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

Question

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?

Answer 1

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 $0$.