Definition of $\omega_1$, comparing it to $2^\mathbb{N}$?

Question

I'm taking an Intro to Topology class, and we just started defining ordinals. We defined finite ordinals as: \begin{align*} 0 & = \varnothing \\ 1 & = \{0\} \\ 2 & = \{0,1\} \\ & \vdots \\ k & = \{0, 1, 2, \ldots, k-1 \} \\ & \vdots \end{align*} Then we defined the first infinite ordinal, $\omega_0$, as the set of all finite ordinals, and clearly $|\omega_0| = |\mathbb{N}|.$

If $\omega_1$ is then defined as the set of all the countable ordinals, how does that imply it is the smallest uncountable ordinal? And, doesn't its definition mean that it is the same size as the power set of $\mathbb{N}$? But that can't be right, because that would mean that $|\omega_1| = |2^\mathbb{N}| \iff |\omega_1| = |\mathbb{R}|$, which is the Continiuum Hypothesis and cannot be proven or disproved.

Answer 1

If $\omega_1$ is countable, then $\omega_1\in\omega_1$ (having the property that it is the set of all countable ordinals), which is impossible because for ordinals $\alpha\notin\alpha$.

And you are absolutely right. The definition doesn't mean that $|\omega_1|=|\Bbb R|$.

Answer 2

$|\omega_1|=\aleph_1$ — that is from the definition. $\omega_1$ — the smallest uncountable ordinal. And $\aleph_1$ is the first cardinal number after $\aleph_0$, which represents countable sets.

And we have $\aleph_0<\aleph_1\leq\mathfrak{c}=2^{\aleph_0}$.
The continuum hypothesis is equivalent to $\aleph_0<\aleph_1=\mathfrak{c}=2^{\aleph_0}$