The set of rational numbers has continuum many subsets that are order-wise inequivalent

Question

How to show that there are continuum many subsets of $\mathbb Q$, no two of which are similar?

Two sets are called similar if there is an order-preserving bijection between them.

Answer 1

There are several possible ways to do this; here's one. Given $f:\mathbb{N}\to\{0,1\}$, consider the ordered set $$S_f=\mathbb{Z}+f(0)+\mathbb{Z}+f(1)+\mathbb{Z}+f(2)+\cdots$$ Here $+$ is the sum of ordered sets (like ordinal addition), and we think of the values $f(n)$ as ordered sets by letting $0$ be the empty set and $1$ be the singleton ordered set. Then each $S_f$ is a countable totally ordered set, and it is not hard to show that $S_f\cong S_g$ iff $f=g$. Since every countable totally ordered set embeds in $\mathbb{Q}$ (alternatively, it is easy to show directly that each $S_f$ embeds in $\mathbb{Q}$), you get continuum many dissimilar subsets of $\mathbb{Q}$.