Under what assumptions on an infinite cardinal $\kappa$ we have $$\kappa^\kappa= 2^\kappa?$$
Please delete this question. I know the answer.
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3Once posting a question, it might be useful for future visitors. If you know the answer it is fine to post an answer on your own. – Asaf Karagila Apr 21 '12 at 15:46
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1I'm pretty sure this is a repeat, and I know I've posted an answer. In fact, if $\lambda$ satisfies $2\leq \lambda \leq 2^{\kappa}$, then $\lambda^{\kappa}=2^{\kappa}$ by the same argument Asaf uses below. – Arturo Magidin Apr 21 '12 at 22:07
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These answers are about $\lambda^\kappa=2^\kappa$ for $2\le\lambda\le (\kappa^+)$: Cardinal equalities: $\aleph_0^\mathfrak c=2^\mathfrak c$, Is $\aleph_0^{\aleph_0}$ smaller than or equal to $2^{\aleph_0}$?. @ArturoMagidin Maybe you meant one of those answers when you wrote that you posted an answer to this before? – Martin Sleziak Jul 10 '19 at 06:50
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Assuming $\kappa$ is an ordinal the answer is always.
The reason is simple: by Cantor's theorem we have $2<\kappa<2^\kappa$, therefore using exponentiation laws: $$2^\kappa\le\kappa^\kappa\le\left(2^\kappa\right)^\kappa=2^{\kappa\cdot\kappa}=2^\kappa$$
Asaf Karagila
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