 Solution

For an infinite cardinal regarded as an ordinal, its cofinality $\theta =\mathrm{cf}(\kappa )$ 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 sethas 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). ConsequentlyThis 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.