Why is $cofinality (\kappa^+) = \kappa^+$ for infinite cardinals? While studying introductory set theory I found this property of cofinality:
$cofinality (\kappa^+) = \kappa^+$ where $\kappa^+ $ is the successor of $\kappa$
I can't seem to understand why, nor find proof for it.
 A: First thing to note is that this result requires the axiom of choice. Without choice there can be singular successor cardinals.
Suppose that $\kappa$ is any cardinal. To show that $\text{cof}(\kappa^+) = \kappa^{+}$, it will clearly suffice to show that $\text{cof}(\kappa^+) > \kappa$. To this end, let $I$ be a set with $|I| \leq \kappa$, and for each $i \in I$, let $S_i$ be a set with $|S_i| \leq \kappa$. Then we have the following chain of inequalities:
$$ \left | \bigcup_{i \in I} S_i \right | \leq \left |  \coprod_{i \in I} \kappa\right | \leq  \left |  \coprod_{i \in \kappa} \kappa\right | = |\kappa \times \kappa| = \kappa < \kappa^+ $$
The first inequality follows by consider the “worst case” cardinality for each $S_i$, with special care taken by making the union disjoint. The second inequality follows from taking the “worst case” cardinality for $I$. The third equality is intuitively easy to justify, but a good to justify by hand. The fourth equality, that $|\kappa \times \kappa| = \kappa$, requires the axiom of choice.
Since the cardinality of this set is less than $\kappa^+$, the result follows.
