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

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.

• Since it contains all countable ordinals, it would have to contain itself if it was countable. All the smaller ordinals are elements of $\omega_1$, hence countable - that's why $\omega_1$ is the smallest uncountable one. And I cannot see how is it the same as a power set of $\mathbb{N}$ . $\omega_1$ contains "just" ordinals - including transitive subsets of $\mathbb{N}$ (which happen to by "only" natural numbers), not arbitrary subsets. – Marcin Łoś Feb 22 '14 at 19:32
• @Marcin: Actually, if you want to be correct, either $\Bbb N$ is not $\omega_0$, or every $k\in\omega_0$ is transitive. In either case I agree that $\mathcal P(\omega_0)\cap\omega_1=\omega_0\cup\{\omega_0\}$. – Asaf Karagila Feb 22 '14 at 20:05
• @AsafKaragila also, what does it mean to say $\omega_0 + 1$ or $3\omega_0 + 2$? My instructor said refers to adding one element to $\omega_0$, or taking $3$ copies of $\omega_0$ and then adding $2$ elements, but what does that actually mean? What elements are you adding in? more ordinals? – justin Feb 22 '14 at 20:25
• @AsafKaragila I'm not sure I understand. Every $k\in \omega_0$ is transitive, being an ordinal, isn't it? I suppose you are referring to my forgetting that $\mathbb{N}$ is also subset of $\mathbb{N}$? – Marcin Łoś Feb 22 '14 at 20:32
• justin, the easiest way looking at it, is to remember that a well-ordered set is isomorphic to a unique ordinal, so $\alpha+\beta$ is the unique ordinal that can be partitioned into an initial segment isomorphic to (in fact equal to) $\alpha$, and an end-segment isomorphic to $\beta$; and $\alpha\cdot\beta$ is the unique ordinal isomorphic to the reverse lexicographic order on $\alpha\times\beta$. But you can also define these operations by transfinite recursion. In either case, remember that the elements of an ordinal are themselves ordinals. So "adding new elements" you in fact add ordinals. – Asaf Karagila Feb 22 '14 at 20:34

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 $|\omega_1|=|\Bbb R|$.
$|\omega_1|=\aleph_1$ — 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 $\aleph_0<\aleph_1=\mathfrak{c}=2^{\aleph_0}$