4
$\begingroup$

I'm reading up on ordinals, and I've come across various definitions for what it means to be an ordinal. The main book that I'm following is Kunen's Set Theory book. He defines an ordinal as a set that is transitive and is well-ordered by $\in$.

I found a couple of exercises in his book that relates his definition of an ordinal to other definitions that I've seen. I'm trying to figure these out to make more sense of the definitions and concepts in Kunen's book and connect his definition with other definitions I've seen. The exercises both assume the Foundation Axiom and are as follows:

$z$ is an ordinal iff $z$ is a transitive set and $\in$ satisfies trichotomy on $z$. Hint: $z \cap ON$ is an ordinal by a previous lemma. Then, use trichotomy to derive a contradiction from an $\in$-minimal element of $z \backslash ON$, where $ON$ denotes the class of all ordinals.

Showing $(\Rightarrow)$ seemed to follow easily since being an ordinal implies transitivity and well-ordering $\in$ implies that $\in$ is well-founded on $z$, implying trichotomy. I'm having trouble setting up the contradiction in $(\Leftarrow)$ by using the hint.

$z$ is an ordinal iff $z$ is a transitive set and all elements of $z$ are transitive sets.

Again, $(\Rightarrow)$ seemed to follow easily since every element of a transitive set is also transitive. As for $(\Leftarrow)$, I'm stuck.

As always, I would really appreciate any help/hints for these exercises.

$\endgroup$
  • $\begingroup$ I didn't read your post thoroughly, but I suspect that you need to use foundation. $\;$ $\endgroup$ – user57159 Sep 10 '13 at 0:31
  • $\begingroup$ @RickyDemer, yes you are correct. I mentioned that both assume Foundation. $\endgroup$ – josh Sep 10 '13 at 0:42
4
$\begingroup$

Suppose that $z$ is transitive, and $\in$ satisfies trichotomy on $z$. Suppose that $z\nsubseteq\mathbf{ON}$, and let $x$ be $\in$-minimal in $z\setminus\mathbf{ON}$. Then $x\subseteq z\cap\mathbf{ON}$ by transitivity of $z$ and $\in$-minimality of $x$. Let $\alpha=z\cap\mathbf{ON}\in\mathbf{ON}$; $x\subseteq\alpha$. Suppose that $\xi\in\alpha$; then $\xi\in z$, so either $\xi\in x$, $\xi=x$, or $x\in\xi$. The last two cases contradict the assumption that $x\notin\mathbf{ON}$, so $\xi\in x$. And $\xi\in\alpha$ was arbitrary, so $\alpha\subseteq x$, and hence $x=\alpha\in\mathbf{ON}$, a contradiction. It follows that $z\subseteq\mathbf{ON}$, so $z$ is well-ordered by $\in$ and therefore an ordinal.

For the second one, proceed in exactly the same fashion up to the point at which we observe that $x\subseteq\alpha$. Now $x$ is a transitive set of ordinals, so ... ?

$\endgroup$
  • $\begingroup$ Since $x$ is a transitive set of ordinals, then $x$ is an ordinal. Wouldn't that lead to a contradiction since we assumed $x \not \in ON$? $\endgroup$ – josh Sep 10 '13 at 1:34
  • $\begingroup$ @josh: Exactly. In outline it’s the same argument as for the first problem; it differs only in the reason for being able to conclude that $x$ must be a transitive set of ordinals. $\endgroup$ – Brian M. Scott Sep 10 '13 at 1:35
  • $\begingroup$ Ahh, that makes sense! I'm developing a growing interest in Set Theory, and I appreciate you helping me along the way. $\endgroup$ – josh Sep 10 '13 at 1:41
  • $\begingroup$ @josh: No problem: set theory’s fun! $\endgroup$ – Brian M. Scott Sep 10 '13 at 1:42

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.