A solutions manual for Set Theory by Thomas Jech
5. The Axiom of Choice and Cardinal Arithmetic
5.1. There exists a set of reals of cardinality \(2^\aleph_0\) without a perfect subset.
\(\quad\)[Let \(\langle P_\alpha :\alpha < 2^{\aleph_0}\rangle\) be an enumeration of all perfect sets of reals. Construct disjoint \(A=\{a_\alpha :\alpha <2\aleph_0\}\) and \(B=\{b_\alpha :\alpha <2^\aleph_0\}\) as follows: Pick \(a_\alpha\) such that \(a_\alpha\notin \{a_\xi :\xi <\alpha \}\cup \{b_\xi :\xi <\alpha \}\), and \(b_\alpha\) such that \(b_\alpha\in P_\alpha -\{a_\xi :\xi\le\alpha \}\). Then \(A\) is the set.]
5.2. If \(X\) is an infinite set and \(S\) is the set of all finite subsets of \(X\), then \(|S| = |X|\).
\(\quad\)[Use \(|X| =\aleph_\alpha\).]
5.3. Let \((P, <)\) be a linear ordering and let \(\kappa\) be a cardinal. If every initial segment of \(P\) has cardinality \(< \kappa\), then \(|P |\le \kappa\).
5.4. If \(A\) can be well-ordered then \(P(A)\) can be linearly ordered.
\(\quad\)[Let \(X < Y\) if the least element of \(X\bigtriangleup Y\) belongs to \(X\).]
5.5. Prove the Axiom of Choice from Zorn’s Lemma.
\(\quad\)[Let \(S\) be a family of nonempty sets. To find a choice function on \(S\), let \(P = \{f : f\) is a choice function on some \(Z\subset S\}\), and apply Zorn’s Lemma to the partially ordered set \((P,\subset)\).]
5.6. The countable AC implies that every infinite set has a countable subset.
\(\quad\)[If \(A\) is infinite, let \(A_n = \{s : s\) is a one-to-one sequence in \(A\) of length \(n\}\) for each \(n\). Use a choice function for \(S = \{A_n : n\in N \}\) to obtain a countable subset of \(A\).]
5.7. Use DC to prove the countable AC.
\(\quad\)[Given \(S = \{An : n\in N\}\), consider the set \(A\) of all choice functions on some \(S_n = \{A_i : i\le n\}\), with the binary relation \(\supset\).]
5.8 (The Milner-Rado Paradox). For every ordinal \(\alpha < \kappa^+\) there are sets \(X_n\subset\alpha (n\in N)\) such that \(\alpha =S X_n\), and for each \(n\) the order-type of \(X_n\) is \(\le \kappa^n\).
\(\quad\)[By induction on \(\alpha\), choosing a sequence cofinal in \(\alpha\).]
5.9. If \(\{Xi :i\in I\}\) and \(\{Yi :i\in I\}\) are two disjoint families such that \(|X_i|=|Y_i|\) for each \(i\in I\), then \(|\bigcup_{i\in I} X_i| = |\bigcup_{i\in I} Y_i|\).
\(\quad\)[Use AC.]
5.10. If \(\{X_i :i\in I\}\) and \(\{Y_i :i\in I\}\) are such that \(|X_i|=|Y_i|\) for each \(i\in I\), then \(|\prod_{i\in I} X_i| = |\prod_{i\in I} Y_i|\).
\(\quad\)[Use AC.]
5.11. \(\prod_{0<n<\omega}n=2^{\aleph_0}\).
5.12. \(\prod_{n<\omega}\aleph_n=\aleph^{\aleph_0}_\omega\).
5.13. \(\prod_{\alpha<\omega+\omega}\aleph_\alpha=\aleph^{\aleph_0}_{ \omega+\omega}\).
5.14. If GCH holds then
\(\quad\)(i) \(2^{<\kappa} =\kappa\) for all \(\kappa\), and
\(\quad\)(ii) \(\kappa^{<\kappa} = \kappa\) for all regular \(\kappa\).
5.15. If \(\beta\) is such that \(2^{\aleph_\alpha} =\aleph_{\alpha +\beta}\) for every \(\alpha\), then \(\beta <\omega\).
\(\quad\)[Let \(\beta\ge\omega\). Let \(\alpha\) be least such that \(\alpha +\beta >\beta\). We have \(0 <\alpha\le\beta\), and \(\alpha\) is limit. Let \(\kappa =\aleph_{\alpha +\alpha}\); since cf \(\kappa=\) cf \(\alpha\le\alpha < \kappa\), \(\kappa\) is singular. For each \(\xi <\alpha\), \(\xi +\beta =\beta\), and so \(2^{\aleph_\alpha +\xi} =\aleph_{\alpha +\xi +\beta} =\aleph_{\alpha +\beta}\). By Corollary 5.17, \(2^\kappa= \aleph_{\alpha +\beta}\), a contradiction, since \(\aleph_{\alpha +\beta} <\aleph_{\alpha +\alpha +\beta}\).]
5.16. \(\prod_{\alpha<\omega_1+\omega}\aleph_\alpha=\aleph^{\aleph_1}_{\omega_1+ \omega}\).
\(\quad\)[\(\aleph^{\aleph_1}_{\omega_1+\omega}\le(\prod^\infty_{n=0} \aleph_{\omega_1+n})^{\aleph_1}=\prod_n\aleph^{\aleph_1}_{\omega_1+n}=\) \(\prod_n(\aleph^{\aleph_1}_{\omega_1}\cdot\aleph_{\omega_1+n})=\) \(\aleph^{\aleph_1}_{\omega_1}\cdot\prod_n\aleph_{\omega_1+n}=\) \(\prod_{\alpha<\omega_1+\omega}\aleph_{\alpha}\).]
5.17. If \(\kappa\) is a limit cardinal and \(\lambda<\) cf \(\kappa\), then \(\kappa^\lambda =\sum_{\alpha <\kappa}|\alpha|^\lambda\).
5.18. \(\aleph^{\aleph_1}_\omega =\aleph^{\aleph_0}_\omega\cdot 2^{ \aleph_1}\).
5.19. If \(\alpha <\omega_1\), then \(\aleph_\alpha^{\aleph_1} = \aleph^{\aleph_0}_\alpha\cdot 2^{\aleph_1}\).
5.20. If \(\alpha <\omega_2\), then \(\aleph_\alpha^{\aleph_2} = \aleph^{\aleph_1}_\alpha\cdot 2^{\aleph_2}\).
5.21. If \(\kappa\) is regular and limit, then \(\kappa^{<\kappa}= 2^{<\kappa}\). If \(\kappa\) is regular and strong limit then \(\kappa^{<\kappa} = \kappa\).
5.22. If \(\kappa\) is singular and is not strong limit, then \(\kappa ^{<\kappa} = 2^{<\kappa} > \kappa\).
5.23. If \(\kappa\) is singular and strong limit, then \(2^{<\kappa}=\kappa\) and \(\kappa <\kappa = \kappa^{\text{cf }\kappa}\).
5.24. If \(2^{\aleph_0} >\aleph_\omega\), then \(\aleph^{\aleph_0}_\omega = 2^{\aleph_0}\).
5.25. If \(2^{\aleph_1}=\aleph_2\) and \(\aleph^{\aleph_0}_\omega> \aleph_{\omega_1}\), then \(\aleph^{\aleph_1}_{\omega_1}= \aleph^{\aleph_0}_{\omega}\).
5.26. If \(2^{\aleph_0}\ge\aleph_{\omega_1}\), then \(\gimel(\aleph_{\omega_1})=2^{\aleph_0}\) and \(\gimel(\aleph_{\omega_1})= 2^{\aleph_1}\).
5.27. If \(2^{\aleph_1}=\aleph_2\), then \(\aleph^{\aleph_0}\neq\aleph_ {\omega_1}\).
5.28. If \(\kappa\) is a singular cardinal and if \(\kappa <\gimel(\lambda)\) for some \(\lambda<\kappa\) such that cf \(\kappa \le\) cf \(\lambda\) then \(\gimel(\kappa )\le\gimel(\lambda)\).
5.29. If \(\kappa\) is a singular cardinal such that \(2^{ \text{cf }\kappa}<\kappa \le \lambda^{\text{cf }\kappa}\) for some \(\lambda<\kappa\), then \(\gimel(\kappa)=\gimel(\lambda)\) where \(\lambda\) is the least \(\lambda\) such that \(\kappa\le\lambda^{\text{cf }\kappa}\).