I know that the standard way of proving that the set of all countable ordinals is uncountable is by stating that if the set is countable, then it incurs Burali-Forti paradox.

Is there other ways of proving this?

  • $\begingroup$ Not Russell's paradox so much as Burali-Forti's paradox, but they're closely related. $\endgroup$ – Zhen Lin Oct 18 '12 at 9:43
  • $\begingroup$ Some proofs are given in this question: Uncountability of countable ordinals. (And you might want to have a look at linked questions, too.) $\endgroup$ – Martin Sleziak Oct 18 '12 at 10:20
  • $\begingroup$ I think this question already has an answer at math.stackexchange.com/questions/38468/…. $\endgroup$ – Timothy Oct 26 '17 at 1:52
  • $\begingroup$ This question has nothing to do with the axiom of choice, so I've removed that tag. $\endgroup$ – Noah Schweber Nov 12 '17 at 22:27

The set of all countable ordinals is the supremum of all countable ordinals, which is simply their unions. If $\omega_1$ were countable, $\omega_1+1$ would be countable too and hence $\omega_1+1<\omega_1$. Hence $\omega_1\in\omega_1+1\in\omega_1$, which shows that the set $\{\omega_1,\omega_1+1\}$ has no $\in$-minimum, which is impossible since the ordinals are well-ordered by $\in$.

  • $\begingroup$ Wait, how is $\omega_1+1 \leq \omega$? Isn't $\omega$ the first infinite countable ordinal? $\endgroup$ – DOT Oct 18 '12 at 10:20
  • $\begingroup$ @DOT Yes. But this is a proof by contradiction. Since the set of all countable ordinals is the supremum of all countable ordinals, and $\omega_1$ being countable implies $\omega_1+1$ we have $\omega_1+1<\omega_1$, which is by definition of the ordering on ordinals $\omega_1+1\in\omega_1$. $\endgroup$ – Michael Greinecker Oct 18 '12 at 10:28
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    $\begingroup$ @Asaf But the proof works even when the axiom of regularity fails. $\endgroup$ – Michael Greinecker Oct 18 '12 at 19:50
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    $\begingroup$ @Timothy There is a theorem by Hartog, provable without AC, that for every set there is an ordinal such that there is no injection from the form into the latter. In particular, uncountable ordinals exist. $\endgroup$ – Michael Greinecker Oct 26 '17 at 7:31
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    $\begingroup$ @Timothy I obviously didn't vote on my own answer. But every textbook discussing ordinals without the axiom of choice will contain the result. More importantly, my argument actually proves the existence of uncountable ordinals. If you have questions understanding the proof, feel free to ask. $\endgroup$ – Michael Greinecker Oct 26 '17 at 20:18

I think that all proofs will be similar in some way to the Burali-Forti paradox. Here is a proof that is slightly different than other proofs on this site, so maybe someone will find it useful. Note that the Axiom of Regularity is not used anywhere.

Definition: An ordinal is a transitive set well-ordered by $\in$.

Fact: The class of ordinals is transitive and well-ordered by $\in$.

Definition: $\omega_1$ is the class of countable ordinals.

Fact: $\omega_1$ is a set.

To see that $\omega_1$ is an ordinal using the above facts, it remains to observe that every element of a countable ordinal is a subset of that ordinal and is therefore countable also.

Now suppose toward a contradiction that $\omega_1$ is countable. Then $\omega_1 \in \omega_1$ by definition. This contradicts the fact that the class of ordinals is well-ordered by $\in$.


In Principia Mathematica, Whitehead & Russell presented Burali-Forti's paradox in a form that can dispense with the axiom of infinity. They eventually dispelled the paradox by showing that in higher types there are greater ordinals than any to be found in lower types.

The paradox:

Let $\alpha$ be any ordinal number, then the ordinal number for $0_r, 1_s, 2_r, 3_r, ... \alpha$ in order of magnitude is $\alpha + \overset{.}{1}$; thus $\alpha +\overset{.}{1}$ exists, and is greater than $\alpha$. But $\alpha$ is similar to the segment of series of ordinals consisting of the predecessors of $\alpha$, and is therefore less than the ordinal of all ordinals. Hence the ordinal of all ordinals is greater than every ordinal, including itself, which is absurd.

In order to dispel the paradox, it is only necessary to make the types explicit. Let $N$ be the series of all ordinals arranged in order of magnitude; when we say "$P$ is well-ordered $\Rightarrow$ $P$ is less than $N$," we mean, by $N$, a series two types above $P$, i.e. $N$'s terms belong to the type of $P$'s ordinal number ($t‘N_0r‘P$). In other words:

$P$ is well-ordered $\Rightarrow$ $P \lt N(P,P)$

Whence $N(P,P)$ is well-ordered $\Rightarrow N(P,P)\lt N\{N(P,P),N(P,P)\}$

Thus, N is no longer less than itself, and the paradox disappears.

Source: Whitehead & Russell, Principia Mathematica V.3. ✳256. Merchant Books, 1913

  • $\begingroup$ Note to self: I have yet been able to come up with a W-O series in which some terms, except for the first one, do not have immediate predecessors. $\endgroup$ – George Chen Dec 29 '14 at 21:02
  • $\begingroup$ Let $P, Q \in \Omega$ and $P$ has no last term, then $P⤉Q \in \Omega$, and $Q$'s first term has no immediate predecessor in $P⤉Q$. $\endgroup$ – George Chen Dec 30 '14 at 5:39

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