A set is an ordinal if it's transitive and well-ordered with respect to $\in$.
I found that $\emptyset$ is very special and that $\emptyset\in \mathbb N$. Is it correct that $a\text{ is an ordinal }\implies\emptyset\in a$?
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.
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 $y$.)