= Solution
For an infinite <cardinal> regarded as an <ordinal>, its <cofinality> $\theta=\operatorname{cf}(\kappa)$ is the least <cardinality> of a <cofinal subset> of $\kappa$. Equivalently there is a family $(\alpha_i)_{i<\theta}$ of <ordinals> below $\kappa$ whose <supremum> is $\kappa$, and no smaller indexed family has that property. Since an infinite <initial ordinal> is a <limit ordinal>, $\theta$ is infinite and in particular nonzero.
Use the diagonal mechanism of <König theorem for cardinal numbers> directly. Suppose $F:\kappa\to{}^{\theta}\kappa$ were <surjective>. For each $i<\theta$, the <set>
$$
E_i=\{F(\alpha)(i):\alpha<\alpha_i\}
$$
has <cardinality> at most $|\alpha_i|<\kappa$. Choose the least $b_i\in\kappa\setminus E_i$. The resulting <function> $b:\theta\to\kappa$ is not in the range: for any $\alpha<\kappa$, <cofinality> supplies $i$ with $\alpha<\alpha_i$, and then $b_i\ne F(\alpha)(i)$.
Thus no such surjection exists. Constant <functions> provide an <injection> $\kappa\to{}^{\theta}\kappa$, and an <injection> in the reverse direction would give a surjection by the same inverse-and-default construction as in part (i). Consequently
$$
\boxed{\kappa^{\operatorname{cf}(\kappa)}>\kappa.}
$$
This argument also handles $\kappa=\aleph_0$, when <cofinal> initial segments are finite; it does not incorrectly assume that smaller <cardinals> themselves must form a <cofinal> sequence in every infinite <cardinal>.
Back to article page
