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Exercise 5.1
Lemma 5.1.A
There are exactly $2^{\aleph_0}$ many open sets, $2^{\aleph_0}$ many closed sets, and $2^{\aleph_0}$ many perfect sets.
Proof
Let $\lbrace (a_\alpha,b_\alpha) : \alpha < \omega\rbrace$ be an enumeration of all open intervals with rational endpoints (possible as $\vert \Q \times \Q\vert = \aleph_0$). Then, for any open set $U \subseteq \R$, we may write $U = \bigcup_{\alpha \in I} (a_\alpha,b_\alpha)$ for some $I \subseteq \omega$. Thus, by $\AC$, there exists an injection from $\lbrace \text{open subsets of $\R$}\rbrace$ to $P(\omega)$, in which the latter has cardinality $2^{\aleph_0}$. On the other hand, there are clearly at least $2^{\aleph_0}$ open sets, e.g. those of the form $(-\infty,r)$ for some $r \in \R$. Thus, there are exactly $2^{\aleph_0}$ open sets.
Since $A$ is open iff $\R - A$ is closed, there is a bijection between the collection of open sets and collection of closed sets, so there are also exactly $2^{\aleph_0}$ closed sets.
Since every perfect set is closed, there are at most $2^{\aleph_0}$ closed sets. On the other hand, all closed $[0,r]$, where $r > 0$, are perfect, so there are at least $2^{\aleph_0}$ perfect sets.
There are exactly $2^{\aleph_0}$ many open sets, $2^{\aleph_0}$ many closed sets, and $2^{\aleph_0}$ many perfect sets.
Proof
Let $\lbrace (a_\alpha,b_\alpha) : \alpha < \omega\rbrace$ be an enumeration of all open intervals with rational endpoints (possible as $\vert \Q \times \Q\vert = \aleph_0$). Then, for any open set $U \subseteq \R$, we may write $U = \bigcup_{\alpha \in I} (a_\alpha,b_\alpha)$ for some $I \subseteq \omega$. Thus, by $\AC$, there exists an injection from $\lbrace \text{open subsets of $\R$}\rbrace$ to $P(\omega)$, in which the latter has cardinality $2^{\aleph_0}$. On the other hand, there are clearly at least $2^{\aleph_0}$ open sets, e.g. those of the form $(-\infty,r)$ for some $r \in \R$. Thus, there are exactly $2^{\aleph_0}$ open sets.
Since $A$ is open iff $\R - A$ is closed, there is a bijection between the collection of open sets and collection of closed sets, so there are also exactly $2^{\aleph_0}$ closed sets.
Since every perfect set is closed, there are at most $2^{\aleph_0}$ closed sets. On the other hand, all closed $[0,r]$, where $r > 0$, are perfect, so there are at least $2^{\aleph_0}$ perfect sets.
$\blacksquare$
Solution
Let $A,B$ be the sets in the hint. Note that such a construction works, as Theorem 4.5 asserts that $\vert P_\alpha\vert = 2^{\aleph_0}$ for all $\alpha$, and so $P_\alpha - \lbrace a_\xi : \xi \leq \alpha\rbrace$ is always non-empty, hence all $b_\alpha$'s can indeed be chosen. The construction of $A$ and $B$ also requires $\AC$, since the $a_\alpha$'s and $b_\alpha$'s have to be chosen. Finally, such an enumeration of the perfect sets is possible by Lemma 5.1.A.
We now show that both $A$ and $B$ contain no perfect subset of reals. This follows from the observation that $A \cap B = \emptyset$, and every perfect set $P_\alpha$ has non-empty intersection with $A$ and $B$.
Let $A,B$ be the sets in the hint. Note that such a construction works, as Theorem 4.5 asserts that $\vert P_\alpha\vert = 2^{\aleph_0}$ for all $\alpha$, and so $P_\alpha - \lbrace a_\xi : \xi \leq \alpha\rbrace$ is always non-empty, hence all $b_\alpha$'s can indeed be chosen. The construction of $A$ and $B$ also requires $\AC$, since the $a_\alpha$'s and $b_\alpha$'s have to be chosen. Finally, such an enumeration of the perfect sets is possible by Lemma 5.1.A.
We now show that both $A$ and $B$ contain no perfect subset of reals. This follows from the observation that $A \cap B = \emptyset$, and every perfect set $P_\alpha$ has non-empty intersection with $A$ and $B$.
$\square$
Remark. Such a set is called a Bernstein Set.
Exercise 5.2
Solution
We have: $$ \begin{align*} S = [X]^{<\omega} = \bigsqcup_{n < \omega} [X]^n \end{align*}$$ Thus, by Lemma 5.7: $$ \begin{align*} \vert S\vert = \sum_{n < \omega} \vert [X]^n\vert = \sum_{n < \omega}\vert X\vert ^n = \sum_{n < \omega} \vert X\vert = \vert X\vert \end{align*}$$ where $\vert X\vert ^n = X$ for all $n \in \omega$ as $X$ is infinite.
We have: $$ \begin{align*} S = [X]^{<\omega} = \bigsqcup_{n < \omega} [X]^n \end{align*}$$ Thus, by Lemma 5.7: $$ \begin{align*} \vert S\vert = \sum_{n < \omega} \vert [X]^n\vert = \sum_{n < \omega}\vert X\vert ^n = \sum_{n < \omega} \vert X\vert = \vert X\vert \end{align*}$$ where $\vert X\vert ^n = X$ for all $n \in \omega$ as $X$ is infinite.
$\square$
Exercise 5.3
Solution
For $\alpha \in \boldsymbol{\mathrm{ORD}}$, define a sequence $\c{x_\alpha : \alpha \in \boldsymbol{\mathrm{ORD}}}$ as follows (which will terminate at some ordinal stage):
Thus, we have: $$ \begin{align*} P = \bigcup_{\alpha < \lambda} \lbrace x \in P : x < x_\alpha\rbrace \end{align*}$$ Then: $$ \begin{align*} \vert P\vert \leq \sum_{\alpha < \lambda} \kappa = \lambda \cdot \kappa = \kappa \end{align*}$$
For $\alpha \in \boldsymbol{\mathrm{ORD}}$, define a sequence $\c{x_\alpha : \alpha \in \boldsymbol{\mathrm{ORD}}}$ as follows (which will terminate at some ordinal stage):
- Let $x_0 \in P$ be arbitrary.
- Given $x_\alpha$, let $x_{\alpha+1}$ be any element in $P$ such that $x_{\alpha+1} > x_\alpha$, if it exists (this is possible due to $\AC$). Otherwise, terminate the recursion.
- If $\alpha$ is limit, then choose $x_\alpha$ to be any element in $P$ such that $x_\alpha > x_\beta$ for all $\beta < \alpha$, if it exists (again, this is possible due to $\AC$). Otherwise, terminate the recursion.
Thus, we have: $$ \begin{align*} P = \bigcup_{\alpha < \lambda} \lbrace x \in P : x < x_\alpha\rbrace \end{align*}$$ Then: $$ \begin{align*} \vert P\vert \leq \sum_{\alpha < \lambda} \kappa = \lambda \cdot \kappa = \kappa \end{align*}$$
$\square$
Exercise 5.4
Solution
We do not use $\AC$ here, for otherwise every set is trivially well-ordered. We present a proof which is essentially equivalent to the hint, but presented differently.
It suffices to show that $2^\alpha$ is linearly ordered for all ordinals $\alpha$, where $2^\alpha$ is the set of all functions $f : \alpha \to 2$. Define a relation $<$ on $2^\omega$ as follows: For two functions $f \neq g$, let $\beta_{f,g} := \min\lbrace \beta < \alpha : f(\beta) \neq g(\beta)\rbrace$. Define: $$ \begin{align*} f < g \iff f(\beta_{f,g}) < g(\beta_{f,g}) \end{align*}$$ Observe that $f(\beta_{f,g}) < g(\beta_{f,g})$ iff $f(\beta_{f,g}) = 0$ and $g(\beta_{f,g}) = 1$. We shall show that this is a linear order. Clearly $f \not< f$ for all $f \in 2^\alpha$, and every $f \neq g$ are comparable. Suppose $f < g$ and $g < h$. We consider cases.
We do not use $\AC$ here, for otherwise every set is trivially well-ordered. We present a proof which is essentially equivalent to the hint, but presented differently.
It suffices to show that $2^\alpha$ is linearly ordered for all ordinals $\alpha$, where $2^\alpha$ is the set of all functions $f : \alpha \to 2$. Define a relation $<$ on $2^\omega$ as follows: For two functions $f \neq g$, let $\beta_{f,g} := \min\lbrace \beta < \alpha : f(\beta) \neq g(\beta)\rbrace$. Define: $$ \begin{align*} f < g \iff f(\beta_{f,g}) < g(\beta_{f,g}) \end{align*}$$ Observe that $f(\beta_{f,g}) < g(\beta_{f,g})$ iff $f(\beta_{f,g}) = 0$ and $g(\beta_{f,g}) = 1$. We shall show that this is a linear order. Clearly $f \not< f$ for all $f \in 2^\alpha$, and every $f \neq g$ are comparable. Suppose $f < g$ and $g < h$. We consider cases.
- Note that we must have $\beta_{f,g} \neq \beta_{g,h}$.
- Suppose $\beta_{f,g} < \beta_{g,h}$. Then $f(\beta_{f,g}) = 0$, $g(\beta_{f,g}) = 1$, $g(\beta_{g,h}) = 0$, $h(\beta_{g,h}) = 1$. Since $\beta_{f,g} < \beta_{g,h}$, we must have $h(\beta_{f,g}) = g(\beta_{f,g}) = 1$. Furthermore, if $f(\beta) \neq h(\beta)$ for some $\beta < \min\lbrace \beta_{f,g},\beta_{g,h}\rbrace$, then we must have $g(\beta) \neq f(\beta)$ or $g(\beta) \neq h(\beta)$, contradicting the minimality of $\beta_{f,g}$ or $\beta_{g,h}$. Thus, $\beta_{f,h} = \beta_{f,g}$, and $f(\beta_{f,h}) < h(\beta_{f,h})$.
- Suppose $\beta_{f,g} > \beta_{g,h}$. Then this implies that $f(\beta_{g,h}) = g(\beta_{g,h}) = 0$, and $h(\beta_{g,h}) = 1$. By similar reasoning as above, we have $f(\beta) = h(\beta)$ for $\beta < \beta_{g,h}$. Thus $\beta_{f,h} = \beta_{g,h}$ and $f(\beta_{f,h}) < h(\beta_{f,h})$.
$\square$
Exercise 5.5
Solution
We follow the hint. We note that $P$ is non-empty, as $\emptyset$ is trivially a choice function on $\emptyset \subseteq S$. Furthermore, if $C \subseteq P$ is a chain, then $\bigcup C$ is clearly an upper bound to this chain. Thus, the premise of Zorn's Lemma is satisfied, so we obtain a maximal element $f \in P$. We shall show that $f$ is a choice function on $S$.
Suppose not, so there exists some $X \in S$ such that $x \notin \dom(f)$. Fix $x \in X$. Then observe that $f \cup \lbrace (X,x)\rbrace$ remains a choice function on $\dom(f) \cup \lbrace X\rbrace \subseteq X$, so $f$ is not maximal, a contradiction.
We follow the hint. We note that $P$ is non-empty, as $\emptyset$ is trivially a choice function on $\emptyset \subseteq S$. Furthermore, if $C \subseteq P$ is a chain, then $\bigcup C$ is clearly an upper bound to this chain. Thus, the premise of Zorn's Lemma is satisfied, so we obtain a maximal element $f \in P$. We shall show that $f$ is a choice function on $S$.
Suppose not, so there exists some $X \in S$ such that $x \notin \dom(f)$. Fix $x \in X$. Then observe that $f \cup \lbrace (X,x)\rbrace$ remains a choice function on $\dom(f) \cup \lbrace X\rbrace \subseteq X$, so $f$ is not maximal, a contradiction.
$\square$
Exercise 5.6
Lemma 5.6.A
Assume the countable $\AC$. Let $\lbrace A_n : n \in \omega\rbrace$ be a countable family of pairwise disjoint sets such that $A_n$ is at most countable for each $n$. Then $A := \bigcup_{n \in \omega} A_n$ is at most countable.
Proof
We define a function $\F$ with domain $\omega$ by stipulating that for each $n < \omega$: $$ \begin{align*} \F(n) := \lbrace f : f : \omega \to A_n \text{ is an onto function}\rbrace \end{align*}$$ Since $A_n$ is at most countable, $\F(n)$ is non-empty for each $n$. Countable $\AC$ endows us with a choice function $F$ on $\ran(\F)$ (which is a set by the Axiom Schema of Replacement). Then we may define an onto function $G : \omega \times \omega \to A$ by: $$ \begin{align*} G(n,m) := F(n)(m) \end{align*}$$ Thus $\vert A\vert \leq \vert \omega \times \omega\vert = \aleph_0$.
Assume the countable $\AC$. Let $\lbrace A_n : n \in \omega\rbrace$ be a countable family of pairwise disjoint sets such that $A_n$ is at most countable for each $n$. Then $A := \bigcup_{n \in \omega} A_n$ is at most countable.
Proof
We define a function $\F$ with domain $\omega$ by stipulating that for each $n < \omega$: $$ \begin{align*} \F(n) := \lbrace f : f : \omega \to A_n \text{ is an onto function}\rbrace \end{align*}$$ Since $A_n$ is at most countable, $\F(n)$ is non-empty for each $n$. Countable $\AC$ endows us with a choice function $F$ on $\ran(\F)$ (which is a set by the Axiom Schema of Replacement). Then we may define an onto function $G : \omega \times \omega \to A$ by: $$ \begin{align*} G(n,m) := F(n)(m) \end{align*}$$ Thus $\vert A\vert \leq \vert \omega \times \omega\vert = \aleph_0$.
$\blacksquare$
Solution
Let $f$ be a choice function on $S$, so $f(A_n) \in A_n$ for all $n$. Now let: $$ \begin{align*} B := \bigcup_{n=0}^\infty \ran(f(A_n)) = \bigcup_{n=0}^\infty \bb{\ran(f(A_{n+1})) - \bigcup_{i<n} \ran(f(A_i))} \end{align*}$$ Clearly $B \subseteq A$. By Lemma 5.6.A, $B$ is at most countable. Furthermore, it can't be finite as for every $n \in \omega$ the map $i \mapsto f(A_n)(i)$ is one-to-one. Thus $B$ is countable.
Let $f$ be a choice function on $S$, so $f(A_n) \in A_n$ for all $n$. Now let: $$ \begin{align*} B := \bigcup_{n=0}^\infty \ran(f(A_n)) = \bigcup_{n=0}^\infty \bb{\ran(f(A_{n+1})) - \bigcup_{i<n} \ran(f(A_i))} \end{align*}$$ Clearly $B \subseteq A$. By Lemma 5.6.A, $B$ is at most countable. Furthermore, it can't be finite as for every $n \in \omega$ the map $i \mapsto f(A_n)(i)$ is one-to-one. Thus $B$ is countable.
$\square$
Exercise 5.7
Solution
Following the hint, we note that each $S_n$ is non-empty. Furthermore, if $f$ is a choice function on $S_n$, then for any $x \in A_{n+1}$, $f \cup \lbrace (A_{n+1},x)\rbrace$ is a choice function on $S_{n+1}$. Thus, the premise of $\DC$ is fulfilled, so there exists a chain of choice functions $f_0 \subseteq f_1 \subseteq \dots$, where $f_i$ is a choice function on $S_i$. Let $f := \bigcup_{i=0}^\infty f_i$, and clearly $f$ is a choice function on $S$.
Following the hint, we note that each $S_n$ is non-empty. Furthermore, if $f$ is a choice function on $S_n$, then for any $x \in A_{n+1}$, $f \cup \lbrace (A_{n+1},x)\rbrace$ is a choice function on $S_{n+1}$. Thus, the premise of $\DC$ is fulfilled, so there exists a chain of choice functions $f_0 \subseteq f_1 \subseteq \dots$, where $f_i$ is a choice function on $S_i$. Let $f := \bigcup_{i=0}^\infty f_i$, and clearly $f$ is a choice function on $S$.
$\square$
Exercise 5.8
Solution
For each ordinal $\alpha < \kappa^+$, we let $\c{X_n^\alpha : n \in \omega}$ denote a sequence of sets satisfying the property in the question.
We induct on $\alpha$. Clearly $X_n^0 = \emptyset$ for all $n$ works. Now suppose $\bigcup X_n^\beta = \beta$ works. Let $X_0^{\beta+1} := \lbrace \beta+1\rbrace$ and let $X_{n+1}^{\beta+1} := X_n^\beta$. Then clearly $\beta + 1 = \bigcup_n X_n^{\beta+1}$ works.
Now suppose $\alpha$ is a limit ordinal. Let $\lbrace \alpha_\gamma : \gamma < \cf(\alpha)\rbrace$ be cofinal. Let $X_0^\alpha := \lbrace 0\rbrace$, and let: $$ \begin{align*} X_{n+1}^\alpha := \bigcup_{\gamma < \cf(\alpha)} (X_n^{\alpha_{\gamma+1}} - \alpha_\gamma) \end{align*}$$ Then: $$ \begin{align*} \bigcup_{n < \omega} X_n^\alpha &= \lbrace \alpha_0\rbrace \cup \bigcup_{n < \omega} \bigcup_{\gamma < \cf(\alpha)} (X_n^{\alpha_{\gamma+1}} - \alpha_\gamma) \\ &= \lbrace \alpha_0\rbrace \cup \bigcup_{\gamma < \cf(\alpha)} \bigcup_{n < \omega} (X_n^{\alpha_{\gamma+1}} - \alpha_\gamma) \\ &= \lbrace \alpha_0\rbrace \cup \bigcup_{\gamma < \cf(\alpha)} (\alpha_{\gamma+1} - \alpha_\gamma) \\ &= \alpha \end{align*}$$ We now show that the order types work. We note that for each $n$, we have $X_n^{\alpha_{n+1}} - \alpha_\gamma \subseteq [\alpha_\gamma,\alpha_{\gamma+1}) = \lbrace \beta : \alpha_\gamma \leq \beta < \alpha_{\gamma+1}\rbrace$. Thus, the $X_n^\alpha$'s are pairwise disjoint, so: $$ \begin{align*} \otp(X_{n+1}^\alpha) &= \otp\bb{\bigcup_{\gamma < \cf(\alpha)} (X_n^{\alpha_{\gamma+1}} - \alpha_\gamma)} \\ &= \sum_{\gamma < \cf(\alpha)} \otp(X_n^{\alpha_{\gamma+1}} - \alpha_\gamma) \\ &\leq \sum_{\gamma < \cf(\alpha)} \otp(X_n^{\alpha_{\gamma+1}}) \\ &\leq \sum_{\gamma < \cf(\alpha)} \kappa^n \\ &= \kappa^n \cdot \cf(\alpha) \\ &\leq \kappa^{n+1} \end{align*}$$ Note that $\otp(X_0^\alpha) = \otp(\lbrace \alpha_0\rbrace) = 1 = \kappa^0$, so $\alpha = \bigcup_n X_n^\alpha$ works.
For each ordinal $\alpha < \kappa^+$, we let $\c{X_n^\alpha : n \in \omega}$ denote a sequence of sets satisfying the property in the question.
We induct on $\alpha$. Clearly $X_n^0 = \emptyset$ for all $n$ works. Now suppose $\bigcup X_n^\beta = \beta$ works. Let $X_0^{\beta+1} := \lbrace \beta+1\rbrace$ and let $X_{n+1}^{\beta+1} := X_n^\beta$. Then clearly $\beta + 1 = \bigcup_n X_n^{\beta+1}$ works.
Now suppose $\alpha$ is a limit ordinal. Let $\lbrace \alpha_\gamma : \gamma < \cf(\alpha)\rbrace$ be cofinal. Let $X_0^\alpha := \lbrace 0\rbrace$, and let: $$ \begin{align*} X_{n+1}^\alpha := \bigcup_{\gamma < \cf(\alpha)} (X_n^{\alpha_{\gamma+1}} - \alpha_\gamma) \end{align*}$$ Then: $$ \begin{align*} \bigcup_{n < \omega} X_n^\alpha &= \lbrace \alpha_0\rbrace \cup \bigcup_{n < \omega} \bigcup_{\gamma < \cf(\alpha)} (X_n^{\alpha_{\gamma+1}} - \alpha_\gamma) \\ &= \lbrace \alpha_0\rbrace \cup \bigcup_{\gamma < \cf(\alpha)} \bigcup_{n < \omega} (X_n^{\alpha_{\gamma+1}} - \alpha_\gamma) \\ &= \lbrace \alpha_0\rbrace \cup \bigcup_{\gamma < \cf(\alpha)} (\alpha_{\gamma+1} - \alpha_\gamma) \\ &= \alpha \end{align*}$$ We now show that the order types work. We note that for each $n$, we have $X_n^{\alpha_{n+1}} - \alpha_\gamma \subseteq [\alpha_\gamma,\alpha_{\gamma+1}) = \lbrace \beta : \alpha_\gamma \leq \beta < \alpha_{\gamma+1}\rbrace$. Thus, the $X_n^\alpha$'s are pairwise disjoint, so: $$ \begin{align*} \otp(X_{n+1}^\alpha) &= \otp\bb{\bigcup_{\gamma < \cf(\alpha)} (X_n^{\alpha_{\gamma+1}} - \alpha_\gamma)} \\ &= \sum_{\gamma < \cf(\alpha)} \otp(X_n^{\alpha_{\gamma+1}} - \alpha_\gamma) \\ &\leq \sum_{\gamma < \cf(\alpha)} \otp(X_n^{\alpha_{\gamma+1}}) \\ &\leq \sum_{\gamma < \cf(\alpha)} \kappa^n \\ &= \kappa^n \cdot \cf(\alpha) \\ &\leq \kappa^{n+1} \end{align*}$$ Note that $\otp(X_0^\alpha) = \otp(\lbrace \alpha_0\rbrace) = 1 = \kappa^0$, so $\alpha = \bigcup_n X_n^\alpha$ works.
$\square$
Exercise 5.9
Solution
Let: $$ \begin{align*} \F := \ss{f : f \text{ is a one-to-one function of $\bigcup_{i \in J} X_i$ onto $\bigcup_{i \in J} Y_i$ for some $J \subseteq I$}} \end{align*}$$ Partially ordered by inclusion. Clearly $\F$ is non-empty, for if $J = \lbrace i\rbrace$ for some $i \in I$ then any one-to-one function $f$ of $X_i$ onto $Y_i$ would be in $\F$. If $C \subseteq \F$ is a chain, then $\bigcup C$ is clearly an upper bound. Thus, let $f \in \F$ be maximal by Zorn's lemma. Then $f$ is a one-to-one function of $\bigcup_{i \in I} X_i$ onto $\bigcup_{i \in I} Y_i$ - otherwise, if $\dom(f) = \bigcup_{i \in J} X_i$ and $i' \notin J$, then let $f_{i'} : X_i \to Y_i$ be one-to-one and onto. Then $f \cup f_{i'}$ would be an element that strictly contains $f$, contradicting maximality. Thus $\mod{\bigcup_{i \in I} X_i} = \mod{\bigcup_{i \in I} Y_i}$.
Let: $$ \begin{align*} \F := \ss{f : f \text{ is a one-to-one function of $\bigcup_{i \in J} X_i$ onto $\bigcup_{i \in J} Y_i$ for some $J \subseteq I$}} \end{align*}$$ Partially ordered by inclusion. Clearly $\F$ is non-empty, for if $J = \lbrace i\rbrace$ for some $i \in I$ then any one-to-one function $f$ of $X_i$ onto $Y_i$ would be in $\F$. If $C \subseteq \F$ is a chain, then $\bigcup C$ is clearly an upper bound. Thus, let $f \in \F$ be maximal by Zorn's lemma. Then $f$ is a one-to-one function of $\bigcup_{i \in I} X_i$ onto $\bigcup_{i \in I} Y_i$ - otherwise, if $\dom(f) = \bigcup_{i \in J} X_i$ and $i' \notin J$, then let $f_{i'} : X_i \to Y_i$ be one-to-one and onto. Then $f \cup f_{i'}$ would be an element that strictly contains $f$, contradicting maximality. Thus $\mod{\bigcup_{i \in I} X_i} = \mod{\bigcup_{i \in I} Y_i}$.
$\square$
Exercise 5.10
Solution
The proof is almost identical to that of Exercise 5.9, but replace $\bigcup$ with $\prod$.
The proof is almost identical to that of Exercise 5.9, but replace $\bigcup$ with $\prod$.
$\square$
Exercise 5.11
Solution
By Lemma 5.9: $$ \begin{align*} \prod_{0 < n < \omega} n = (\sup_n n)^{\aleph_0} = (\aleph_0)^{\aleph_0} = 2^{\aleph_0} \end{align*}$$
By Lemma 5.9: $$ \begin{align*} \prod_{0 < n < \omega} n = (\sup_n n)^{\aleph_0} = (\aleph_0)^{\aleph_0} = 2^{\aleph_0} \end{align*}$$
$\square$
Exercise 5.12
Solution
By Lemma 5.9: $$ \begin{align*} \prod_{n < \omega} \aleph_n = (\sup_n \aleph_n)^{\aleph_0} = \aleph_\omega^{\aleph_0} \end{align*}$$
By Lemma 5.9: $$ \begin{align*} \prod_{n < \omega} \aleph_n = (\sup_n \aleph_n)^{\aleph_0} = \aleph_\omega^{\aleph_0} \end{align*}$$
$\square$
Exercise 5.13
Solution
By Lemma 5.9: $$ \begin{align*} \prod_{\alpha < \omega + \omega} \aleph_\alpha = (\sup_{\alpha<\omega+\omega} \aleph_\alpha)^{\aleph_0} = \aleph_{\omega+\omega}^{\aleph_0} \end{align*}$$
By Lemma 5.9: $$ \begin{align*} \prod_{\alpha < \omega + \omega} \aleph_\alpha = (\sup_{\alpha<\omega+\omega} \aleph_\alpha)^{\aleph_0} = \aleph_{\omega+\omega}^{\aleph_0} \end{align*}$$
$\square$
Exercise 5.14
Exercise 5.14(i)
Solution
If $\kappa = \aleph_{\alpha+1}$ then: $$ \begin{align*} \aleph_{\alpha+1} = 2^{\aleph_\alpha} = \sup_{\beta<\alpha+1} 2^{\aleph_\beta} = 2^{<\aleph_{\alpha+1}} \end{align*}$$ If $\kappa = \aleph_\alpha$, and $\alpha$ is a limit ordinal, then: $$ \begin{align*} \aleph_\alpha = \sup_{\beta<\alpha} \aleph_\beta = \sup_{\beta<\alpha} \aleph_{\beta+1} = \sup_{\beta<\alpha} 2^{\aleph_\beta} = 2^{<\aleph_\alpha} \end{align*}$$
If $\kappa = \aleph_{\alpha+1}$ then: $$ \begin{align*} \aleph_{\alpha+1} = 2^{\aleph_\alpha} = \sup_{\beta<\alpha+1} 2^{\aleph_\beta} = 2^{<\aleph_{\alpha+1}} \end{align*}$$ If $\kappa = \aleph_\alpha$, and $\alpha$ is a limit ordinal, then: $$ \begin{align*} \aleph_\alpha = \sup_{\beta<\alpha} \aleph_\beta = \sup_{\beta<\alpha} \aleph_{\beta+1} = \sup_{\beta<\alpha} 2^{\aleph_\beta} = 2^{<\aleph_\alpha} \end{align*}$$
$\square$
Exercise 5.14(ii)
Solution
By Theorem 5.15: $$ \begin{align*} \kappa^{<\kappa} = \sup_{\lambda<\kappa} \kappa^\lambda = \sup_{\lambda<\kappa} \kappa = \kappa \end{align*}$$
By Theorem 5.15: $$ \begin{align*} \kappa^{<\kappa} = \sup_{\lambda<\kappa} \kappa^\lambda = \sup_{\lambda<\kappa} \kappa = \kappa \end{align*}$$
$\square$
Exercise 5.15
Solution
We fill in some details of the hint.
We fill in some details of the hint.
- $\alpha \leq \beta$, as for all $\alpha > \beta$ we have $\alpha + \beta \geq \beta + \beta > \beta$.
- If $\alpha + \beta = \beta$, then $\alpha + 1 + \beta = \alpha + \beta = \beta$ as $\beta \geq \omega$. Thus $\alpha$ is a limit ordinal.
- $\beta < \alpha + \beta \implies \alpha + \beta < \alpha + \alpha + \beta$, so $\aleph_{\alpha+\beta} < \aleph_{\alpha+\alpha+\beta}$.
$\square$
Exercise 5.16
Solution
Alternatively, observe that one may easily generalise Lemma 5.9 such that for ordinals $\alpha$, we have: $$ \begin{align*} \prod_{i < \alpha} \kappa_i = \bb{\sup_i \kappa_i}^{\vert \alpha\vert } \end{align*}$$ In which then this exercise follows immediately.
Alternatively, observe that one may easily generalise Lemma 5.9 such that for ordinals $\alpha$, we have: $$ \begin{align*} \prod_{i < \alpha} \kappa_i = \bb{\sup_i \kappa_i}^{\vert \alpha\vert } \end{align*}$$ In which then this exercise follows immediately.
$\square$
Exercise 5.17
Solution
On one hand we have $\sum_{\alpha<\kappa} \vert \alpha\vert ^\lambda \leq \sum_{\alpha<\kappa} \kappa^\lambda = \kappa$. On the other hand, since $\lambda < \cf(\kappa)$ we have that all subsets in $[\kappa]^\lambda$ are bounded, so: $$ \begin{align*} [\kappa]^\lambda = \bigcup_{\alpha < \kappa} [\alpha]^\lambda \end{align*}$$ Therefore: $$ \begin{align*} \kappa^\lambda = \vert [\kappa]^\lambda\vert = \mod{\bigcup_{\alpha<\kappa} [\alpha]^\lambda} \leq \sum_{\alpha<\kappa} \vert [\alpha]^\lambda\vert = \sum_{\alpha<\kappa} \vert \alpha\vert ^\lambda \end{align*}$$
On one hand we have $\sum_{\alpha<\kappa} \vert \alpha\vert ^\lambda \leq \sum_{\alpha<\kappa} \kappa^\lambda = \kappa$. On the other hand, since $\lambda < \cf(\kappa)$ we have that all subsets in $[\kappa]^\lambda$ are bounded, so: $$ \begin{align*} [\kappa]^\lambda = \bigcup_{\alpha < \kappa} [\alpha]^\lambda \end{align*}$$ Therefore: $$ \begin{align*} \kappa^\lambda = \vert [\kappa]^\lambda\vert = \mod{\bigcup_{\alpha<\kappa} [\alpha]^\lambda} \leq \sum_{\alpha<\kappa} \vert [\alpha]^\lambda\vert = \sum_{\alpha<\kappa} \vert \alpha\vert ^\lambda \end{align*}$$
$\square$
Exercise 5.18
Lemma 5.18.A
For $\beta < \omega_{\alpha+1}$, we have: $$ \begin{align*} \aleph_\beta^{\aleph_{\alpha+1}} = \aleph_\beta^{\aleph_\alpha} \cdot 2^{\aleph_{\alpha+1}} \end{align*}$$
Proof
Note that clearly $\aleph_\beta^{\aleph_{\alpha+1}} \geq \aleph_\beta^{\aleph_\alpha} \cdot 2^{\aleph_{\alpha+1}}$, so it suffices to prove $\leq$. We induct on $\beta$. For $\beta \leq \alpha + 1$, by Theorem 5.20(i) we have: $$ \begin{align*} \aleph_\beta^{\aleph_{\alpha+1}} = 2^{\aleph_{\alpha+1}} = 2^{\aleph_\alpha} \cdot 2^{\aleph_{\alpha+1}} = \aleph_\beta^{\aleph_\alpha} \cdot 2^{\aleph_{\alpha+1}} \end{align*}$$ Note that $2^{\aleph_\alpha} \leq \aleph_{\alpha+1}^{\aleph_\alpha} \leq (2^{\aleph_\alpha})^{\aleph_\alpha} = 2^{\aleph_\alpha}$. Now assume that $\beta > \alpha + 1$. Suppose $\aleph_\beta^{\aleph_{\alpha+1}} = \aleph_\beta^{\aleph_\alpha} \cdot 2^{\aleph_{\alpha+1}}$. Then, by the Hausdorff formula: $$ \begin{align*} \aleph_{\beta+1}^{\aleph_{\alpha+1}} &= \aleph_\beta^{\aleph_{\alpha+1}} \cdot \aleph_{\beta+1} \\ &= \aleph_\beta^{\aleph_\alpha} \cdot 2^{\aleph_{\alpha+1}} \cdot \aleph_{\beta+1} \\ &= \aleph_\beta^{\aleph_\alpha} \cdot \aleph_{\beta+1} \cdot 2^{\aleph_{\alpha+1}} \\ &= \aleph_{\beta+1}^{\aleph_\alpha} \cdot 2^{\aleph_{\alpha+1}} \end{align*}$$ Now suppose $\beta$ is a limit ordinal, and for all $\gamma < \beta$ we have $\aleph_\gamma^{\aleph_{\alpha+1}} = \aleph_\gamma^{\aleph_\alpha} \cdot 2^{\aleph_{\alpha+1}}$. By Theorem 5.20, there are three possible cases ((i) is not possible as $\beta > \alpha + 1$).
For $\beta < \omega_{\alpha+1}$, we have: $$ \begin{align*} \aleph_\beta^{\aleph_{\alpha+1}} = \aleph_\beta^{\aleph_\alpha} \cdot 2^{\aleph_{\alpha+1}} \end{align*}$$
Proof
Note that clearly $\aleph_\beta^{\aleph_{\alpha+1}} \geq \aleph_\beta^{\aleph_\alpha} \cdot 2^{\aleph_{\alpha+1}}$, so it suffices to prove $\leq$. We induct on $\beta$. For $\beta \leq \alpha + 1$, by Theorem 5.20(i) we have: $$ \begin{align*} \aleph_\beta^{\aleph_{\alpha+1}} = 2^{\aleph_{\alpha+1}} = 2^{\aleph_\alpha} \cdot 2^{\aleph_{\alpha+1}} = \aleph_\beta^{\aleph_\alpha} \cdot 2^{\aleph_{\alpha+1}} \end{align*}$$ Note that $2^{\aleph_\alpha} \leq \aleph_{\alpha+1}^{\aleph_\alpha} \leq (2^{\aleph_\alpha})^{\aleph_\alpha} = 2^{\aleph_\alpha}$. Now assume that $\beta > \alpha + 1$. Suppose $\aleph_\beta^{\aleph_{\alpha+1}} = \aleph_\beta^{\aleph_\alpha} \cdot 2^{\aleph_{\alpha+1}}$. Then, by the Hausdorff formula: $$ \begin{align*} \aleph_{\beta+1}^{\aleph_{\alpha+1}} &= \aleph_\beta^{\aleph_{\alpha+1}} \cdot \aleph_{\beta+1} \\ &= \aleph_\beta^{\aleph_\alpha} \cdot 2^{\aleph_{\alpha+1}} \cdot \aleph_{\beta+1} \\ &= \aleph_\beta^{\aleph_\alpha} \cdot \aleph_{\beta+1} \cdot 2^{\aleph_{\alpha+1}} \\ &= \aleph_{\beta+1}^{\aleph_\alpha} \cdot 2^{\aleph_{\alpha+1}} \end{align*}$$ Now suppose $\beta$ is a limit ordinal, and for all $\gamma < \beta$ we have $\aleph_\gamma^{\aleph_{\alpha+1}} = \aleph_\gamma^{\aleph_\alpha} \cdot 2^{\aleph_{\alpha+1}}$. By Theorem 5.20, there are three possible cases ((i) is not possible as $\beta > \alpha + 1$).
- If $\aleph_\beta^{\aleph_{\alpha+1}} = \aleph_\gamma^{\aleph_{\alpha+1}}$ for some $\gamma < \beta$, then: $$ \begin{align*} \aleph_\beta^{\aleph_{\alpha+1}} = \aleph_\gamma^{\aleph_{\alpha+1}} = \aleph_\gamma^{\aleph_\alpha} \cdot 2^{\aleph_{\alpha+1}} \leq \aleph_\beta^{\aleph_\alpha} \cdot 2^{\aleph_{\alpha+1}} \end{align*}$$
- If $\cf(\aleph_\beta) > \aleph_{\alpha+1}$, then: $$ \begin{align*} \aleph_\beta^{\aleph_{\alpha+1}} = \aleph_\beta \leq \aleph_\beta^{\aleph_\alpha} \cdot 2^{\aleph_{\alpha+1}} \end{align*}$$
- If $\cf(\aleph_\beta) \leq \aleph_{\alpha+1}$, then since $\aleph_{\alpha+1} < \aleph_\beta$: $$ \begin{align*} \aleph_\beta^{\aleph_{\alpha+1}} = \aleph_\beta^{\cf(\aleph_\beta)} \leq \aleph_\beta^{\aleph_\alpha} \leq \aleph_\beta^{\aleph_\alpha} \cdot 2^{\aleph_{\alpha+1}} \end{align*}$$
$\blacksquare$
Exercise 5.21
Solution
Suppose $\kappa$ is regular and limit. Clearly $\kappa^{<\kappa} \geq 2^{<\kappa}$. On the other hand, let $\lambda < \kappa$. Since $\kappa$ is regular, we have: $$ \begin{align*} \kappa^\lambda = \sum_{\mu < \kappa} \mu^\lambda \leq \sum_{\mu < \kappa} 2^{\mu^+} = 2^{<\kappa} \end{align*}$$ Suppose $\kappa$ is regular and strong limit. Clearly $\kappa^{<\kappa} \geq 2^{<\kappa}$. On the other hand, by above we have $\kappa^{<\kappa} = 2^{<\kappa}$. Since $\kappa$ is strong limit, we have $2^\lambda < \kappa$ for all $\lambda < \kappa$. Thus $2^{<\kappa} \leq \kappa$.
Suppose $\kappa$ is regular and limit. Clearly $\kappa^{<\kappa} \geq 2^{<\kappa}$. On the other hand, let $\lambda < \kappa$. Since $\kappa$ is regular, we have: $$ \begin{align*} \kappa^\lambda = \sum_{\mu < \kappa} \mu^\lambda \leq \sum_{\mu < \kappa} 2^{\mu^+} = 2^{<\kappa} \end{align*}$$ Suppose $\kappa$ is regular and strong limit. Clearly $\kappa^{<\kappa} \geq 2^{<\kappa}$. On the other hand, by above we have $\kappa^{<\kappa} = 2^{<\kappa}$. Since $\kappa$ is strong limit, we have $2^\lambda < \kappa$ for all $\lambda < \kappa$. Thus $2^{<\kappa} \leq \kappa$.
$\square$
Exercise 5.22
Solution
Again, clearly $\kappa^{<\kappa} \geq 2^{<\kappa}$. On the other hand, let $\lambda < \kappa$. Since $\kappa$ is not a strong limit, there exists $\mu < \kappa$ such that $\kappa \leq 2^\mu$. Then $\kappa^\lambda \leq (2^\mu)^{\lambda} = 2^{\mu \cdot \lambda} \leq 2^{<\kappa}$, so $\kappa^{<\kappa} \leq 2^{<\kappa}$.
For the second inequality, it follows from Corollary 5.14 that $\kappa < \kappa^{\cf(\kappa)} \leq \kappa^{<\kappa}$.
Again, clearly $\kappa^{<\kappa} \geq 2^{<\kappa}$. On the other hand, let $\lambda < \kappa$. Since $\kappa$ is not a strong limit, there exists $\mu < \kappa$ such that $\kappa \leq 2^\mu$. Then $\kappa^\lambda \leq (2^\mu)^{\lambda} = 2^{\mu \cdot \lambda} \leq 2^{<\kappa}$, so $\kappa^{<\kappa} \leq 2^{<\kappa}$.
For the second inequality, it follows from Corollary 5.14 that $\kappa < \kappa^{\cf(\kappa)} \leq \kappa^{<\kappa}$.
$\square$
Exercise 5.23
Solution
Since $\kappa$ is a strong limit: $$ \begin{align*} \kappa = \sup\lbrace \lambda : \lambda < \kappa\rbrace \leq \underbrace{\sup\lbrace 2^\lambda : \lambda < \kappa\rbrace}_{=2^{<\kappa}} \leq \kappa \end{align*}$$ For the second equality, we note that for $\lambda < \kappa$ and $\mu < \kappa$, we have $\mu^\lambda \leq 2^{\mu^+} < \kappa$. Thus for $\lambda \geq \cf(\kappa)$ we have $\kappa^\lambda = \kappa^{\cf(\kappa)}$ by Theorem 5.20. This implies that $\kappa^{<\kappa} = \kappa^{\cf(\kappa)}$.
Since $\kappa$ is a strong limit: $$ \begin{align*} \kappa = \sup\lbrace \lambda : \lambda < \kappa\rbrace \leq \underbrace{\sup\lbrace 2^\lambda : \lambda < \kappa\rbrace}_{=2^{<\kappa}} \leq \kappa \end{align*}$$ For the second equality, we note that for $\lambda < \kappa$ and $\mu < \kappa$, we have $\mu^\lambda \leq 2^{\mu^+} < \kappa$. Thus for $\lambda \geq \cf(\kappa)$ we have $\kappa^\lambda = \kappa^{\cf(\kappa)}$ by Theorem 5.20. This implies that $\kappa^{<\kappa} = \kappa^{\cf(\kappa)}$.
$\square$
Exercise 5.24
Solution
$$ \begin{align*} 2^{\aleph_0} \leq \aleph_\omega^{\aleph_0} \leq (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0} \end{align*}$$
$$ \begin{align*} 2^{\aleph_0} \leq \aleph_\omega^{\aleph_0} \leq (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0} \end{align*}$$
$\square$
Exercise 5.25
Solution
By Exercise 5.18: $$ \begin{align*} \aleph_\omega^{\aleph_0} \leq \aleph_{\omega_1}^{\aleph_1} \leq (\aleph_\omega^{\aleph_0})^{\aleph_1} = \aleph_\omega^{\aleph_1} = \aleph_\omega^{\aleph_0} \cdot 2^{\aleph_1} = \aleph_\omega^{\aleph_0} \cdot \aleph_2 = \aleph_\omega^{\aleph_0} \end{align*}$$
By Exercise 5.18: $$ \begin{align*} \aleph_\omega^{\aleph_0} \leq \aleph_{\omega_1}^{\aleph_1} \leq (\aleph_\omega^{\aleph_0})^{\aleph_1} = \aleph_\omega^{\aleph_1} = \aleph_\omega^{\aleph_0} \cdot 2^{\aleph_1} = \aleph_\omega^{\aleph_0} \cdot \aleph_2 = \aleph_\omega^{\aleph_0} \end{align*}$$
$\square$
Exercise 5.26
Solution
We have $\gimel(\aleph_\omega) = \aleph_\omega^{\aleph_0}$, and: $$ \begin{align*} 2^{\aleph_0} = \aleph_\omega^{\aleph_0} \leq \aleph_{\omega_1}^{\aleph_0} \leq (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0} \end{align*}$$ For second part, we have $\gimel(\aleph_{\omega_1}) = \aleph_{\omega_1}^{\aleph_1}$, and: $$ \begin{align*} 2^{\aleph_1} = \aleph_{\omega_1}^{\aleph_1} \leq (2^{\aleph_0})^{\aleph_1} = 2^{\aleph_1} \end{align*}$$
We have $\gimel(\aleph_\omega) = \aleph_\omega^{\aleph_0}$, and: $$ \begin{align*} 2^{\aleph_0} = \aleph_\omega^{\aleph_0} \leq \aleph_{\omega_1}^{\aleph_0} \leq (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0} \end{align*}$$ For second part, we have $\gimel(\aleph_{\omega_1}) = \aleph_{\omega_1}^{\aleph_1}$, and: $$ \begin{align*} 2^{\aleph_1} = \aleph_{\omega_1}^{\aleph_1} \leq (2^{\aleph_0})^{\aleph_1} = 2^{\aleph_1} \end{align*}$$
$\square$
Exercise 5.27
Solution
If $2^{\aleph_1} = \aleph_2$ then by Exercise 5.18 we have $\aleph_\omega^{\aleph_0} = \aleph_\omega^{\aleph_1}$. Thus, if $\aleph_\omega^{\aleph_1} = \aleph_{\omega_1}$ then: $$ \begin{align*} \cf(\aleph_\omega^{\aleph_1}) = \cf(\aleph_{\omega_1}) = \omega_1 \end{align*}$$ contradicting Corollary 5.13.
If $2^{\aleph_1} = \aleph_2$ then by Exercise 5.18 we have $\aleph_\omega^{\aleph_0} = \aleph_\omega^{\aleph_1}$. Thus, if $\aleph_\omega^{\aleph_1} = \aleph_{\omega_1}$ then: $$ \begin{align*} \cf(\aleph_\omega^{\aleph_1}) = \cf(\aleph_{\omega_1}) = \omega_1 \end{align*}$$ contradicting Corollary 5.13.
$\square$
Exercise 5.28
Solution
$$ \begin{align*} \gimel(\kappa) = \kappa^{\cf(\kappa)} \leq (\lambda^{\cf(\lambda)})^{\cf(\kappa)} \leq (\lambda^{\cf(\lambda)})^{\cf(\lambda)} = \lambda^{\cf(\lambda)} = \gimel(\lambda) \end{align*}$$
$$ \begin{align*} \gimel(\kappa) = \kappa^{\cf(\kappa)} \leq (\lambda^{\cf(\lambda)})^{\cf(\kappa)} \leq (\lambda^{\cf(\lambda)})^{\cf(\lambda)} = \lambda^{\cf(\lambda)} = \gimel(\lambda) \end{align*}$$
$\square$
Exercise 5.29
Solution
By Theorem 5.20(ii), we have that $\gimel(\kappa) = \kappa^{\cf(\kappa)} = \lambda^{\cf(\kappa)}$. If $\mu < \lambda$ and $\mu^{\cf(\kappa)} \geq \lambda$, then $\mu^{\cf(\kappa)} = \lambda^{\cf(\kappa)}$, contradicting minimality of $\lambda$. Thus, we must have that $\mu^{\cf(\kappa)} < \lambda$ for all $\mu < \lambda$.
Now observe that we can't have $\cf(\lambda) > \cf(\kappa)$ - otherwise, by Theorem 5.20(iii) we have $\lambda^{\cf(\kappa)} = \lambda < \kappa$, a contradiction. Thus, by the same theorem, $\cf(\lambda) \leq \cf(\kappa)$ implies that $\lambda^{\cf(\kappa)} = \lambda^{\cf(\lambda)} = \gimel(\lambda)$, as desired.
By Theorem 5.20(ii), we have that $\gimel(\kappa) = \kappa^{\cf(\kappa)} = \lambda^{\cf(\kappa)}$. If $\mu < \lambda$ and $\mu^{\cf(\kappa)} \geq \lambda$, then $\mu^{\cf(\kappa)} = \lambda^{\cf(\kappa)}$, contradicting minimality of $\lambda$. Thus, we must have that $\mu^{\cf(\kappa)} < \lambda$ for all $\mu < \lambda$.
Now observe that we can't have $\cf(\lambda) > \cf(\kappa)$ - otherwise, by Theorem 5.20(iii) we have $\lambda^{\cf(\kappa)} = \lambda < \kappa$, a contradiction. Thus, by the same theorem, $\cf(\lambda) \leq \cf(\kappa)$ implies that $\lambda^{\cf(\kappa)} = \lambda^{\cf(\lambda)} = \gimel(\lambda)$, as desired.
$\square$