0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ @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 :) $\endgroup$ – Le Anh Dung Jul 7 '18 at 3:48
2
$\begingroup$

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 $y$.)

$\endgroup$
  • $\begingroup$ 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 $x=\emptyset$. $\endgroup$ – Le Anh Dung Jul 7 '18 at 4:15

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.