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The only reasoning I've seen given for this is that it's uncountable because it can't include itself an element. I'm a little unconvinced and was looking for a more proper formal proof demonstrating the equality: $\omega_1 = \aleph_1$.

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I feel that I may have answered this once or twice before on this site. –  Asaf Karagila Sep 4 '12 at 12:41
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$\omega_1$ is usually defined to be the the least uncountable cardinal. So it is uncountable by definition. Thus every ordinal in $\omega_1$ is countable. Moreover any countable ordinal $\alpha$ cannot be larger than or equal to $\omega_1$ and so $\alpha \in \omega_1$. Thus $\omega_1$ is the set of countable ordinals. –  Levon Haykazyan Sep 4 '12 at 12:42
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3 Answers 3

up vote 3 down vote accepted

Following Henning's suggestion, posting my comment as an answer.

$\omega_1$ is usually defined to be the the least uncountable cardinal. So it is uncountable by definition. Thus every ordinal in $\omega_1$ is countable. Moreover any countable ordinal $\alpha$ cannot be larger than or equal to $\omega_1$ and so $\alpha \in \omega_1$. Thus $\omega_1$ is the set of countable ordinals.

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I'm a little confused. If $\omega_1$ is defined to be the least uncountable cardinal, I can't see/understand exactly how that gives you as a consequence that every ordinal in $\omega_1$ is countable (you used the word 'thus'). Can you please be more clear? I assume you're not using the hypothesis in the question there that every countable ordinal is in $\omega_1$ because then it becomes circular reasoning to prove the hypothesis using the the hypothesis. –  Acid2 Sep 4 '12 at 18:06
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@Bakhtiar: For ordinals it holds that $\alpha<\beta\iff\alpha\in\beta$. If $\omega_1$ is the first uncountable ordinal it means that $\alpha<\omega_1$ implies that $\alpha$ is not uncountable (else it would be the first, preceding $\omega_1$). –  Asaf Karagila Sep 5 '12 at 0:12
    
@AsafKaragila: Okay, I see, we're viewing the least uncountable cardinal as an ordinal (the first uncountable ordinal) and then it follows every ordinal in it ($\omega_1$) is countable... makes sense. Thanks. –  Acid2 Sep 5 '12 at 12:54

The reasoning is correct: The set $\omega_1$ of all countable ordinals (here always meant as: including finite ones) is an ordinal, hence we have $\omega_1\notin\omega_1$ (even in set theories that allow Quine atoms), hence it is not a countable ordinal. Since it is definitely an ordinal, it must be an uncountable ordinal.

EDIT: removed bad reason for $\le$ argument

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$\omega_1\setminus\omega$ is a proper subset of $\omega_1$ whose cardinailty if $\aleph_1$. –  Henning Makholm Sep 4 '12 at 12:43

Why is this not a reasonable reason?

Ordinals are sets which are transitive and well-ordered by $\in$. $\omega$ is defined to be the first infinite cardinal, $\omega_1$ is defined to be the first ordinal which is larger than $\omega$ not in bijection with any of its members.

The definition of $\omega_1$, then, implies that $(1)$ There is no bijection between $\omega_1$ and $\omega$, that is to say that $\omega_1$ is uncountable; and $(2)$ that every ordinal below $\omega_1$ has to be countable. Since $\in$ is well-founded we have that $\omega_1\notin\omega_1$.

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