König's theorem and cofinality

Question

I am trying to prove that if $\kappa \geq \aleph_{0}$, then $\kappa < \kappa^{cf(\kappa)}$, where $cf(\kappa)$ is cofinality of $\kappa$.

Wikipedia is using König's theorem and proves that,

choosing a strictly increasing $cf(κ)$-sequence of ordinals approaching $κ$ and obtaining that each of them is less than $κ$, one notices that their sum (which is $κ$) is less than the product of $cf(κ)$ copies of $κ$.

But I don't get what exactly they mean by $cf(κ)$-sequence neither how the rest holds...

Answer 1

By definition, $\operatorname{cf}(\kappa)$ is the least cardinal $\lambda$ such that there exists an unbounded subset $S\subseteq\kappa$ of cardinality $\lambda$. It turns out that you can choose this $S$ to have order-type $\lambda$, so there is an increasing unbounded map $f:\lambda\to\kappa$ whose image is $f$; this is what is meant by an "increasing $\operatorname{cf}(\kappa)$-sequence of ordinals approaching $\kappa$". In any case, for this argument you don't actually need $S$ to have order-type $\lambda$. König's theorem tells us that $$\sum_{\alpha\in S}|\alpha|<\prod_{\alpha\in S}\kappa=\kappa^{|S|}=\kappa^\lambda$$ since $|\alpha|<\kappa$ for each $\alpha\in S$. To conclude $\kappa<\kappa^\lambda$, it thus suffices to show that $$\kappa\leq \sum_{\alpha\in S}|\alpha|.$$ For this you have to consider two cases. If $\kappa$ is a successor cardinal, then $\operatorname{cf}(\kappa)=\kappa$, so $|S|=\kappa$ so $\sum_{\alpha\in S}|\alpha|\geq\kappa$ since there are $\kappa$ terms in the sum (and at most one of them is $0$). If $\kappa$ is a limit cardinal, then since $S$ is unbounded in $\kappa$ the cardinalities of elements of $S$ must also be unbounded in $\kappa$. So we have $$\sum_{\alpha\in S}|\alpha|\geq\sup_{\alpha\in S}|\alpha|=\kappa.$$ Thus either way, we get that $$\kappa\leq\sum_{\alpha\in S}|\alpha|<\kappa^\lambda.$$ (Note that in fact, $\kappa=\sum_{\alpha\in S}|\alpha|$, since each term is less than $\kappa$ and there are $\lambda$ terms so the sum is at most $\lambda\cdot\kappa=\kappa$.)

Answer 2

The idea is to take a "short increasing sequence". How short? As short as possible: $\operatorname{cf}(\kappa)$-short. It is helpful to divide the case where $\kappa=\lambda^+$, in which case $\kappa$ is regular so $\kappa^{\operatorname{cf}(\kappa)}=\kappa^\kappa=2^\kappa>\kappa$, and the case where $\kappa$ is a limit cardinal, in which case let $\kappa_i$ for $i<\operatorname{cf}(\kappa)$ be our sequence. And we can assume that $\kappa_i$ is a cardinal. Now we can apply Koenig's theorem:

$$\kappa=\sum_{i<\operatorname{cf}(\kappa)}\kappa_i<\prod_{i<\operatorname{cf}(\kappa)}\kappa_{i+1}\leq\prod_{i<\operatorname{cf}(\kappa)}\kappa=\kappa^{\operatorname{cf}(\kappa)}.$$