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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.

Andrés E. Caicedo
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ashpool
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    Ask Google: first uncountable ordinal immediately gives this, see also this – t.b. Oct 11 '11 at 14:59
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    But you were just told that this is not an adequate construction of all ordinals, much less a "standard way" to construct them. – hmakholm left over Monica Oct 11 '11 at 15:38
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    Of course there are only countably many ordinals with names, since there are countably many names :) – Ben Millwood Jun 02 '12 at 18:05
  • I think for some audiences, none of the answers to this question are satisfactory and this question needs an answer that breaks the intuition that because you can never think of uncountably many ordinal numbers as you think of stronger systems of ordinal numbers, all ordinal numbers are countable. Maybe somebody can write an answer that breaks that intuition just like Asaf Karagila's answer to "ZF — Sets that can be proven to exist" breaks the intuition that there are only countably many sets. – Timothy Apr 16 '18 at 02:14
  • Does my answer at https://matheducators.stackexchange.com/questions/689/how-can-i-familiarize-elementary-school-students-with-infinities-larger-than-a/14983#14983 answer your question? – Timothy Jan 15 '19 at 06:16
  • The answer at https://math.stackexchange.com/questions/71726/uncountability-of-countable-ordinals/71730#71730 appears to have solved your problem. Maybe you just think it solved your problem when it really didn't. If you just assume the set of all ordinal numbers exists, you can derive the Burali-Forti paradox. From this, we can show in ZF that there is no set of all ordinal numbers. We can't assume there is a set of all countable ordinal numbers either. I know there is a way to show that there is a set of all countable ordinal numbers. If you don't prove that, you could extend ZF into a – Timothy Oct 21 '19 at 20:01
  • variation of NBG that doesn't assume the axiom of limitation of size making it impossible to prove the axiom of choice or axiom of global choice. If you then you even further to a stronger system that can refer to all conglomerates of classes, then you would need to ask yourself, can I show that there is no bijective class function from the natural numbers to the class of all ordinal numbers. This is at least true if you think of NBG the way I once thought of it. Now I know that NBG is not intended to be interpreted as a real theory and is meant to be used only because it has been proven that – Timothy Oct 21 '19 at 20:07
  • a statement describable in ZFC is provable in ZFC if and only if it is provable in NBG and then people look for a proof of a statement describable in ZFC in NBG because they know that if they find one, a proof of the same statement in ZFC exists. If you thought NBG was intended as a real theory, then it might be tempting to add an additional assumption that a proper class is something and once you show that a specific class has a certain property, you can deduce from it that something has that property even if ZFC disproves that something has that property making the system contradictory. – Timothy Oct 21 '19 at 20:11

5 Answers5

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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.

mercio
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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
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    @kahen, the ordinals are well-ordered by $\in$, so every ordinal smaller than $\alpha$ is a member of $\alpha$ and thus, by definition, countable. – hmakholm left over Monica Oct 11 '11 at 15:36
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    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 01:44
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    The problem with this argument is that you didn't prove that there is a set of all countable ordinal numbers. You could apply the same argument to the set of all ordinal numbers to derive the Burali-Forti paradox. See https://math.stackexchange.com/questions/216229/other-ways-of-proving-that-the-set-of-all-countable-ordinals-is-uncountable. For those who have already learned that it has been proven without the axiom of choice that uncountable ordinal numbers exist, this answer might be good enough. – Timothy Oct 29 '17 at 02:12
  • Sir if there is infinite descending sequence of ordinals then what? – Akash Patalwanshi Sep 22 '19 at 16:02
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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...

Asaf Karagila
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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
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    ${{\varnothing}}={1}$ 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
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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.

weux082690
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  • You have to consider the finite ordinals too, otherwise this logic fails. 2) From this explanation it's absolutely not clear why $f(n)<\alpha$: for example, ${4,5}$ is order-isomorphic to ${0,1}$, but I still don't believe that $4<2$...
    • – Alexander Kuleshov Sep 21 '19 at 21:51
  • (1) The finite oridinals only add an $\omega$ to the front of the ordering. Since $\alpha > \omega$, this disappears, i.e., $\omega + \alpha = \alpha$. (2) You can define the next ordinal to be the order type of all the ordinals so far. Since this list is supposed to contain all the countable ordinals, its order type will be the next ordinal after them. If this is countable, it also needs to be on the list though. Which gives the contradiction. – weux082690 Sep 26 '19 at 00:17
  • Don't get it at all: (1) $\omega + 1 > \omega$, but $\omega+\omega+1 \ne \omega + 1$. (2) You have only the fact, that your ordinal $\alpha$ is order-isomorphic to $\alpha^\prime$: the well-ordered set of all countable(and infinite!) ordinals. So by now it can happen that $f(1)=\omega, f(2)=\omega+1...; \alpha=\omega$. To show that $x<\alpha$ for each $x\in \alpha^\prime$ you need to add all the finite ordinals to $\alpha^\prime$, then show that $\alpha^\prime$ is an ordinal! It is exactly what mercio suggested in his answer. And why then do you need ordering on $\mathbb{N}$ and $f(n)$ at all? – Alexander Kuleshov Sep 27 '19 at 01:01