# Do we need Axiom of Regularity to prove that $a=\{x,\{x\}\}\text{ is an ordinal }\implies x=\emptyset$?

A set is an ordinal if it's transitive and well-ordered with respect to $\in$.

1. I found that $\emptyset$ is very special and that $\emptyset\in \mathbb N$. Is it correct that $a\text{ is an ordinal }\implies\emptyset\in a$?

2. Do we need Axiom of Regularity to prove that $a=\{x,\{x\}\}\text{ is an ordinal }\implies\text{ x}=\emptyset$?

Here is my attempt for 2nd question:

It's clear that $a$ is well-ordered in respect to $\in$. $a$ is transitive set $\iff x\subsetneq a$ and $\{x\}\subsetneq a\iff x\subsetneq a$ and $x\in a\iff x\subsetneq a$.

If $x=\emptyset\implies x\subsetneq a\implies a$ is transitive set. (satisfied)

If $x\neq\emptyset\implies x\in x$ or $\{x\}\in x$. To make this case impossible to happen, we must appeal to Axiom of Regularity.

• @EricWofsey, In my definition: a set is an ordinal if it's transitive and well-ordered with respect to $\in$. I will added this definition to my post :) – Le Anh Dung Jul 7 '18 at 3:48

The answer to both questions is no. For the first question, notice that $\emptyset$ is an ordinal.
For the second question, note that if $a=\{x,\{x\}\}$ well-ordered by $\in$, it has a least element $y$ with respect to $\in$. Then $z\not\in y$ for all $z\in a$. But since $a$ is transitive, every element of $y$ is in $a$, so $y$ cannot have any elements and so $y=\emptyset$. Since $\{x\}$ is not empty, we must have $y=x$ and thus $x=\emptyset$.
(This argument shows more generally that any nonempty ordinal must have $\emptyset$ as an element. We need $a$ to be nonempty to get the existence of a least element $y$.)
• Thank you so much! I got your point. From your solution, I figure another way to do this: It's clear that $x$ is the least element of $a$ under $\in$. If $x\neq\emptyset$, then $\exists t\in x$. Hence $t\in x\in a$ by the transitivity of $a$. This contradicts the minimality of $x$. Thus $x=\emptyset$. – Le Anh Dung Jul 7 '18 at 4:15