Why is $\Omega_0=[1,\omega_1)$ called the set of countable ordinals

Let $\omega_1$ be the first uncountable ordinal. In some book, the set $\Omega_0:=[1,\omega_1)=[1,\omega_1]\backslash\{\omega_1\}$ is called the set of countable ordinals. Why? It is obvious that it is an uncountable set, because $[1,\omega_1]$ is uncountable. The most possible reason I think is that for any $x\prec \omega_1$, the set $[1,x)$ is countable.

• An ordinal is a set. Sets are either countable or uncountable. The countable ordinals are precisely those $<\omega_1$. – Lord Shark the Unknown Aug 16 '17 at 20:14
• $0$ is an ordinal too.. and $\Omega_0 = [0,\omega_1)$. – Henno Brandsma Aug 16 '17 at 20:28
• @HennoBrandsma In the book I read, it uses "1" as the first element. The book is the "infinite dimensional analysis: A Hitchhiker’s Guide" – Yuhang Aug 16 '17 at 20:30
• Topologically it's the same (just one extra isolated point at the start) but set-theoretically it's wrong, I'd say. – Henno Brandsma Aug 16 '17 at 20:31

It is just like $\omega$ being the set of all finite ordinals. Every member of $\omega$ is finite but $\omega$ itself is infinite. Similarly, $\Omega_0$ is uncountable but all its members are (finite or) countable.
Every ordinal is a well-ordered set, and the first one that is uncountable is by definition $\omega_1$. So this set which is uncountable because there are uncountably many different ways to well-order a countable set, is thus called "the set of all countable ordinals". By definition, all $\alpha < \omega_1$ are countable sets, as otherwise they'd contradict the minimality of $\omega_1$, analogous, to $\omega$ being the set of finite ordinals (despite not being finite itself).
$[0,\omega_1]$, including the maximal element $\omega_1$, is the unique compactification of $\omega_1$, also denoted $\omega_1 + 1$. It has one element more than $\omega_1$ so is also uncountable. By the usual definition, every ordinal is the set of all strictly smaller ordinals.
The set $\Omega=[0,\omega_1)$ is a set of countable ordinals because every element of $\Omega$ is a countable ordinal. To see this, suppose that $x\in\Omega$ is not a countable ordinal. Since $x$ is an ordinal, it follows that $x$ is an uncountable ordinal; but $x\lt\omega_1,$ contradicting the fact that $\omega_1$ is the first uncountable ordinal.
In fact, $\Omega_0=\omega_1$ (EDIT: with $0$ removed): remember that each ordinal is the set of all smaller ordinals. So in particular, since $\omega_1$ is the smallest uncountable ordinal, it is also the set of all countable ordinals (since each countable ordinal is smaller than $\omega_1$).