# 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.

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Ask Google: first uncountable ordinal immediately gives this, see also this – t.b. Oct 11 '11 at 14:59
But you were just told that this is not an adequate construction of all ordinals, much less a "standard way" to construct them. – Henning Makholm Oct 11 '11 at 15:38
Of course there are only countably many ordinals with names, since there are countably many names :) – Ben Millwood Jun 2 '12 at 18:05

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.

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I think you have to do a little bit more work to say that $\alpha$ is the smallest uncountable ordinal. The argument above only shows that it's uncountable. – kahen Oct 11 '11 at 15:13
@kahen, the ordinals are well-ordered by $\in$, so every ordinal smaller than $\alpha$ is a member of $\alpha$ and thus, by definition, countable. – Henning Makholm Oct 11 '11 at 15:36
This answer and the (essentially equivalent) one by Asaf Karagila rely on conventions unknown to Cantor. But Cantor proved that the set of all countable ordinals is uncountable. There should be a way to do this that doesn't require encoding ordinals as von Neumann ordinals. – Michael Hardy Oct 12 '11 at 1:44

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...

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I'm confused by the fact: is $\{ \{ \emptyset \} \}$ not downwards closed? Meaning it has a least element with respect to $\in$, or what does downwards closed mean? – Rudy the Reindeer Jan 17 '12 at 19:29
$\{\{\varnothing\}\}=\{1\}$ is not downwards closed. $A$ is downwards closed if $\beta\in A,\alpha<\beta\rightarrow\alpha\in A$. In the case of ordinals, which are transitive sets this is the same as to say $\beta\in A\rightarrow\beta\subseteq A$. – Asaf Karagila Jan 17 '12 at 19:31
Ah, I didn't know what downwards closed means. Thanks, Asaf! – Rudy the Reindeer Jan 17 '12 at 19:33

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.

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