Uncountability of countable ordinals According to Wikipedia, there are uncountably many countable ordinals. What is the easiest way to see this? If I construct ordinals in the standard way,
$$1,\ 2,\ \ldots,\ \omega,\ \omega +1,\ \omega +2,\ \ldots,\ \omega\cdot 2,\ \omega\cdot 2 +1,\ \ldots,\ \omega^{2},\ \ldots,\ \omega^{3},\ \ldots\ \omega^{\omega},\ \ldots,\ \omega^{\omega^{\omega}},\ \ldots, \epsilon_{0},\ \ldots$$
I seem to get only countably many countable ordinals.
 A: Fact: If $A$ is a set of ordinals which is downwards closed, then $A$ is an ordinal.
Now consider the following set: $A=\{\alpha\mid\exists f\colon\alpha\to\omega,\ f \text{ injective}\}$, this is the set of all countable ordinals.
If $\alpha\in A$ then clearly $\beta<\alpha$ implies $\beta\in A$, simply because $\beta\subseteq\alpha$. We have, if so, that $A$ is itself an ordinal. If $A$ was a countable ordinal then $A\in A$, which is a contradiction. Therefore $A$ is uncountable, in fact $A$ is the least uncountable ordinal, also known as $\omega_1$.
There are just many ordinals which you cannot describe nicely. It just shows you that you can well order a countable set in so many ways...
A: I have searched and searched for an answer to this question that makes intuitive sense and have yet to find one.  So, after some thought of my own, this is what I came up with.
Suppose that the countable ordinals were countable.  Let f be a one-to-one correspondence between the natural numbers and every well-ordering of the natural numbers.  For instance:
1 <--> 1 < 3 < 5 <... 2 < 4 < 6 <...
2 <--> 1 < 2 < 3 < 4 <...
3 <--> 1 < 2 < 4 < 8 <... 3 < 6 <... 5 < 10 <...
...

Then, you only need to show there is a well-ordering of the natural numbers that is not on this list.  
For each natural number $n$, let $f(n)$ be the order type of the ordering corresponding to $n$.  Following the example list above, $f(1) = \omega*2$, $f(2) = \omega$, $f(3) = \omega^2$, and so on.  Define an ordering on the natural numbers from $m < n$ iff $f(m) < f(n)$.  This ordering of the natural numbers is a well-ordering since the ordering of the ordinals is a well-ordering.  Therefore, it has some order type, call it $\alpha$.  For all $n$, $f(n) < \alpha$, which follows from each ordinal being order-isomorphic to the ordered set of ordinals less than it.  We are assuming $\alpha$ is countable, so it must be somewhere in our list, say $f(n) = \alpha$.  But $f(n) < \alpha$ also, which is the contradiction we want.  Therefore, $\alpha$ is nowhere on the list.  Therefore, the countable ordinals are uncountable.
As far as whether the countable ordinals are the first uncountable ordinal, use again the fact that each ordinal is order-isomorphic to the ordered set of ordinals less than it.  The order type of the countable ordinals must be the first uncountable ordinal, because all ordinals less than it are countable, from the order-isomorphism with the countable ordinals.
A: I will prove without assuming the axiom of choice that there are uncountably many countable ordinals. First, I will show that there is a set of all countable ordinals.
Take the set of all well-ordering relations on the set of all natural numbers. Let $f$ be the function that assigns to each of the relations in that set its order type. By the axiom of replacement, there is a set of all images of that function, and that's the set of all infinite countable ordinals. By the axiom of union, the union of that set and the set of all finite ordinals is also a set, and that's the set of all countable ordinals.
By definition, $\omega_1$ is the smallest ordinal such that the set of all ordinal numbers smaller than it is uncountable if such an ordinal exists. The supremum of the set of all countable ordinals is $\omega_1$ so $\omega_1$ exists. Therefore, the set of all countable ordinals is uncountable.
A: Let $\alpha$ be the set of all countable ordinals.
It is an ordinal : if $\beta \in \alpha$, then $\beta \subset \alpha$ because the elements of $\beta$ are countable ordinals.
It is uncountable : if it were countable, $\alpha$ would be a member of itself, so there would be an infinite descending sequence of ordinals.
Therefore, $\alpha$, the set of all countable ordinals, is the smallest uncountable ordinal.
A: Not sure, if people aren't making this too difficult by looking at a generalized case.
If you look at the successor cardinal of Omega, it is the union of all ordinals lower than itself. If there only were countably many, then it would be a union of countably many countable sets, hence countable itself.
