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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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    Your argument is fine. As you say, if $\omega_1$ is countable, then it belongs to the set of all countable ordinals, so $\omega_1\in\omega_1$. This contradicts that ordinals are well-ordered. The reason why you may be a little unconvinced is that you have not yet verified that there is a set of all countable ordinals, to begin with. (If there is such a set, it should not be hard for you to check that it is indeed an ordinal.) Once you argue that the collection of countable ordinals indeed forms a set, we are done.2014-10-01

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

$\omega_1$ is usually defined to be the the least uncountable ordinal number. 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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    @AsafKaragila You're right, I just read that cardinals are *defined as* ordinals after posting my comment... ::foot in mouth:: Still, making the edit would help clear the confusion. However I disagree that it is "overreaching." In my opinion, a question that can be improved should, no matter how old, especially for users who so happen to stumble upon it years later.2014-10-03
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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 $\aleph_1$.2012-09-04
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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 $\omega_1\notin\omega_1$.