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Asaf Karagila
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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.

No.

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

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.

Source Link
Asaf Karagila
  • 39.8k
  • 8
  • 135
  • 283

No.

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