# Set theory - Chapter 5: The Axiom of Choice and Cardinal Arithmetic

## 5.1

We saw in exercise 4.6 that the set of all open sets has cardinality $2^{{\aleph_{0}}}$. The operation ”taking the complement of a set” puts this set in bijection with the set of all closed sets. Since every perfect set is closed, there are at most $2^{{\aleph_{0}}}$ many such sets. Moreover, every nonempty closed interval of reals $[a,b]$ is perfect and so there are exactly $2^{{\aleph_{0}}}$ perfect sets of reals. We consider an enumeration $\langle P_{\alpha},\alpha<2^{{\aleph_{0}}}\rangle$ of all perfect sets.

Now, we construct sequences $\langle a_{\alpha},\alpha<2^{{\aleph_{0}}}\rangle$ and $\langle b_{\alpha},\alpha<2^{{\aleph_{0}}}\rangle$ by induction by picking for each $\alpha<2^{{\aleph_{0}}}$ two elements $a_{\alpha}\in\mathbb{R}\backslash\left(\{a_{{\xi}}:\xi<\alpha\}\cup\{b_{{\xi}}% :\xi<\alpha\}\right)$ and $b_{{\alpha}}\in P_{\alpha}\backslash\{a_{{\xi}}:\xi\leq\alpha\}$. This choice is possible because $|\mathbb{R}|=|P_{\alpha}|=2^{{\aleph_{0}}}>|\alpha|$ (recall that any perfect set has cardinality $2^{{\aleph_{0}}}$).

Now let $S=\{a_{\alpha}:\alpha<2^{{\aleph_{0}}}\}$. The $a_{\alpha}$ are distinct by definition and so $|S|=2^{{\aleph_{0}}}$. Let $a<2^{{\aleph_{0}}}$. For all $\alpha<\xi<2^{{\aleph_{0}}}$, $a_{\xi}\neq b_{{\alpha}}$ by definition of $a_{{\xi}}$. For all $\xi\leq\alpha$, $b_{{\alpha}}\neq a_{{\xi}}$ by definition of $b_{{\alpha}}$. Hence $b_{{\alpha}}\notin S$ and because $b_{{\alpha}}\in P_{\alpha}$ we have $P_{\alpha}\nsubseteq S$.

Finally, $S$ is a set of reals of cardinality $2^{{\aleph_{0}}}$ without a perfect subset.

## 5.2

Let $X$ be an infinite set and $\aleph_{\alpha}=|X|$ its aleph and $S$ the set of all finite subsets of $X$. Then we have

 $\displaystyle|S|$ $\displaystyle=|{[X]}^{{<\omega}}|$ $\displaystyle=\sum_{{n<\omega}}\aleph_{\alpha}^{n}$ $\displaystyle=\sum_{{n<\omega}}\aleph_{\alpha}$ $\displaystyle=\aleph_{0}\aleph_{\alpha}$ $\displaystyle=\aleph_{\alpha}$ $\displaystyle=|X|$

## 5.3

Let $P$ a linearly ordered set and $\kappa$ a cardinal such that for all $x\in P$, the initial segment $P_{x}=\{y\in P:y is of size less than $\kappa$.

If $M$ is a greatest element of $P$, then clearly $|P|=|P_{M}\cup\{M\}|=1+|P_{M}|\leq\kappa$ since $|P_{M}|<\kappa$.

Otherwise, let $x_{0} an enumeration of $P$ of size $\lambda=|P|$. Then for all $\alpha<\lambda$, $|\alpha|=|P_{{x_{\alpha}}}|<\kappa$ thus $\alpha<\kappa$. Since $P$ does not have a greatest element, $\lambda$ is limit i.e. $\lambda=\sup_{{\alpha<\lambda}}\alpha\leq\kappa$. Again $|P|\leq\kappa$.

## 5.4

I actually discovered and proved the statement of this exercise when I was in high school, but here is a more straightforward proof:

Let $A$ be a well-ordered set. For all subsets $X\neq Y$, $X\Delta Y$ is non empty. Define $X if the least element of $X\Delta Y$ belongs to $X$. Note that by definition $<$ is irreflexive and assymetric. If the order is well defined, then it is clearly total (if $X\neq Y$, the least element of $X\Delta Y$ is either in $X$ or in $Y$).

It remains to prove that $<$ is transitive. Let $X,Y,Z$ such that $X and $Y. We define and have:

1. (1)

$m_{1}=\min(X\Delta Y)\in X\backslash Y$

2. (2)

$m_{2}=\min(Y\Delta Z)\in Y\backslash Z$

3. (3)

$m_{3}=\min(X\Delta Z)$

We want to prove that $m_{3}\in X$. Suppose the contrary i.e. $m_{3}\in Z\backslash X$.

First, suppose $m_{3}\in Y$. Then $m_{3}\in Y\backslash X$ and by (1), we have $m_{1}\in X$ and $m_{1}. By (3) we get $m_{1}\in Z$. By (1), we get $m_{1}\in Z\backslash Y$. By (2) we get $m_{2} and $m_{2}\in Y$. By (1) we get $m_{2}\in X$. By (2) we get $m_{2}\in X\backslash Z$. By (3) we get $m_{3}. Finally, we have obtained $m_{3}. A contradiction.

Now, suppose instead that $m_{3}\notin Y$. Then $m_{3}\in Z\backslash Y$ so by (2) we have $m_{2} and $m_{2}\notin Z$. By (3) we get $m_{2}\notin X$ and still $m_{2} and moreover by (2) we have $m_{2}\in Y$. By (1), we get $m_{1} and $m_{1}\in X$. From $m_{1} and $m_{1}\in X$ we get by (3) that $m_{1}\in Z$ and we still have $m_{1}. By (2), we get $m_{1}\in Y$. But this contradicts (1).

## 5.5

Let $S$ be a family of nonempty sets. Let $P=\{f:f\text{ is a choice function on some }Z\subset S\}$, ordered by inclusion. It is clearly non empty since there is a choice function on any finite $Z\subset S$. If we have a chain of choice functions $f_{\alpha},\alpha<\lambda$ then $f=\cup_{{\alpha<\lambda}}f_{\alpha}$ is a choice function on $\cup_{{\alpha<\lambda}}{\operatorname{dom}(f_{\alpha})}$ and so is an upper bound of the chain. By Zorn’s Lemma, $P$ has a maximal element $f$. Moreover $\operatorname{dom}(f)=S$ for otherwise, we could extended $f$ and it would not be maximal. So $f$ is a choice function on $S$.

## 5.6

Let $A$ be an infinite set and define for all $n\in\mathbb{N}$, the set $A_{n}$ of one-to-one sequences in $A$ of length $n$. Each $A_{n}$ is nonempty so assuming the countable AC, there exists a choice function $f$ on $S=\{A_{n}:n\in\mathbb{N}\}$. Define for all $n\in\mathbb{N}$, $s_{n}=f(A_{n})$ and $B=\cup_{{n\in\mathbb{N}}}\operatorname{ran}(s_{n})\subseteq A$. Clearly $B$ is infinite. For all $b\in B$, we define $n_{b}$ to be the least natural integer $n$ such $b\in\operatorname{ran}(s_{{n}})$. Then the function $g:B\rightarrow\mathbb{N}\times\mathbb{N}$ given by $g(b)=(n_{b},s_{{n_{b}}}^{{-1}}(b))$ is one-to-one and so $B$ is countable.

## 5.7

Let $S=\{A_{n}:n\in N\}$ a countable family of non-empty set. Let $A$ be the set of all choices functions on some $S_{n}=\{A_{i}|i\leq n\}$. Let $f\in A$ is a choice function on $S_{n}$. $A_{{n+1}}$ is non-empty so we can pick $x\in A_{{n+1}}$ and define the function $\tilde{f}$ on $S_{{n+1}}$ by $\tilde{f}_{{|S_{n}}}=f$ and $\tilde{f}(A_{{n+1}})=x$. Then $\tilde{f}\in A$ and $\tilde{f}\supset f$. Assuming DC, there is a sequence ${(g_{n})}_{{n\in\mathbb{N}}}$ of elements of $A$ such that for all $n\in\mathbb{N}$, $g_{{n+1}}\supset g_{{n}}$. Hence we can define $g=\cup_{{n\in\mathbb{N}}}g_{n}$. This is a choice function on $S$.

## 5.8

Let $\kappa$ be a cardinal. Let’s show that for all $\alpha<\kappa^{+}$ there are sets $X_{n}^{\alpha}\subset\alpha$ ($n\in\mathbb{N}$) such that $\alpha=\bigcup_{n}X_{n}^{\alpha}$ and for each $n$, the order-type of $X_{n}^{\alpha}$ is $\leq\kappa^{n}$.

First for any limit ordinal $\alpha<\kappa^{+}$, we define $A_{\alpha}$ the set of increasing sequences of elements $\alpha$ of length $|\alpha|$. Using AC, we consider a choice function $f$ on $S=\{A_{\alpha}:\alpha<\kappa^{+}\}$. We set ${(\alpha_{\xi})}_{{\xi<|\alpha|}}=f(A_{\alpha})$.

We shall prove the result by induction on $\alpha<\kappa^{+}$. More precisely, we will construct by induction on $\alpha<\kappa^{+}$, countable families of sets $X_{n}^{{\alpha}}$ and associated increasing function $f_{n}^{{\alpha}}:X_{n}^{{\alpha}}\rightarrow\kappa^{n}$.

1. (1)

If $\alpha=0$, we can just take $X_{n}^{{\alpha}}=\emptyset$ and $f_{n}^{{\alpha}}=\emptyset$ for all $n\in\mathbb{N}$.

2. (2)

Suppose $\alpha=\beta+1$ is a successor ordinal. In particular, $0\in\kappa$. If the induction hypothesis is true for $\beta$ then we have

 $\alpha=\beta\cup\{\beta\}=\left(\bigcup_{{\xi<\beta}}X_{n}^{\beta}\right)\cup% \{\beta\}$

and we can take:

• $X_{0}^{\alpha}=\{\beta\}$ and $f_{0}^{\alpha}=\{(\beta,0)\}$

• For all $n\in\mathbb{N}$, $X_{{n+1}}^{\alpha}=X_{n}^{\beta}$ and $f_{{n+1}}^{\alpha}(x)=f_{{n}}^{{\beta(x)}}\times\{0\}$.

3. (3)

If $\alpha$ is a limit cardinal and the induction hypothesis is true below $\alpha$ then

 $\alpha=\bigcup_{{\xi<|\alpha|}}\alpha_{\xi}=\bigcup_{{\xi<|\alpha|}}\left(% \alpha_{\xi}\backslash{\bigcup_{{\xi^{{\prime}}<\xi}}\alpha_{{\xi^{{\prime}}}}% }\right)=\bigcup_{{\xi<|\alpha|}}\left({\left(\bigcup_{{n\in\mathbb{N}}}X_{n}^% {{\alpha_{\xi}}}\right)}\backslash{\bigcup_{{\xi^{{\prime}}<\xi}}\alpha_{{\xi^% {{\prime}}}}}\right)=\bigcup_{{n\in\mathbb{N}}}\left(\bigcup_{{\xi<|\alpha|}}% \left(X_{n}^{{\alpha_{\xi}}}\backslash\bigcup_{{\xi^{{\prime}}<\xi}}\alpha_{{% \xi^{{\prime}}}}\right)\right)$

and we can take

• $X_{0}^{{\alpha}}=\emptyset$, $f_{0}^{{\alpha}}=\emptyset$.

• For all $n\in\mathbb{N}$, $X_{{n+1}}^{{\alpha}}=\bigcup_{{\xi<\alpha}}Y_{n}^{{\alpha_{\xi}}}$ where $Y_{n}^{{\alpha_{\xi}}}=X_{n}^{{\alpha_{\xi}}}\backslash\bigcup_{{\xi^{{\prime}% }<\xi}}\alpha_{{\xi^{{\prime}}}}\subseteq X_{n}^{{\alpha_{\xi}}}\subseteq% \alpha_{{\xi}}$. We define for all $x\in X_{{n+1}}^{\alpha}$, $f_{{n+1}}^{{\alpha}}(x)=(\xi_{x},f_{n}^{{\alpha_{{\xi_{x}}}}}(x))\in\kappa^{{n% +1}}$ where $\xi_{x}$ is the least ordinal such that $x\in Y_{n}^{{\alpha_{{\xi_{x}}}}}$. By construction $f$ is an increasing function.

## 5.9

Let $\{X_{i}:i\in I\}$ and $\{Y_{i}:i\in I\}$ be two disjoint families such that $|X_{i}|=|Y_{i}|$. For all $i\in I$, let $A_{i}$ be the set of all bijection between $X_{i}$ and $Y_{i}$. $A_{i}$ is nonempty by assumption. Suppose AC and let’s consider a choice function $A_{i}\mapsto f_{i}$ on $S=\{A_{i}:i\in I\}$. Let $X=\cup_{{i\in I}}X_{i}$ and $Y=\cup_{{i\in I}}Y_{i}$. For all $x\in X$ denote by $i_{x}$ the unique index such that $x\in X_{i}$ (no need for AC here). Then $x\mapsto f_{{i_{x}}}(x)$ is a bijection between $X$ and $Y$.

## 5.10

Same hypothesis as exercise 5.9, without assuming that the two families contain disjoint sets. We define $f_{i}$ the same way, using AC. Then we have a bijection between $\prod_{{i\in I}}X_{i}$ and $\prod_{{i\in I}}Y_{i}$ given by ${(x_{i})}_{{i\in I}}\mapsto{(f_{i}(x_{i}))}_{{i\in I}}$.

## 5.11

 $\prod_{{0

## 5.12

 $\prod_{{n<\omega}}\aleph_{n}={\sup_{{n<\omega}}\aleph_{n}}^{{\aleph_{0}}}={% \aleph_{\omega}}^{{\aleph_{0}}}$

## 5.13

 $\prod_{{\alpha<\omega+\omega}}\aleph_{\alpha}={\sup_{{\alpha<\omega+\omega}}% \aleph_{\alpha}}^{{|\omega+\omega|}}={\aleph_{{\omega+\omega}}}^{{\aleph_{0}}}$

## 5.14

Suppose GCH holds, then:

1. (1)

$2^{{<\kappa}}=\sup_{{\mu<\kappa}}2^{\mu}=\sup_{{\mu<\kappa}}\mu^{+}=\kappa$

2. (2)

If $\kappa$ is regular, $\kappa^{\mu}=\kappa$ for all $\mu<\kappa$, so $\kappa^{{<\kappa}}=\sup_{{\mu<\kappa}}\kappa^{\mu}=\sup_{{\mu<\kappa}}\kappa=\kappa$.

## 5.15

Let $\beta$ be such that for every $\alpha,2^{{\aleph_{\alpha}}}=\aleph_{{\alpha+\beta}}$ and suppose that $\beta\geq\omega$.

We have $0+\beta=\beta$ and $\beta+\beta>\beta$ so we can define $\alpha$ to be the least $\alpha$ such that $\alpha+\beta>\beta$ and moreover $0<\alpha\leq\beta$.

If $\alpha=\alpha^{{\prime}}+1$, let $\delta$ be such that $\beta=\omega+\delta$. We have $\alpha+\beta=\alpha^{{\prime}}+1+\omega+\delta=\alpha^{{\prime}}+\omega+\delta% =\alpha^{{\prime}}+\beta<\beta$. A contradiction.

Hence $\alpha$ is limit and $\kappa=\aleph_{{\alpha+\alpha}}$ is a limit cardinal. We have $\operatorname{cf}(\kappa)=\operatorname{cf}(\alpha)\leq\alpha\leq\aleph_{{% \alpha}}<\aleph_{{\alpha+\alpha}}=\kappa$. Thus $\kappa$ is singular.

Now, for all $\xi<\alpha$, we have $\xi+\beta=\beta$ and

 $2^{{\aleph_{{\alpha+\xi}}}}=\aleph_{{\alpha+\xi+\beta}}=\aleph_{{\alpha+\beta}}$

an by corollary 5.17,

 $2^{\kappa}=2^{{\aleph_{{\alpha+\alpha}}}}=\aleph_{{\alpha+\beta}}<\aleph_{{% \alpha+(\alpha+\beta)}}=2^{{\aleph_{{\alpha+\alpha}}}}=2^{\kappa}$

As a consequence, $\beta<\omega$.

## 5.16

 $\displaystyle\aleph_{{\omega_{1}+\omega}}^{{\aleph_{1}}}$ $\displaystyle\leq{\left(\prod_{{n=0}}^{{+\infty}}\aleph_{{\omega_{1}+n}}\right% )}^{{\aleph_{1}}}$ $\displaystyle=\prod_{{n=0}}^{{+\infty}}\aleph_{{\omega_{1}+n}}^{{\aleph_{1}}}$

But by repeated application of the Hausdorff formula, we have $\aleph_{{\omega_{1}+n}}^{{\aleph_{1}}}=\aleph_{{\omega_{1}}}^{{\aleph_{1}}}{% \prod_{{i=1}}^{{n}}\aleph_{{\omega_{1}+i}}}=\aleph_{{\omega_{1}}}^{{\aleph_{1}% }}\aleph_{{\omega_{1}+n}}$, so

 $\displaystyle\prod_{{n=0}}^{{+\infty}}\aleph_{{\omega_{1}+n}}^{{\aleph_{1}}}$ $\displaystyle=\aleph_{{\omega_{1}}}^{{\aleph_{1}\aleph_{0}}}{\prod_{{n=0}}^{{+% \infty}}\aleph_{{\omega_{1}+n}}}$ $\displaystyle=\aleph_{{\omega_{1}}}^{{\aleph_{1}}}{\prod_{{n=0}}^{{+\infty}}% \aleph_{{\omega_{1}+n}}}$ $\displaystyle={\left({\sup_{{\alpha<\omega_{1}}}\aleph_{\alpha}}\right)}^{{% \aleph_{1}}}{\prod_{{n=0}}^{{+\infty}}\aleph_{{\omega_{1}+n}}}$ $\displaystyle={\prod_{{\alpha<\omega_{1}}}\aleph_{{\alpha}}}{\prod_{{n=0}}^{{+% \infty}}\aleph_{{\omega_{1}+n}}}$ $\displaystyle={\prod_{{\alpha<\omega_{1}+\omega}}\aleph_{{\alpha}}}$

Moreover, we have

 $\displaystyle\prod_{{\alpha<\omega_{1}+\omega}}\aleph_{{\alpha}}$ $\displaystyle\leq\prod_{{\alpha<\omega_{1}+\omega}}\aleph_{{\omega_{1}+\omega}}$ $\displaystyle\leq\aleph_{{\omega_{1}+\omega}}^{{\aleph_{1}}}$

and finally

 $\prod_{{\alpha<\omega_{1}+\omega}}\aleph_{{\alpha}}=\aleph_{{\omega_{1}+\omega% }}^{{\aleph_{1}}}$

## 5.17

Let $\kappa$ be a limit cardinal and $1\leq\lambda<\operatorname{cf}\kappa$. Then every function $\kappa^{\lambda}$ is bounded and so $\kappa^{\lambda}=\bigcup_{{\alpha<\kappa}}\alpha^{\lambda}$. Hence we have

 $\displaystyle\kappa^{\lambda}$ $\displaystyle=\left|\bigcup_{{\alpha<\kappa}}\alpha^{\lambda}\right|$ $\displaystyle\leq\kappa\sup_{{\alpha<\kappa}}{|\alpha|}^{\lambda}$ $\displaystyle=\sum_{{\alpha<\kappa}}{|\alpha|}^{\lambda}$ $\displaystyle\leq\sum_{{\alpha<\kappa}}\kappa^{\lambda}$ $\displaystyle=\kappa\kappa^{\lambda}$ $\displaystyle=\kappa^{\lambda}$

## 5.18

Obviously, $\aleph_{\omega}^{{\aleph_{1}}}$ is at least $\aleph_{\omega}^{{\aleph_{0}}}2^{{\aleph_{1}}}$. Let’s prove the equality.

By theorem 5.20, it is either ${\aleph_{n}}^{{\aleph_{0}}}$ for some $n<\omega$ or (since $\operatorname{cf}(\aleph_{\omega})=\aleph_{0}\leq\aleph_{1}$) it is $\aleph_{\omega}^{{\aleph_{0}}}$. In the latter case, using Hausdorff formula we get ${\aleph_{n}}^{{\aleph_{0}}}=\aleph_{1}^{{\aleph_{1}}}{\aleph_{2}...\aleph_{n}}% \leq 2^{{\aleph_{1}}}\aleph_{\omega}^{{\aleph_{0}}}$.

## 5.19

As in 5.18, one inequality is clear. We show the other direction by induction on $\alpha<\omega_{1}$. For $\alpha=0,1$, we have $\aleph_{\alpha}^{{\aleph_{1}}}=2^{{\aleph_{1}}}$. If the equality is true below some $1<\alpha<\omega_{1}$ then $\aleph_{\alpha}^{{\aleph_{1}}}$ can only take the values $\aleph_{\beta}^{{\aleph_{1}}}$ ($\beta<\alpha$), $\aleph_{\alpha}$ or $\aleph_{\alpha}^{{\operatorname{cf}{\aleph_{\alpha}}}}$ (theorem 5.20). The first is not more than the expect result, by induction hypothesis. The second is less than the third, which is itself at most $\aleph_{\alpha}^{{\aleph_{0}}}$ because $\operatorname{cf}{\aleph_{\alpha}}=\operatorname{cf}{\alpha}\leq\alpha<\omega_% {1}$. Finally $\aleph_{\alpha}^{{\aleph_{1}}}=2^{{\aleph_{1}}}\aleph_{\alpha}^{{\aleph_{0}}}$

## 5.20

We proceed as in 5.19: one equality is clear and we show the result by induction on $\alpha<\omega_{2}$. For $\alpha=0,1,2$, we have $\aleph_{\alpha}^{{\aleph_{2}}}=2^{{\aleph_{2}}}$. If the equality is true below some $2<\alpha<\omega_{1}$ then $\aleph_{\alpha}^{{\aleph_{1}}}$ can only take the values $\aleph_{\beta}^{{\aleph_{2}}}$ ($\beta<\alpha$), $\aleph_{\alpha}$ or $\aleph_{\alpha}^{{\operatorname{cf}{\aleph_{\alpha}}}}$ (theorem 5.20). The first is not more than the expect result, by induction hypothesis. The second is less than the third, which is itself at most $\aleph_{\alpha}^{{\aleph_{1}}}$ because $\operatorname{cf}{\aleph_{\alpha}}=\operatorname{cf}{\alpha}\leq\alpha<\omega_% {2}$. Finally $\aleph_{\alpha}^{{\aleph_{2}}}=2^{{\aleph_{2}}}\aleph_{\alpha}^{{\aleph_{1}}}$

## 5.21

Let $\kappa$ a regular limit cardinal. Clearly, $2^{{<\kappa}}\leq\kappa^{{<\kappa}}$. For the other direction, we not that by theorem 5.20 and because $\kappa$ is regular, for all $\mu$ infinite $\kappa^{\mu}$ is either $\lambda^{\mu}$ (for some $\lambda<\kappa$) or $\kappa$. But $\forall\mu<\kappa,2^{{<\kappa}}\geq 2^{\mu}=\mu^{\mu}$ and a fortiori $\forall\mu,\lambda<\kappa,2^{{<\kappa}}\geq\lambda^{\mu}$ (consider the maxium of $\lambda$ and $\mu$ and use the previous inequality). Moreover, $2^{{<\kappa}}\geq\sup_{{\mu<\kappa}}2^{\mu}\geq\sup_{{\mu<\kappa}}\mu^{{+}}\geq\kappa$. Thus $2^{{<\kappa}}\geq\kappa^{{<\kappa}}$ and finally Thus $2^{{<\kappa}}=\kappa^{{<\kappa}}$.

If moreover $\kappa$ is strong limit, then again theorem 5.20 shows that $\kappa^{{<\mu}}=\kappa$ for $\mu<\kappa$ and so $2^{{<\kappa}}=\kappa^{{<\kappa}}=\kappa$

## 5.22

Let $\kappa$ singular but not strong limit. Now, for all $\mu<\kappa$, $\kappa^{\mu}$ can take the values $\lambda^{\mu}$ (for some $\lambda<\kappa$), $\kappa$ or $\kappa^{{\operatorname{cf}(\kappa)}}$. We have already seen that $2^{{<\kappa}}\geq\lambda^{\mu}$ and we have $\kappa<\kappa^{{\operatorname{cf}(\kappa)}}$. Since $\kappa$ is not strong limit, we can find $\mu,\lambda<\kappa$ such that $\lambda^{\mu}>\kappa$ and thus $2^{{<\kappa}}\geq\lambda^{{\mu\operatorname{cf}(\kappa)}}>\kappa^{{% \operatorname{cf}(\kappa)}}$. So finally $2^{{<\kappa}}=\kappa^{{<\kappa}}\geq\kappa^{{\operatorname{cf}(\kappa)}}>\kappa$

## 5.23

Let $\kappa$ is strong limit and singular. Now, for all $\mu<\kappa$, $\kappa^{\mu}$ can take the values $\kappa$ or $\kappa^{{\operatorname{cf}(\kappa)}}$. In particular $\kappa^{{<\kappa}}=\kappa^{{\operatorname{cf}(\kappa)}}$.

Moreover $2^{{<\kappa}}=\sup_{{\mu<\kappa}}2^{\mu}\leq\sup_{{\mu<\kappa}}\kappa=\kappa$ and $2^{{<\kappa}}\geq\sup_{{\mu<\kappa}}2^{\mu}\geq\sup_{{\mu<\kappa}}\mu^{{+}}\geq\kappa$. Hence $2^{{<\kappa}}=\kappa$.

## 5.24

Suppose that $2^{{\aleph_{0}}}\geq\aleph_{\omega}$ then $\aleph_{\omega}^{{\aleph_{0}}}\geq 2^{{\aleph_{0}}}\geq{(2^{{\aleph_{0}}})}^{{% \aleph_{0}}}\geq\aleph_{\omega}^{{\aleph_{0}}}$

## 5.25

Suppose $2^{{\aleph_{1}}}=\aleph_{2}$ and $\aleph_{\omega}^{{\aleph_{0}}}>\aleph_{{\omega_{1}}}$. Then $\aleph_{\omega}^{{\aleph_{0}}}\geq\aleph_{{\omega_{1}}}\geq\aleph_{2}\geq 2^{{% \aleph_{1}}}$ so by exercise 5.18 we have $\aleph_{\omega}^{{\aleph_{0}}}=\aleph_{\omega}^{{\aleph_{1}}}$. Finally, $\aleph_{\omega}^{{\aleph_{0}}}=\aleph_{\omega}^{{\aleph_{1}}}=\aleph_{\omega}^% {{\aleph_{1}\aleph_{0}}}\geq\aleph_{{\omega_{1}}}^{{\aleph_{1}}}$. The other sense is obvious so $\aleph_{\omega}^{{\aleph_{0}}}=\aleph_{{\omega_{1}}}^{{\aleph_{1}}}$.

## 5.26

If $2^{{\aleph_{0}}}\geq\aleph_{{\omega_{1}}}$ then $\operatorname{\gimel}(\aleph_{\omega})=\aleph_{\omega}^{{\aleph_{0}}}\geq 2^{{% \aleph_{0}}}\geq 2^{{\aleph_{0}\aleph_{0}}}\geq\aleph_{{\omega_{1}}}^{{\aleph_% {0}}}\geq\aleph_{{\omega}}^{{\aleph_{0}}}$ and $\operatorname{\gimel}(\aleph_{{\omega_{1}}})=\aleph_{{\omega_{1}}}^{{\aleph_{1% }}}\leq{2^{{\aleph_{0}}}}^{{\aleph_{1}}}\leq 2^{{\aleph_{1}}}\leq\aleph_{{% \omega_{1}}}^{{\aleph_{1}}}$.

## 5.27

Suppose $2^{{\aleph_{1}}}=\aleph_{2}$. If we have ${\aleph_{\omega}}^{{\aleph_{0}}}=\aleph_{{\omega_{1}}}$ then by exercise 5.18 ${\aleph_{\omega}}^{{\aleph_{1}}}=\aleph_{\omega}^{{\aleph_{0}}}2^{{\aleph_{1}}% }=\aleph_{{\omega_{1}}}\aleph_{2}=\aleph_{{\omega_{1}}}$. This is not possible since $\operatorname{cf}(\aleph_{{\omega_{1}}})=\aleph_{1}<\operatorname{cf}({\aleph_% {\omega}}^{{\aleph_{1}}})$.

## 5.28

With the hypothesis of the exercise, $\operatorname{\gimel}(\kappa)=\kappa^{{\operatorname{cf}(\kappa)}}\leq{% \operatorname{\gimel}(\lambda)}^{{\operatorname{cf}(\kappa)}}\leq\lambda^{{% \operatorname{cf}(\lambda)\operatorname{cf}(\kappa)}}\leq\lambda^{{% \operatorname{cf}(\lambda)}}\leq\operatorname{\gimel}(\lambda)$.

## 5.29

Let $\kappa$ be a singular cardinal such that $2^{{\operatorname{cf}(\kappa)}}<\kappa$. Suppose there exists some $\lambda<\kappa$ such that $\kappa\leq\lambda^{{\operatorname{cf}(\kappa)}}$. Let’s consider the smallest such $\lambda$. Clearly, we have $\operatorname{\gimel}(\lambda)\leq\operatorname{\gimel}(\kappa)$. Moreover $\kappa^{{\operatorname{cf}(\kappa)}}\leq\lambda^{{\operatorname{cf}(\kappa)% \operatorname{cf}(\kappa)}}=\lambda^{{\operatorname{cf}(\kappa)}}\leq\kappa^{{% \operatorname{cf}(\kappa)}}$ so $\operatorname{\gimel}(\kappa)=\lambda^{{\operatorname{cf}(\kappa)}}$.

If $\operatorname{cf}(\kappa)\leq\operatorname{cf}(\lambda)$, then $\kappa<\operatorname{\gimel}(\kappa)=\lambda^{{\operatorname{cf}(\kappa)}}\leq% \lambda^{{\operatorname{cf}(\lambda)}}$ and we can apply exercise 5.28: $\operatorname{\gimel}(\kappa)\leq\operatorname{\gimel}(\lambda)$ and so $\operatorname{\gimel}(\kappa)=\operatorname{\gimel}(\lambda)$. Suppose instead that $\operatorname{cf}(\kappa)>\operatorname{cf}(\lambda)$ and let’s use Theorem 5.20 to compute $\operatorname{\gimel}(\kappa)=\lambda^{{\operatorname{cf}(\kappa)}}$. First, $\lambda\leq\operatorname{cf}(\kappa)$ is impossible for otherwise $2^{\lambda}\leq 2^{{\operatorname{cf}(\kappa)}}<\kappa\leq\lambda^{{% \operatorname{cf}(\kappa)}}$ by assumption while Theorem 5.20 implies that $\lambda^{{\operatorname{cf}(\kappa)}}=2^{\lambda}$. Next, the existence of some $\mu<\kappa$ such that $\mu^{{\operatorname{cf}(\kappa)}}=\lambda^{{\operatorname{cf}(\kappa)}}$ would contradict the minimality of $\lambda$. Finally, because $\operatorname{cf}(\lambda)\leq\operatorname{cf}(\kappa)$ Theorem 5.20 implies that $\operatorname{\gimel}(\kappa)=\lambda^{{\operatorname{cf}(\kappa)}}=\lambda^{{% \operatorname{cf}(\lambda)}}=\operatorname{\gimel}(\lambda)$.