If $\lambda$ is a limit cardinal then $\kappa^\lambda = (\kappa^{<\lambda})^{cf \lambda}$ for each cardinal $\kappa$.

Question

I'm reading Thomas Jech's "Set Theory" and there is a theorem about properties of continuum function and cardinals:

Theorem 5.16:

1. If $\kappa < \lambda$ then $2^\kappa \leq 2^\lambda$.

2. $\operatorname{cf} 2^\kappa > \kappa$.

3. If $\kappa$ is a limit cardinal then $2^\kappa = (2^{< \kappa})^{\operatorname{cf} \kappa}$.

The proof of (3) is following: Let $\kappa = \sum_{i < \operatorname{cf} \kappa} \kappa_i$ where $\kappa_i < \kappa$. Then
$$2^\kappa = 2^{\sum_{i < \operatorname{cf} \kappa} \kappa_i} = \prod_{i < \operatorname{cf} \kappa} 2^{\kappa_i} \leq \prod_{i < \operatorname{cf} \kappa} 2^{< \kappa} = (2^{< \kappa})^{\operatorname{cf} \kappa} \leq (2^\kappa)^{\operatorname{cf} \kappa} = 2^\kappa$$

The proof relies only on properties $\kappa < \lambda$ implies $2^\kappa \leq 2^\lambda$, $\kappa < \lambda$ implies $\kappa^\mu \leq \lambda^\mu$, and connections between infinite sums and products of cardinals. But all such properties are still satisfied if we replace $2$ by some arbitrary cardinal $\kappa$. So, following same proof, we can deduce a stronger proposition: If $\lambda$ is a limit cardinal then $\kappa^\lambda = (\kappa^{<\lambda})^{cf \lambda}$ for each cardinal $\kappa$.

Am I missing something? I was given such problem as an exercise, and it seems so simple that I started to doubt if I'm getting everything right.

Thanks!

Answer 1

No, you aren't missing anything. (And note also that $\lambda$ doesn't even need to be a limit cardinal for the formula to hold, just an infinite cardinal, although the case where it is a successor cardinal is less interesting.)