# Weak cardinal powers and singular cardinals

Suppose $\kappa > \operatorname{cf}(\kappa)$. Show that:

i) if $\kappa$ strong limit then $\kappa^{<\kappa} = \kappa^{\operatorname{cf}(\kappa)}$

ii) if $\kappa$ not strong limit then $2^{<\kappa} = \kappa^{<\kappa} > \kappa$.

My thoughts:

for i), write $\kappa = \sum_{i < \operatorname{cf}(\kappa)} \kappa_i$ with $\kappa_i < \kappa$ for each i. Then, $\kappa^{<\kappa} = \sum_{\lambda < \kappa} \kappa^{\lambda} = \sum_{i<\operatorname{cf}(\kappa)}\kappa^{\kappa_i}$. Now, we know that $\kappa^{\lambda} \leq (\sum_{\alpha < \kappa} \alpha^{\lambda})^{\operatorname{cf}(\kappa)}$, so $\sum_{i<\operatorname{cf}(\kappa)}\kappa^{\kappa_i} \leq \sum_{i<\operatorname{cf}(\kappa)}(\sum_{\alpha < \kappa} \alpha^{\kappa_i})^{\operatorname{cf}(\kappa)}$, but I'm not sure what to do now.

for ii) I guess an argument similar to the above shows that $\kappa^{<\kappa}= \kappa^{\operatorname{cf}(\kappa)} > \kappa$, but I'm stuck on showing $2^{<\kappa} = \kappa^{<\kappa}$. How does merely knowing the fact that for some $\mu < \kappa$ $2^{\mu} \geq \kappa$ let us deduce the equivalence?

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For (ii), if $\lambda>\mu$, then $2^\lambda=(2^\mu)^\lambda\geq\kappa^\lambda$, so $2^{<\kappa}\geq\kappa^{<\kappa}$. –  Miha Habič Jan 4 '13 at 19:19

For (i), notice that you have $\alpha^{\kappa_i}<\kappa$ in the notation of your last inequality. This is because $\kappa$ is a strong limit, so $\alpha^{\kappa_i}\leq (2^\alpha)^{\kappa_i}<\kappa$. But then, following your inequality, $$\kappa^\lambda\leq \sum_{i<\mathrm{cf}(\kappa)}\left(\sum_{\alpha<\kappa}\alpha^{\kappa_i}\right)^{\mathrm{cf}(\kappa)} \leq \sum_{i<\mathrm{cf}(\kappa)}\kappa^{\mathrm{cf}(\kappa)} = \kappa^{\mathrm{cf}(\kappa)}$$ Sending $\lambda$ to $\kappa$ we get $\kappa^{<\kappa}\leq \kappa^{\mathrm{cf}(\kappa)}$ and the reverse inequality is obvious.

For (ii), Brian and I have already argued why $2^{<\kappa}=\kappa^{<\kappa}$. Of course, since $\kappa$ is assumed to not be a strong limit, $2^{<\kappa}>\kappa$ must clearly hold.

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You beat me to it. :-) –  Brian M. Scott Jan 4 '13 at 20:01
@BrianM.Scott It's good to flex the old cardinal arithmetic muscle now and again. –  Miha Habič Jan 4 '13 at 20:05
@MihaHabič: Sorry, I might be being obtuse, but I don't see how we have $\alpha^{\kappa_i} < \kappa$. Your argument seems to only work if we already assume that $\alpha^{\kappa_i} < \kappa$, as strong limit only says that if $\lambda < \kappa$ then $2^{\lambda} < \kappa$, so $\alpha^{\kappa_i} < 2^{\alpha^{\kappa_i}}$, but we can't conclude $2^{\alpha^{\kappa_i}} < \kappa$ immediately –  Kris Jan 4 '13 at 20:27
@Kris The estimate is $\alpha^{\kappa_i}\leq (2^{\alpha})^{\kappa_i}<\kappa$, using the fact that $\kappa$ is a strong limit in the second step. There is just the one layer of exponents, not two as in your comment. –  Miha Habič Jan 4 '13 at 21:06
@MihaHabič: oh, I see, sorry, I misread your comment. Thanks! –  Kris Jan 4 '13 at 21:48

(ii) If $\kappa$ is not a strong limit, then $\kappa\le 2^\lambda$ for some $\lambda<\kappa$. Thus, if $\lambda\le\mu<\kappa$, then

$$2^\mu=\left(2^\lambda\right)^\mu\ge\kappa^\mu\ge 2^\mu\;,$$

and it follows that $2^{<\kappa}=\kappa^{<\kappa}$.

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