I am self-learning about ordinal numbers using Hrbacek and Jech's book. On page 123 of the second edition, exercise 5.11 asks:
Find a set $A$ of rational numbers such that $(A,\leq_{\mathbb{Q}})$ is isomorphic to $(\alpha,\leq)$ where
- $\alpha=\omega+1$,
- $\alpha=\omega\cdot2$,
- $\alpha=\omega\cdot3$,
- $\alpha=\omega^{\omega}$,
- $\alpha=\varepsilon$.
[Hint:$\{n-1/m\,|\,n,m\in\mathbb{N}-\{0\}\}$ is isomorphic to $\omega^{2}$, etc]
I have solved parts 1-3. Indeed, the set $\{1-\frac{1}{m}\,|\,m\in\mathbb{N}-\{0\}\}\cup\{1\}$ is isomorphic to $\omega+1$, and for $k\in\mathbb{N}-\{0\}$, the set $$S_{k}=\bigcup_{i=1}^{k}\left\{i-\frac{1}{m}\,|\,n\in\mathbb{N}-\{0\}\right\}$$ is isomorphic to $\omega\cdot k$. Also, the set $\bigcup_{k\in\mathbb{N}}S_{k}$ is simply the set provided in the hint and it is isomorphic to $\omega^{2}$.
How do I use the hint provided in order to solve parts 4 and 5?
I edited the question to add a possible solution. I post it now as an answer in order to remove the question from the unanswered list.
For part 4, let $A_{2}$ denote the set provided in the hint. Now, for $k\ge 1$ let $f_{k}:[0,\infty)\rightarrow[k-1,k)$ be the bijection defined by $x\mapsto k+\frac{1}{1+x}$. Then the set $A_{3}=\bigcup_{k\ge 1}f_{k}(A_{2})$ is isomorphic to $\omega^{3}$. More generally, for $r\ge 3$, the set $A_{r}=\bigcup_{k\ge 1}f_{k}(A_{r-1})$ is isomorphic to $\omega^{r}$. Therefore, the set $B_{1}=\bigcup_{r\ge 2}A_{r}$ will be isomorphic to $\omega^{2}+\omega^{3}+\cdots\omega^{n}+\cdots = \omega^{\omega}$.
Now, we would like to construct a set $B_{2}\subset\mathbb{Q}$ isomorphic to $\omega^{\omega^{\omega}}$. Indeed, let $C_{1}^{1}=B_{1}$, and for $r\ge 2$ let $C_{r}^{1}=\bigcup_{k\ge 1}f_{k}(C_{r-1}^{1})$. The set $C_{1}^{2} = \bigcup_{r\ge 1}C_{r}^{1}$ will be isomorphic to $\omega^{\omega\cdot 2}$. Continuing, we can construct for each $m\ge 1$, a set $C_{1}^{m}$ isomorphic to $\omega^{\omega\cdot m}$. Then, the set $D_{1} = \bigcup_{m\ge 1}C_{1}^{m}$ will be isomorphic to $\omega^{\omega^{2}}$. In a similar fashion, we could keep iterating the process in order to obtain sets $D_{k}$, each of which will be isomorphic to $\omega^{\omega^{k+1}}$, $k\geq 1$. Then the set $B_{2}$ will simply be $\bigcup_{k\ge 1}D_{k}$.
We could continue, obtaining for each $r\ge 1$ a set $B_{r}$ isomorphic to $\omega\uparrow\uparrow (r+1)=\underbrace{\omega^{\omega^{\,.\,^{.\,^{.\,^{\omega}}}}}}_{r+1}$. Therefore, since $1+\omega+\omega^{\omega}+\omega^{\omega^{\omega}}+\cdots=\varepsilon_{0}$, the set $\bigcup_{r\ge 1}B_{r}$ will be isomorphic to $\varepsilon_{0}$.
Something like this seems what the authors may have intended, since the Cantor Normal Form in not discussed until the following section.