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Exercise 2.1.
Solution. Suppose $f : P \to Q$ and $g : Q \to R$ are two isomorphisms.Reflexive: The identity $\id : P \to P$ is clearly an isomorphism.
Symmetry: By definition of an isomorphism, it's easy to see if $f : P \to Q$ is an isomorphism then so is $f^{-1} : Q \to P$.
Transitive: Composition of isomorphisms are isomorphisms, as they remain bijective and:
$$ \begin{align*}
x < y \implies f(x) < f(y) \implies g(f(x)) < g(f(y))
\end{align*}$$
$\square$
Exercise 2.2.
Solution. Note that $\beta < \alpha$ iff $\beta + 1 \leq \alpha$.$\implies$: Suppose $\beta \in \alpha$. Since $\alpha = \bigcup \alpha$, $\beta \in \gamma$ for some $\gamma \in \alpha$. Then $\beta + 1 \subseteq \gamma$. Since $\gamma \in \alpha$, $\beta + 1 \in \alpha$.
$\impliedby$: If $\beta \in \bigcup \alpha$, then $\beta \in \gamma$ for some $\gamma \in \alpha$. By transitivity of ordinals, $\gamma \subseteq \alpha$, so $\beta \in \alpha$. Thus, $\bigcup \alpha \subseteq \alpha$ (we have not used the hypothesis).
On the other hand, let $\beta \in \alpha$. By hypothesis, $\beta + 1 \in \alpha$, so $\beta + 1 \subseteq \bigcup\alpha$. Hence $\beta \in \bigcup\alpha$.
$\square$
Exercise 2.3.
Solution. We shall show that if $A$ and $B$ are inductive, then so is $A \cap B$. Since $\boldsymbol{\mathrm{ORD}}$ is an inductive class, it follows that $X \cap \boldsymbol{\mathrm{ORD}}$ is also inductive.Clearly $\emptyset \in A \cap B$. Let $x \in A \cap B$. Then since $A$ and $B$ are inductive, $x \cup \lbrace x\rbrace \in A$ and $x \cup \lbrace x\rbrace \in B$, so $x \cup \lbrace x\rbrace \in A \cap B$. Thus $A \cap B$ is inductive.
Since $\boldsymbol{N}$ is the smallest inductive set, $\boldsymbol{N} = \boldsymbol{N} \cap \boldsymbol{\mathrm{ORD}}$, i.e. $\boldsymbol{N} \subseteq \boldsymbol{\mathrm{ORD}}$. Since $\boldsymbol{N}$ is transitive by Exercise 1.3, and is well-ordered, it is an ordinal.
Finally, to see that $\boldsymbol{N}$ is a limit ordinal, suppose not, so $\boldsymbol{N} = n \cup \lbrace n\rbrace$ for some $n \in \boldsymbol{N}$. By the inductive property of $\boldsymbol{N}$, we have that $\boldsymbol{N} = n \cup \lbrace n\rbrace \in \boldsymbol{N}$, contradicting the Axiom of Regularity.
$\square$
Exercise 2.4.
Solution. (i) $\implies$ (ii): If $X$ is inductive, then $\boldsymbol{N} \subseteq X$. Then there exists an injective map $\boldsymbol{N} \to X$, which is the inclusion map. Then $\vert \boldsymbol{N}\vert \leq \vert X\vert $, so $X$ is infinite.(ii) $\implies$ (iii): Let $X$ be an infinite set. Consider the map on $[X]^{<\omega}$ (the set of all finite subsets of $X$) defined by:
$$ \begin{align*}
Y \mapsto \vert Y\vert
\end{align*}$$ This is well-defined, as $\vert Y\vert $ is well-defined for finite sets without Axiom of Infinity. By Axiom of Replacement, the image of this map is a set. Since $X$ is infinite, we have that for all $n \in \omega$, there exists $Y \subseteq X$ such that $\vert Y\vert = n$. Then $Y \mapsto n$, the image of this map is $\omega$. Hence $\omega$ is a set.
(iii) $\implies$ (i): $\omega$ is inductive by definition.
$\square$
Exercise 2.5.
Solution. If such a sequence exists, then $\lbrace a_n : n \in \boldsymbol{N}\rbrace \subseteq \boldsymbol{N}$ has no minimal element, contradicting the definition of well-order.$\square$
Exercise 2.6.
Solution. For any ordinal $\alpha$, consider the sequence $\alpha_0 := \alpha$ and $\alpha_{n+1} := \alpha_n + 1$. Let $\beta := \lim_{n \to \omega} \alpha_n$. Then $\beta$ is a limit of a (strictly) increasing sequence of ordinals, and is hence a limit ordinal (as $\bigcup\beta = \beta$).$\square$
Exercise 2.7.
Solution. For any $\alpha_0 \in \boldsymbol{\mathrm{ORD}}$, define $\alpha_{n+1} = \gamma_{\alpha_n}$ as in the hint. Let $\alpha = \lim_{n \to \omega} \alpha_n$. Then, by normality of the sequence:$$ \begin{align*}
\gamma_\alpha = \lim_{n \to \omega} \gamma_{\alpha_n} = \lim_{n \to \omega} \alpha_{n+1} = \alpha
\end{align*}$$
$\square$
Exercise 2.8.
Lemma 2.8.A. If $\beta$ is a limit ordinal, then so are $\alpha + \beta$, $\alpha \cdot \beta$ and $\alpha^\beta$.
Proof. Let $F(\alpha,\beta)$ denote $\alpha + \beta$, $\alpha \cdot \beta$ or $\alpha^\beta$. In all three cases, we have $F(\alpha,\beta) = \lim_{\xi \to \beta} F(\alpha,\xi)$. Since $F(\alpha,\beta)$ is a limit ordinal iff $F(\alpha,\beta) = \lim_{\delta \to F(\alpha,\beta)} \delta$, it suffices to show that for all $\delta < F(\alpha,\beta)$, there exists some $\xi < \beta$ such that $\delta \leq F(\alpha,\xi$). But this is obviously true, for if $\delta > F(\alpha,\xi)$ for all $\xi < \beta$ then $\delta \geq \lim_{\xi \to \beta} F(\alpha,\xi) = F(\alpha,\beta)$ by definition.
$\blacksquare$
Exercise 2.8(i).
Solution. We induct on $\gamma$. For $\gamma = 0$, we have:$$ \begin{align*}
\alpha \cdot (\beta + 0) = \alpha \cdot \beta = \alpha \cdot \beta + \alpha \cdot 0
\end{align*}$$ Suppose $\gamma = \delta + 1$ and $\alpha \cdot (\beta + \delta) = \alpha \cdot \beta + \alpha \cdot \delta$. Then:
$$ \begin{align*}
\alpha \dot (\beta + (\delta + 1)) &= \alpha \dot (\beta + (\delta + 1)) \\
&= \alpha \cdot ((\beta + \delta) + 1) \\
&= \alpha \cdot (\beta + \delta) + \alpha \\
&= (\alpha \cdot \beta + \alpha \cdot \delta) + \alpha && \text{by induction hypothesis} \\
&= \alpha \cdot \beta + (\alpha \cdot \delta + \alpha) \\
&= \alpha \cdot \beta + \alpha \cdot (\delta + 1)
\end{align*}$$ If $\gamma$ is a limit ordinal, then note that $\beta + \gamma$ is also a limit ordinal by Lemma 2.8.A. Thus:
$$ \begin{align*}
\alpha \cdot (\beta + \gamma) &= \alpha \cdot \lim_{\xi \to \gamma} (\beta + \xi) \\
&= \lim_{\xi \to \gamma} \alpha \cdot (\beta + \xi) \text{} && \text{as $\beta + \gamma$ is limit} \\
&= \lim_{\xi \to \gamma} (\alpha \cdot \beta + \alpha \cdot \xi) && \text{by induction hypothesis} \\
&= \alpha \cdot \beta + \lim_{\xi \to \gamma} \alpha \cdot \xi &&\text{as $\alpha \cdot \gamma$ is limit, by Lemma 2.8.A} \\
&= \alpha \cdot \beta + \alpha \cdot \gamma
\end{align*}$$
$\square$
Exercise 2.8(ii).
Solution. We induct on $\gamma$. For $\gamma = 0$, we have:$$ \begin{align*}
\alpha^\beta \cdot \alpha^0 = \alpha^\beta \cdot 1 = \alpha^\beta = \alpha^{\beta + 0}
\end{align*}$$ Suppose $\gamma = \delta + 1$ and $\alpha^{\beta+\gamma} = \alpha^\beta \cdot \alpha^\gamma$. Then:
$$ \begin{align*}
\alpha^{\beta + (\delta + 1)} &= \alpha^{(\beta + \delta) + 1} \\
&= \alpha^{\beta + \delta} \cdot \alpha \\
&= (\alpha^\beta \cdot \alpha^\delta) \cdot \alpha && \text{by induction hypothesis} \\
&= \alpha^\beta \cdot (\alpha^\delta \cdot \alpha) \\
&= \alpha^\beta \cdot \alpha^{\delta+1}
\end{align*}$$ Suppose $\gamma$ is a limit ordinal. Then $\alpha^\gamma$ and $\alpha^{\beta+\gamma}$ is a limit ordinal, so:
$$ \begin{align*}
\alpha^{\beta+\gamma} &= \lim_{\xi \to \gamma} \alpha^{\beta+\xi} \\
&= \lim_{\xi \to \gamma} \alpha^\beta \cdot \alpha^\xi && \text{by induction hypothesis} \\
&= \alpha^\beta \cdot \lim_{\xi \to \gamma} \alpha^\xi && \text{as $\alpha^\gamma$ is a limit ordinal} \\
&= \alpha^\beta \cdot \alpha^\gamma
\end{align*}$$
$\square$
Exercise 2.8(iii).
Solution. We induct on $\gamma$. If $\gamma = 0$, then:$$ \begin{align*}
\alpha^{\beta \cdot 0} = \alpha^0 = 1 = (\alpha^\beta)^0
\end{align*}$$ If $\gamma = \delta + 1$, then:
$$ \begin{align*}
(\alpha^\beta)^{\delta + 1} &= (\alpha^\beta)^\delta \cdot \alpha^\beta \\
&= \alpha^{\beta \cdot \delta} \cdot \alpha^\beta && \text{by induction hypothesis} \\
&= \alpha^{\beta \cdot \delta + \beta} && \text{by Exercise 2.8(ii)} \\
&= \alpha^{\beta \cdot (\delta + 1)}
\end{align*}$$ If $\gamma$ is a limit ordinal, then:
$$ \begin{align*}
(\alpha^\beta)^\gamma &= \lim_{\xi \to \gamma} (\alpha^\beta)^\xi \\
&= \lim_{\xi \to \gamma} \alpha^{\beta\cdot\xi} && \text{by induction hypothesis} \\
&= \alpha^{\beta \cdot \gamma} &&\text{as $\beta \cdot \gamma$ is limit, and therefore so is $\alpha^{\beta \cdot \gamma}$}
\end{align*}$$
$\square$
Exercise 2.9.
Exercise 2.9(i).
Solution. We first note that:$$ \begin{align*}
1 + \omega = \sup_{n < \omega} (1 + n) = \omega
\end{align*}$$ Now we have $(\omega + 1) \cdot 2 = \omega + 1 + \omega + 1 = \omega + \omega + 1 = \omega \cdot 2 + 1$. By Lemma 2.25(i), we have $\omega \cdot 2 + 1 < \omega \cdot 2 + 2$.
$\square$
Exercise 2.9(ii).
Solution. Observe that:$$ \begin{align*}
2 \cdot \omega = \sup_{n < \omega} 2 \cdot n = \omega
\end{align*}$$ We have $(\omega \cdot 2)^2 = \omega \cdot 2 \cdot \omega \cdot 2 = \omega \cdot \omega \cdot 2 = \omega^2 \cdot 2$. By Lemma 2.25(iii), we have that $\omega^2 \cdot 2 < \omega^2 \cdot 4$.
$\square$
Exercise 2.10.
Solution. We induct on $\gamma$.$\alpha + \gamma \leq \beta + \gamma$: If $\gamma = 0$ then this follows from $\alpha < \beta$. If $\gamma = \delta + 1$, then by Lemma 2.25(i) we have that:
$$ \begin{align*}
\alpha + \delta \leq \beta + \delta \implies \alpha + \delta + 1 \leq \beta + \delta + 1 \implies \alpha + \gamma \leq \beta + \gamma
\end{align*}$$ If $\gamma$ is a limit ordinal, then:
$$ \begin{align*}
\alpha + \gamma &= \lim_{\xi \to \gamma} (\alpha + \xi) \\
&\leq \lim_{\xi \to \gamma} (\beta + \xi) && \text{by induction hypothesis} \\
&= \beta + \gamma
\end{align*}$$ $\alpha \cdot \gamma \leq \beta \cdot \gamma$: If $\gamma = 0$ then $\alpha \cdot 0 = 0 = \beta \cdot 0$. If $\gamma = \delta + 1$, then:
$$ \begin{align*}
\alpha \cdot (\delta + 1) &= \alpha \cdot \delta + \alpha \\
&\leq \beta \cdot \delta + \alpha && \text{by above and induction hypothesis} \\
&\leq \beta \cdot \delta + \beta && \text{by $\alpha < \beta$ and Lemma 2.25(i)} \\
&= \beta \cdot (\delta + 1)
\end{align*}$$ If $\gamma$ is a limit ordinal, then:
$$ \begin{align*}
\alpha \cdot \gamma &= \lim_{\xi \to \gamma} \alpha \cdot \xi \\
&\leq \lim_{\xi \to \gamma} \beta \cdot \xi && \text{by induction hypothesis} \\
&= \beta \cdot \gamma
\end{align*}$$ $\alpha^\gamma \leq \beta^\gamma$: If $\gamma = 0$ then $\alpha^0 = 1 = \beta^0$. If $\gamma = \delta + 1$ then by above and Lemma 2.25:
$$ \begin{align*}
\alpha^{\delta+1} &= \alpha^\delta \cdot \alpha \\
&\leq \beta^\delta \cdot \alpha && \text{by above and induction hypothesis} \\
&\leq \beta^\delta \cdot \beta && \text{by $\alpha < \beta$ and Lemma 2.25(iii)} \\
&= \beta^{\delta+1}
\end{align*}$$ If $\gamma$ is a limit ordinal, then:
$$ \begin{align*}
\alpha^\gamma &= \lim_{\xi \to \gamma} \alpha^\xi \\
&\leq \lim_{\xi \to \gamma} \beta^\xi && \text{by induction hypothesis} \\
&= \beta^\gamma
\end{align*}$$
$\square$
Exercise 2.11.
Exercise 2.11(i).
Solution. As we saw in the proof of Exercise 2.9(i), $0 < 1$ but $0 + \omega = \omega = 1 + \omega$.$\square$
Exercise 2.11(ii).
Solution. As we saw in the proof of Exercise 2.9(ii), $1 < 2$ but $1 \cdot \omega = \omega = 2 \cdot \omega$.$\square$
Exercise 2.11(iii).
Solution. $2 < 3$ but:$$ \begin{align*}
2^\omega = \sup_{n < \omega} 2^n = \omega = \sup_{n < \omega} 3^n = 3^\omega
\end{align*}$$
$\square$
Exercise 2.12.
Solution. We first show that $\omega^{\varepsilon_0} = \varepsilon_0$. Indeed:$$ \begin{align*}
\omega^{\varepsilon_0} = \lim_{n \to \omega} \omega^{\alpha_n} = \lim_{n \to \omega} \alpha_{n+1} = \varepsilon_0
\end{align*}$$ Now suppose $\varepsilon$ is another ordinal such that $\omega^\varepsilon = \varepsilon$. Note that $\varepsilon \neq 0$, as $\omega^0 = 1 \neq 0$. Thus, $\varepsilon \geq \omega \alpha_0$. If $\varepsilon \geq \alpha_n$, then by Exercise 2.10:
$$ \begin{align*}
\alpha_{n+1} = \omega^{\alpha_n} \leq \omega^{\varepsilon} = \varepsilon
\end{align*}$$ So $\varepsilon \geq \alpha_n$ for all $n$. Thus $\varepsilon$ is an upper bound of the sequence $\c{\alpha_n : n \in \omega}$, so $\varepsilon_0 \leq \varepsilon$.
$\square$
Exercise 2.13.
Solution. Call those definitions (i), (ii) and (iii) respectively.(i) $\implies$ (ii): Let $\alpha < \gamma$. By Lemma 2.25(ii), there exists some $\delta$ such that $\alpha + \delta = \gamma$. Since $\alpha$ is indecomposable, $\delta \geq \gamma$. If $\delta > \gamma$, then by Lemma 2.25(i) and Exercise 2.10 we have $\alpha + \delta > \alpha + \gamma \geq \gamma$. Thus $\delta = \gamma$.
(ii) $\implies$ (i): For $\alpha < \gamma$ and $\beta < \gamma$, by Lemma 2.25(i) we have $\alpha + \beta < \alpha + \gamma = \gamma$.
(ii) $\implies$ (iii): By Cantor's Normal Form (Theorem 2.26), we may write:
$$ \begin{align*}
\gamma = \omega^{\beta_1} \cdot k_1 + \dots + \omega^{\beta_n} \cdot k_n
\end{align*}$$ where $\gamma \geq \beta_1 > \cdots > \beta_n$, and $k_1,\dots,k_n$ are non-zero natural numbers. Observe that:
$$ \begin{align*}
\gamma + \gamma &= \omega^{\beta_1} \cdot k_1 + \dots + \omega^{\beta_n} \cdot k_n + \gamma \\
&= \omega_{\beta_1} + (\omega^{\beta_1} \cdot (k_1 - 1) + \dots + \omega^{\beta_n} \cdot k_n + \gamma) \\
&= \omega^{\beta_1} + \gamma
\end{align*}$$ as $\omega^{\beta_1} \cdot (k_1 - 1) + \dots + \omega^{\beta_n} \cdot k_n < \gamma$ by uniqueness of Cantor's Normal Form. Thus $\gamma = \omega^{\beta_1}$ (i.e. $n = 1$ and $k_1 = 1$).
(iii) $\implies$ (ii): We induct on $\beta$ where $\gamma = \omega^\beta$. If $\beta = 0$, then $\omega^0 = 1$, so the only $\alpha < \gamma$ is $\alpha = 0$, in which then of course $0 + \omega^\beta = \omega^\beta$.
If $\beta = \delta + 1$, then for $\alpha < \gamma$ we write its Cantor's Normal Form $\alpha = \omega^{\beta_1} \cdot k_1 + \dots + \omega^{\beta_n} \cdot k_n$. Since $\alpha < \gamma$, $\beta_1 < \delta + 1$ so $\beta_1 \leq \delta$. Then:
$$ \begin{align*}
\alpha + \omega^{\delta+1} &= \omega^{\beta_1} \cdot k_1 + \dots + \omega^{\beta_n} \cdot k_n + \omega^\delta \cdot \omega \\
&\leq \omega^\delta \cdot (k_1 + \dots + k_n) + \omega^\delta \cdot \omega \\
&= \omega^\delta \cdot (k_1 + \dots + k_n + \omega) \\
&= \omega^\delta \cdot \omega \\
&= \omega^{\delta+1}
\end{align*}$$ If $\beta$ is a limit ordinal, then for $\alpha < \omega^\beta$ we have that $\alpha < \omega^\xi$ for some $\xi < \beta$. By induction hypothesis, $\alpha + \omega^\xi = \omega^\xi$. Thus we have that $\omega^\beta = \lim_{\xi \to \beta} (\alpha + \omega^\xi) = \lim_{\xi \to \beta} \omega^\xi = \omega^\beta$ for $\alpha < \omega^\beta$.
$\square$
Exercise 2.14.
Solution. Suppose such a sequence exists. Consider $X := \lbrace a_n : n \in \boldsymbol{N}\rbrace \subseteq P$. Since $E$ is well-founded, $X$ has some $E$-minimal element, which is $a_n$ for some $n \in \boldsymbol{N}$. Yet $a_n \; E \; a_{n+1}$, a contradiction.$\square$
Exercise 2.15.
Solution. The proof of Theorem 2.15 can be mimicked. Define the function $F$ such that $F(x) = X$ iff there is a sequence $\c{a_y : y \; E \; x}$ (where $a_x := G(\c{a_y : y \; E \; x})$) such that:- $(\forall y \; E \; x) \, a_y = F(\c{a_z : z \; E \; y})$.
- $x = G(\c{a_y : y \; E \; x})$.
$\square$