, 2021 at 16:44
  • 4
    $\begingroup$ It literally says $2^{\aleph_0}=\dots=\aleph_3$, so yes, this assumes that CH fails. $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 29, 2021 at 16:47
  • $\begingroup$ Yes, but how can you miss the opportunity of raising you hand and pointing it out with a shiny comment? $\endgroup$
    – Ivan Di Liberti
    Commented Dec 29, 2021 at 18:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .