If $ \kappa $ is strongly inaccessible, then $ \kappa^{<\kappa} = \kappa $

Question

I'm trying to show the following about a cardinal $ \kappa $

If $ \kappa $ is strongly inaccessible, then $ \kappa^{<\kappa} = \kappa $

Strongly inaccessible means that $ \kappa $ is uncountable, limit regular and for all $ \lambda < \kappa $ it holds that $ 2^\lambda < \kappa $

I think that it is obvious that $ \kappa^{<\kappa} \geq \kappa $ for infinite cardinals, so I'm going to pursue the other inequality.

$ \kappa^{<\kappa} = \sum\limits_{\lambda < \kappa}|[\kappa]^\lambda| = \sum\limits_{\lambda < \kappa}\kappa^{\lambda} = \sup\limits_{\lambda < \kappa}\kappa^{\lambda}\cdot\kappa$

So it looks like it's enough to show that $ \rho = \sup\limits_{\lambda < \kappa}\kappa^{\lambda} \leq \kappa $ and for this it would seem appropriate to use the fact that $ 2^{\lambda} < \kappa$ - but I can't figure out how

I would appreciate some help

Answer 1

We want to count the set of maps from $\lambda$ to $\kappa$ (specifically, we want to show that this set has size $\kappa$).

First, let's count the bounded maps: that is, the set of $f:\lambda\rightarrow \kappa$ such that $im(f)\subseteq\alpha<\kappa$ for some $\alpha$. It's a good exercise to show that a map from $\lambda$ into $\alpha$ can be identified with a subset of $\mu=\lambda\times\alpha$, which itself is $<\kappa$; since $\mu<\kappa$, we have $2^\mu<\kappa$, so for each $\lambda, \alpha<\kappa$ there are fewer than $\kappa$-many maps from $\lambda$ to $\alpha$. So the total number of bounded maps from an ordinal $<\kappa$ into $\kappa$ is $\kappa$.

But what about the unbounded maps? HINT: $\kappa$ is regular . . .

Answer 2

If $\lambda<\kappa$, every function from $\lambda$ to $\kappa$ has bounded range. For each cardinal $\mu$ such that $\lambda\le\mu<\kappa$ we have

$$\left|{^\lambda\mu}\right|=\mu^\lambda\le\mu^\mu=2^\mu<\kappa\;,$$

so

$$\left|{^\lambda\kappa}\right|\le\sum_{\lambda\le\mu=|\mu|<\kappa}\left|{^\lambda\mu}\right|<\kappa\cdot\kappa=\kappa\;.$$