Subsets of the rationals order isomorphic to transfinite ordinals

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Let $\langle\mathbb Q,\lt\rangle$ be the set of rational numbers with the usual $\lt$ ordering.

Find subsets $A$ and $B$ of $\mathbb Q$ which, with the usual ordering, are order isomorphic to the following ordinal arthimetic expressions:

(i) $\omega^2+\omega$

(ii) $2.\omega^3$

Idea: So I think I need to find order preserving bijections between the expressions and subsets of Q. I can see that $Z=\{\frac1n:n \in \mathbb N\}$ is order isomorphic to $\omega$, but I don't see how to deal with isomorphism to powers of $\omega$. I'm aware that a way of thinking of $\omega^2$ is as $\omega$ copies of $\omega$

Solution: (i) So I'm looking for $\omega$ many increasing sequences following on the idea in the comments.

Let $F:[0,\infty) \rightarrow [0,1)$ be a linearly piecewise map such that

$F:[0,1] \rightarrow [0,\frac 12)$

$F:[1,2] \rightarrow [\frac 12,\frac 23)$ and

$F:[n,n+1] \rightarrow [{n \over n+1},{n+1 \over n+2})$

Now let $A^i$={$i- \frac 1n: n \in \mathbb N$} for $i \ge 1$

So each $A^i$ is an increasing sequence of order type $\omega$

Let $Z= \bigcup_{i \in \mathbb N}F"(A^i)$

Then Z has order tupe $ \omega^2$ where $F"(A^i)$ is the range of $F$ on $A^i$.

So $A = Z \cup$ {$k:k \ge 1 \land k \in \mathbb N$} is order isomorphic to $\omega^{2}+\omega$

(ii) Let $C^l$ = {$k+l:k \in B,l \in$ {odds}}, then $\bigcup_{k \in {odds}}C^k$ has order type $\omega^3$

and $\bigcup_{k \in {odds}}C^k \cup \bigcup_{k \in {evens}}C^k$ is order isomorphic to $2.\omega^3$

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HINT: If $\omega$ is isomorphic to an increasing sequence, and $\omega+\omega=2\cdot\omega$ is two increasing sequences one after another, and $\omega^2$ is $\omega$ copies of $\omega$, what would that be?

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$\{\frac{1}{n}\}$ is not order isomorphic to $\omega$. Rather $\mathbb{N}\subseteq\mathbb{Q}$ is order isomorphic to $\omega$. For $(i)$, you can choose, for each $n\in \mathbb{N}$ an increasing sequence $a_{ni}$ of rationals that are contained in $(-\frac{1}{n}, -\frac{1}{n+1})$ and converge to $-\frac{1}{n+1}$ (the last assumption is not necessary but let's make our order-isomorphism continuous). Then $$ \{a_{ni}:\,\,i,n\in\mathbb{N}\}\cup \{-\frac{1}{n}:\,n\in\mathbb{N}\}\cup\mathbb{N_0} $$ can do the job. Probably you can construct a set for $(ii)$ using similar ideas.