Solutions to selected exercises for Week 5

4. For a family of sets $(X_i : i \in I)$ we define the sum (or "disjoint union") $\sum_{i \in I} X_i = \{(i,x) : i \in I \text{ and } x \in X_i\}$. Prove that the following statements are equivalent for any infinite cardinal $\kappa$.
   1. $\kappa$ is singular.
   2. $\kappa = \bigcup_{i \in I} X_i$ for some family of sets $(X_i : i \in I)$ with $|I| \lt \kappa$ and each $|X_i| \lt \kappa$.
   3. $\kappa = \left| \sum_{i \in I} X_i \right|$ for some family of sets $(X_i : i \in I)$ with $|I| \lt \kappa$ and each $|X_i| \lt \kappa$.

   Solution:

   (a) $\implies$ (b): Let $I = \text{cf}(\kappa)$ and let $f : I \to \kappa$ be a cofinal function. Let $X_i = f(i)$ for $i \in I$. Then the family $(X_i : i \in I)$ has the desired property.

   (b) $\implies$ (c) Let $(X_i : i \in I)$ be a family as in (b). We always have $\left|\bigcup_{i \in I} X_i\right| \le \left| \sum_{i \in I} X_i \right|$ because there is a surjection from the sum to the union. Therefore $\kappa \le \left| \sum_{i \in I} X_i \right|$. On the other hand, $\left| \sum_{i \in I} X_i \right| \le \left| \sum_{i \in I} \kappa \right| = |I \times \kappa| \le |\kappa \times \kappa| = \kappa$.

   (c) $\implies$ (a) Let $(X_i : i \in I)$ be a family as in (c). Let $f : \sum_{i \in I} X_i \to \kappa$ be a surjection. If $f \restriction (\{i\} \times X_i)$ is cofinal in $\kappa$ for some $i \in I$ then $\kappa$ is singular, because $|\{i\} \times X_i| = |X_i| \lt \kappa$. If not, then $f \restriction (\{i\} \times X_i)$ is bounded in $\kappa$ for all $i \in I$. Then define $g(i) = \sup \text{ran}(f \restriction (\{i\} \times X_i)) \lt \kappa$. The function $g:I \to \kappa$ is cofinal because $f$ is a surjection, so again $\kappa$ is singular.