3
$\begingroup$

The countable ordinals are themselves either countable or uncountable. They cannot be countable since that would involve a set with itself as an element, so they are uncountable.

If they are uncountable, then either they form a consistent totality or they are a proper class. If they form a consistent totality, then a least uncountable ordinal exists, otherwise not.

So how is it determined that the sequence of countable ordinals is or is not a consistent totality? My problem is that it cannot be said they form a consistent totality because there is an uncountable limit ordinal since that would be circular. Therefore, how exactly is this resolved?

Edit, as requested: My question is not equivalent to the question, "What axioms are you using? ZFC?" Nor is it answered by (I haven't figured out how to format the symbols on this site so I'll just spell it out): "The set of all countable ordinals is the set of those elements x of the cardinal number two to the aleph-null such that x is a countable ordinal", which seems to be a way of saying x is a countable ordinal if it is a member of the set of countable ordinals, except two to the aleph-null is a number: it is not a set but the cardinality of a set, so it doesn't actually have elements. I suppose the intention of that answer could be taken as that all countable sets are elements of the power set of N, except the assertion isn't true. Nor is my question addressed by comments to that answer. So far as I can tell, the answerer did not define omega plus one, nor have I ever seen two to the aleph-null described as an ordinal. Besides which, the relation of the cardinal number two to the aleph-one to the limit ordinal omega-one is outside ZFC. So I honestly don't see what any of this has to do with my question or why I was asked to edit it accordingly.

$\endgroup$
  • 1
    $\begingroup$ What axioms are you using? ZFC? $\endgroup$ – Ross Millikan Mar 16 '15 at 4:40
  • $\begingroup$ The set of all countable ordinals is $\{ x \in 2^{\aleph_0} : x \text{ is a countable ordinal}\}$, which exists via various axioms. $\endgroup$ – Yuval Filmus Mar 16 '15 at 4:43
  • $\begingroup$ @YuvalFilmus Is $2^{\aleph_0}$ the power set of $\aleph_0$? Then your definition is not correct, you just defined $\omega+1$. Perhaps you meant to take the collection of subsets of $\aleph_0$ that code a well-ordering. To get ordinals you then need to map these well-orderings to their unique order type (and apply replacement). $\endgroup$ – Andrés E. Caicedo Mar 16 '15 at 5:17
  • $\begingroup$ @Andres: I suspect that Yuval meant that $2^{\aleph_0}$ is the least ordinal of that cardinality. Of course this appeals to the axiom of choice in a way that we know is unnecessary. $\endgroup$ – Asaf Karagila Mar 16 '15 at 7:28
  • $\begingroup$ @Andres I knew I should have left it to the experts... $\endgroup$ – Yuval Filmus Mar 16 '15 at 13:48
9
$\begingroup$

Recall the definition of $\omega_1$. It is the set of all finite and countably infinite ordinals. You are asking why is this is a set. Which is an equivalent question to "why does $\aleph_1$ exists?".

But let me give some clarifications to your edit.

  1. $2^{\aleph_0}$ is a cardinal number. Working in $\sf ZFC$, this means that this is an uncountable ordinal, which one we cannot say, but we can say that it is an uncountable ordinal.

    But once you know that some uncountable ordinal exists, with or without choice, it is just an application of the axiom [schema] of separation, to pick the set of all countable ordinals.

  2. We need the axiom [schema] of replacement to show that $\omega_1$ is a set, regardless to the above. Separation (which appear in some axiomatizations of set theory) is insufficient, so we really use replacement and the power of $\sf ZF$. Assuming $\sf ZF$ every well-ordered set is isomorphic to a von Neumann ordinal, and we can prove there is an uncountable well-ordered set, simply by noting that in $\mathcal P(\omega\times\omega)$ includes all the countable well-orders, so mapping each relation which is a well-ordering of its domain to the order type (and those which are not well-ordered to $0$), we get a definable function whose domain is a set, and by the replacement axioms the range is a set as well.

    But every countable ordinal can be realized as a well-ordering of $\omega$. So the set of countable ordinals exists.

  3. We also need the power set axiom to show that $\omega_1$ exists. And in fact, we need the power set axiom in order to show that there are uncountable sets at all. Recall that every set has a transitive closure, namely $\operatorname{tc}(X)$ is the smallest set $Y$ which is transitive (every element of $Y$ is also a subset of $Y$), and $X\subseteq Y$.

    It turns out that $\mathrm{HC}=H(\aleph_1)=\{X\mid\operatorname{tc}(\{X\})\text{ is countable}\}$ satisfies all the axioms of $\sf ZFC$ except the power set axiom, and of course if $X\in\mathrm{HC}$ then there is an injection from $X$ into $\omega$ in $\mathrm{HC}$ as well. So $(\mathrm{HC},\in)$ satisfies the axiom "Every set is countable", as well $\sf ZFC-P$.

  4. Without the axiom of replacement, but with power set and separation at least, we still cannot prove this set exists. The reason is that if you look at the von Neumann hierarchy of the universe, the $V_\alpha$'s then for any limit ordinal, $V_\alpha$ is a model of all the axioms of $\sf ZF$ except, perhaps replacement and if $\alpha=\omega$ then infinity fails. So taking $V_{\omega+\omega}$ is a model for the separation axioms, but there is no set of countable ordinals. Even $V_{\omega_1}$ is still a model of these axioms, but every von Neumann ordinal is countable there.

$\endgroup$
  • $\begingroup$ Thank you. This clears up a lot. I need one bit of additional help, at least for now: I know that certain limit ordinals are associated in some way with distinct cardinalities but I did not know that the cardinal IS an ordinal. How is this identity justified? I find the idea confusing because while it makes sense to say a set has a size, I can't see why a size is a set. For the rest, I'll need some time to digest your comments on the von Neumann hierarchy. $\endgroup$ – Nightspore Mar 17 '15 at 10:44
  • $\begingroup$ Assuming the axiom of choice, every set can be well-ordered. In particular, $2^{\aleph_0}$ is a cardinal of a well-ordered set, so it is equal to some $\aleph_\alpha$ for some ordinal $\alpha$. Then we can look it as the least ordinal with this cardinality. $\endgroup$ – Asaf Karagila Mar 17 '15 at 10:47
  • $\begingroup$ (To my devoted downvoter, I don't expect you to give any criticism, because I don't believe "I HATE YOUU!!!!11oneone" is a valid criticism, and I think you know that. But thank you for taking the time to follow my posts!!!) $\endgroup$ – Asaf Karagila Mar 17 '15 at 10:49
  • $\begingroup$ Thanks for posting this. I was about to ask the same question, and this answer was everything I could have hoped for. $\endgroup$ – MJD Nov 28 '16 at 1:55
  • $\begingroup$ @MJD: You're very welcome! $\endgroup$ – Asaf Karagila Nov 28 '16 at 5:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.