My question is that how to construct an ordered set which has order-type as large as possible in Zermelo set theory Z? For instance, how to construct $\epsilon_0$ or even some uncountable order-type in Z?

I know basic result about ordinals and we couldn't construct $\omega + \omega$ in Z but rather an ordered set isomorphic to it. I'm not very sure if this question is related to Hartogs numbers, since the construction of Hartogs numbers needs ZF rather than Z.

This question is related to another question I asked: The different between Z and ZF

  • 2
    $\begingroup$ Until the end of the second paragraph I was confused and thought tthat Z stood for $\Bbb Z$ and not for $\mathsf{Z}$ ... $\endgroup$ – Hagen von Eitzen Jun 13 '17 at 16:33
  • $\begingroup$ A somewhat related question and answer: math.stackexchange.com/questions/1402271/…. $\endgroup$ – Noah Schweber Jun 13 '17 at 16:54
  • $\begingroup$ Also related: math.stackexchange.com/a/2303064 $\endgroup$ – Eric Wofsey Jun 13 '17 at 17:12
  • $\begingroup$ @EricWofsey I'd clarify to say that your answer, specifically, is relevant to the OP; I think the question as a whole might not be. $\endgroup$ – Noah Schweber Jun 13 '17 at 17:13
  • $\begingroup$ @NoahSchweber: Good point. I've modified the link to go straight to the answer. $\endgroup$ – Eric Wofsey Jun 13 '17 at 17:13

We can do much better than $\epsilon_0$ in Z - we can construct uncountable well-ordered sets! The key observation is that, while building long ordinals uses Replacement, building long well-ordered sets is really a job for just Powerset and Separation.

It's a good exercise to verify the following argument:

  • Z proves that the set $\mathcal{R}$ of all binary relations on $\omega$ exists.

  • Z then proves that the subset $\mathcal{W}\subseteq\mathcal{R}$ of all well-founded binary relations on $\omega$ exists.

  • Z then proves that the set $\mathcal{W}'$ of equivalence classes of elements of $\mathcal{W}'$ under the relation "are order-isomorphic" exists.

  • Finally, Z proves that the relation $\mathcal{L}\subseteq\mathcal{W}'\times\mathcal{W}'$ given by "is longer than" exists.

But now the pair $\mathfrak{O}=(\mathcal{W}', \mathcal{L})$ is an uncountable well-ordered set, provably in Z (by the usual argument). So we have:

Even though the ordinal $\omega_1$ doesn't provably exist in Z, we can emulate its construction by working with sets of natural numbers.

We can go much further: the above argument shows that for any set $A$, we can form the well-order $Ord(A)$ of ordertypes of well-ordered relations on subsets of $A$. In particular, we can apply this to $\mathfrak{O}$ to build a version of $\omega_2$, and so on. In fact, we can prove in Z that for every $n$, a version of $\omega_n$ exists (of course we have to phrase this rather carefully); conversely, it's consistent with Z that every well-ordered set is order-isomorphic to a subset of one of these versions, so in a sense Z "reaches up to" the ordinal $\omega_\omega$.

(Towards interpreting the previous sentence, it's a good exercise to show that in $(V_{\omega+\omega})^L$, the smallest transitive model of Z, the lengths of the well-ordered sets are exactly the ordinals $<\omega_\omega^L$. If you're not familiar with the "[stuff]$^L$" notation, then work with $V_{\omega+\omega}$ instead and assume GCH.)

  • $\begingroup$ I can't see the reason of why $\omega_\omega$ is reachable(you only describe that $\omega_n$ is reachable). Even so, it is enough to solve my question... $\endgroup$ – Minghui Ouyang Jun 16 '17 at 12:28
  • $\begingroup$ @MinghuiOuyang $\omega_\omega$ is not reachable - we can reach up to it (that is, we can reach everything below it) but we can't reach it itself. I guess that was unclear. $\endgroup$ – Noah Schweber Jun 16 '17 at 13:06
  • $\begingroup$ Oh, I see. I didn't understand "reach up to" before. $\endgroup$ – Minghui Ouyang Jun 16 '17 at 14:19
  • $\begingroup$ @MinghuiOuyang I noticed you un-accepted this answer - may I ask what further questions you have? $\endgroup$ – Noah Schweber Jun 17 '17 at 16:59
  • $\begingroup$ Oh... Misoperation. So sorry. $\endgroup$ – Minghui Ouyang Jun 18 '17 at 1:35

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.