Exercises for Week 8

Given a family of sets $(A_i : i \in I)$ we define the sum (disjoint union) by \[\sum_{i \in I} A_i = \bigcup_{i\in I} (\{i\} \times A_i)\] and the product by \[\prod_{i\in I}A_i = \{ f : f \text{ is a function with domain $I$ and $f(i) \in A_i$ for all $i \in I$}\}.\]

1. Prove that $\left| \bigcup_{i \in I} A_i \right| \le \left| \sum_{i \in I} A_i \right|$.
2. 1. Find families of sets $(A_i : i \in I)$ and $(B_i : i \in I)$ such that $|A_i| < |B_i|$ for all $i\in I$, but the inequality $\left| \sum_{i \in I} A_i \right| < \left| \sum_{i \in I} B_i \right|$ fails.
   2. Find families of sets $(A_i : i \in I)$ and $(B_i : i \in I)$ such that $|A_i| < |B_i|$ for all $i\in I$, but the inequality $\left| \prod_{i \in I} A_i \right| < \left| \prod_{i \in I} B_i \right|$ fails.
3. Recall that the $\aleph$ sequence is defined recursively for all ordinals by $\aleph_0 = \omega$, $\aleph_{\alpha+1} = \aleph_\alpha^+$, and $\aleph_\lambda = \sup_{\alpha<\lambda} \aleph_\alpha$ if $\lambda$ is a limit ordinal. ($\aleph_\alpha$ is a synonym for $\omega_\alpha$.) We say that a cardinal $\kappa$ is weakly inaccessible if it is a limit cardinal ($\alpha^+ < \lambda$ for all $\alpha < \lambda$) and it is regular. Show that if $\kappa$ is a weakly inaccessible cardinal, then it is an $\aleph$ fixed point ($\aleph_\kappa = \kappa$.) Hint: consider the least $\alpha$ with $\aleph_\alpha \ge \kappa$.
4. The $\beth$ sequence is defined similarly to the $\aleph$ sequence by $\beth_0 = \omega$, $\beth_{\alpha+1} = |2^{\beth_\alpha}|$, and $\beth_\lambda = \sup_{\alpha<\lambda} \beth_\alpha$ if $\lambda$ is a limit ordinal. Prove that $\aleph_\alpha \le \beth_\alpha$ for all ordinals $\alpha$. (The hypothesis that they are equal for all $\alpha$ is known as the Generalized Continuum Hypothesis, and we can't prove or disprove this.)