Let $\omega_0$ be the first countable ordinal and $\omega_1$ be the first uncountable ordinal. Use $[0, \omega_1)$ to denote the set of all countable ordinals and hence $\omega_1 = [0, \omega_1)$. Could someone construct a set (other than the set of ordinals) that is order isomorphic to the set $[\omega_0, \omega_1)$. In this post, Arthur (in the top post) provided an example, which is the following:
For each $\alpha\in[0, \omega_1)$, let $X_{\alpha}$ be the set of subsets of $\mathbb{Q}$ , each of which is order isomorphic to $\alpha$, and define $X_{\alpha} < X_{\beta}$ iff $\alpha < \beta$. Then it is claimed that the following set $A$, equipped with the order "<"
$$A = \{X_{\alpha}\,\vert\,\alpha\in[0, \omega_1)\}\hspace{1cm}(\subseteq\mathcal{P}[\mathcal{P}(\mathbb{Q})])$$
is order isomorphic to $[0, \omega_1)$. If we can find a uncountable family of sets (ideally not subsets of the set of ordinals) so that the supremum is not in the family and no countable union of elements will be equal to the supremum, will that work for this question?
The map $f(x) = \omega_0 + x$ is an order isomorphism $[0, \omega_1) \to [\omega_0, \omega_1)$.
First, let's prove that it's a bijection. Note that in general, if we have ordinals $a \geq b$, then there exists a unique $c$ such that $a = b + c$. So for all ordinals $a \in [\omega_0, \omega_1)$, there exists a unique $c$ such that $\omega_0 + c = a$. This proves that $f$ is a bijection.
From here, we must show that $f$ is an order isomorphism. Suppose that $x \leq y$. Then write $y = x + z$. Then $f(y) = \omega_0 + y = \omega_0 + x + z = f(x) + z$, so $f(x) \leq f(y)$. The fact that both are total orders proves it's an order isomorphism and not just order-preserving.
Obviously, this result generalises a lot.