# Does there exist a set theory where the following non well-founded definition of the ordinals is possible?

I was wondering if there is a non well-founded set theory such that the following implementation of the ordinals is possible.

• $0 = \emptyset$
• $1 = \{1\}$
• $2 = \{1,2\}$
• ...
• $\omega = \{1,2,3,\cdots\}$
• $\omega+1 = \{1,2,3,\cdots, \omega+1\}$
• $\omega+2 = \{1,2,3,\cdots, \omega+1,\omega+2\}$

So the least element of any non-empty ordinal $\alpha$ is $1$, and the greatest element is always $\alpha$. The limit ordinals, which are precisely those ordinals that fail to have a greatest element, are also precisely those that fail to be elements of themselves, nor indeed of any other ordinal.

Does there exist such a set theory?

• Is there any reason to do this other than aesthetics? I think you could probably do this with the anti-foundation axiom. In NF, one of the set theories without well-foundedness, ordinals are coded directly as isomorphism classes of well-ordered sets. – Zhen Lin Feb 22 '13 at 9:48
• Yes, there's a reason: it prevents "avalanches." Consider the following. "0,1,2,3... okay that's 4 fruit." The ordinals mentioned so far form a new ordinal, namely 5. Okay but now the ordinals mentioned so far form a new ordinal, namely 6. etc. But consider the alternative. "1,2,3,4... okay that's 4 fruit." The ordinals mentioned so far form a new ordinal, namely... oh, it's just 4. – goblin Feb 22 '13 at 10:06
• For finite ordinals/cardinals you might have that problem. But there are very few finite ordinals. – user642796 Feb 22 '13 at 10:24
• Is there a reason that you don't want $0$ and $\omega$ to contain themselves? – Trevor Wilson Feb 22 '13 at 19:27
• Yes. If they were elements of themselves, they would be their own greatest element. – goblin Feb 22 '13 at 23:10

## 1 Answer

Let me play devil's advocate: I accept your definition, and then I assert that you have $1=2$. You may object that these two sets, namely $\{1\}$ and $\{1,2\}$ are obviously different, but I answer that they are not different at all, they have exactly the same members, because the seemingly extra member $2$ in $\{1,2\}$ is merely a repetition of the other member, $1$. Can you convince me, using your definitions, that $1\neq2$?

I wouldn't be surprised if at least some versions of the anti-foundaion axiom would actually let you prove that $1=2$ under your definitions.

The difficulty, of course, concerns not only $1$ and $2$, but all the ordinals you mentioned except $0$.

• The most popular anti-foundation axiom (Aczel's) indeed implies $1=2$ under these definitions. – Andrés E. Caicedo Feb 22 '13 at 20:37
• I was worried about this sort of thing. Asserting that every ordinal is distinct from its successor should do the trick, though. – goblin Feb 22 '13 at 23:06