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On Page 60, Set Theory Jech(2006)

(Show that)if $\kappa$ is regular and limit, then $\kappa^{<\kappa}=2^{<\kappa}$.

It's not difficult to show that $\kappa^{<\kappa}\geq2^{<\kappa}$. But I don't know how to show the other way around.

I rewrite $\kappa^{<\kappa}$ as $sup_{\lambda<\kappa} \{sup_{\alpha<\kappa}{\alpha^\lambda}\}$. It looks promising, if I can switch two $sup$ operators. But I'm afraid it's generally not legitimate which I'm not sure. I don't find the convergence issue here as in Real Analysis.

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Let $\lambda<\kappa$, and consider $2^\lambda$. Either $2^\lambda\le \kappa$, or not. In the latter case, $2^\lambda\le \kappa^\lambda\le (2^\lambda)^\lambda=2^\lambda$, so in fact $2^\lambda=\kappa^\lambda$.

There are now two cases: Either $2^\lambda\le\kappa$ for all $\lambda<\kappa$, or $2^\lambda>\kappa$ for all sufficiently large $\lambda<\kappa$, say $\lambda\ge\lambda_0$. In the second case, we have $\kappa^{<\kappa}=\sup_{\lambda_0\le \lambda<\kappa}\kappa^\lambda=\sup_{\lambda_0\le\lambda<\kappa}2^\lambda=2^{<\kappa}$.

In the first case, $2^{<\kappa}=\kappa$. We just need to check that then we also have $\kappa^{<\kappa}=\kappa$. But if $\rho<\kappa$, and $f:\rho\to\kappa$, then $f$ is bounded, by regularity of $\kappa$, so $f:\rho\to\tau$ for some cardinal $\tau<\kappa$. Let $\lambda=\max\{\rho,\tau\}$. Then $\tau^\rho\le\lambda^\lambda=2^\lambda\le\kappa$. We are done, because we just showed that $${}^{<\kappa}\kappa=\bigcup_{\tau<\kappa}\bigcup_{\rho<\kappa}{}^\rho\tau,$$ and the right hand side has size at most $\kappa\times\kappa\times\kappa$. Since the left hand side has size at least $\kappa$, we have that $\kappa^{<\kappa}=\kappa$, and we are done.

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Thank you very much. It's like a jigsaw puzzle. With the help of your answer, I pinpoint the part of picture I failed to asemble. –  Metta World Peace Dec 3 '12 at 7:29

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