1

Possible Duplicate:
Uncountability of countable ordinals

How does one prove that there are uncountable number of countable ordinals? Obviously, there are equal to or more than countable number of ordinals, but not sure how to prove there are uncountable number of countable numbers. Maybe, should I just assume that as uncountable ordinal has uncountable elements as a set, there should be uncountable number of countable ordinals?

Community
  • 1
John Natawabe
  • 249
  • 1
    The first uncountable ordinal is the set of all lesser ordinals, that is all countable ordinals. Perhaps you want to know what guarantees the existence of the first uncountable ordinal? – Alex Becker Sep 17 '12 at 06:03

2 Answers2

5

If $A$ is a countable set of countable ordinals, let $\beta=\sup\{\alpha+1:\alpha\in A\}$; then $\beta$ is countable, and $\alpha<\beta$ for every $\alpha\in A$. Thus, the set of countable ordinals cannot be countable.

Added: To see that $\beta$ is countable, just note that $\alpha+1$ is clearly countable if $\alpha$ is, so $\beta=\bigcup_{\alpha\in A}(\alpha+1)$ is a countable union of countable sets. This actually also covers the existence of $\beta$: given the set $\{\alpha+1:\alpha\in A\}$, whose existence follows from the axiom schema of replacement and the fact that $A$ is a set, the union axiom ensures the existence of $\sup\beta$.

Brian M. Scott
  • 616,228
  • You should mention why $\beta$ is countable? – William Sep 17 '12 at 06:08
  • @William: I thought that it was obvious to anyone likely to be asking the question, but perhaps you’re right. – Brian M. Scott Sep 17 '12 at 06:09
  • and/or why the sup exists :-) – Mariano Suárez-Álvarez Sep 17 '12 at 06:09
  • @MarianoSuárez-Alvarez if you are saying that the existance of the sup needs to be added to the proof, then I would agree. If you are asking what is sup. given a subset S of a totally or partially ordered set T, the supremum (sup) of S, if it exists, is the least element of T that is greater than or equal to every element of S. – yiyi Sep 17 '12 at 06:12
  • Well, I know why sup exists... but I guess that might be useful to include. Not sure. And obviously why countable set of countable numbers are countable - I know that. :) – John Natawabe Sep 17 '12 at 06:14
  • This argument has the technical disadvantage of using choice (otherwise, we cannot prove that the supremum of a countable set of ordinals is countable). The existence of $\omega_1$ can be established without choice, and this gives the result. For details, see: http://math.stackexchange.com/questions/46833/how-do-we-know-an-aleph-1-exists-at-all/46836#46836 – Andrés E. Caicedo Sep 17 '12 at 06:14
3

Define $\Omega$ to be the smallest uncountable ordinal. It exists because uncountable ordinals exist, and ordinals are well-ordered (any nonempty class of ordinals has a smallest member).

Now $\Omega$ is the set of all countable ordinals, but it is itself uncountable by construction.

Harald Hanche-Olsen
  • 31,960
  • Of course, one has to first demonstrate the existence of uncountable ordinals... – Zhen Lin Sep 17 '12 at 07:23
  • @ZhenLin: Of course. And that requires the existence of some uncountable set, together with the ability to well-order it. But if we want to pick nits, Brian's answer does leave open the possibility that all the countable ordinals taken together does not constitute a set. Nothing wrong with that, but then he should not have referred to the set of countable ordinals either. He still has a perfectly good proof that no countable set can contain all countable ordinals. – Harald Hanche-Olsen Sep 17 '12 at 08:30