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 $\kappa=\lambda$?



Does $|\kappa^{<\kappa}|=|\lambda^{<\lambda}|$ imply $\kappa=\lambda$?
No.


Consider when $2^{\aleph\_0}=2^{\aleph\_1}=2^{\aleph\_2}=\aleph\_3$, with $\kappa=\aleph\_1$ and $\lambda=\aleph\_2$.




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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.


