Is the following claim true: "every ordinal has the empty set as one of its elements"

Question

Since we can encode every element as a set, suppose that we consider only sets in which every element is also a set. My question is, is it true that every ordinal has the empty set as one of its elements?

The proof I was thinking of: Suppose not. Let A be an ordinal which does not contain the empty set as an element. Take $b_1 \in A$. $b_1 \neq \emptyset$ Then by definition of an ordinal we should have $b_2 \in b_1$ and by the same argument we will have $b_3 \in b_2$,...,$b_{n-1} \in b_n$,... so we get an infinite decreasing sequence of elements in $A$, contradicting the fact that $A$ is a well ordered set. am I right? seems suspicious to me since I haven't encountered such a claim...

Thank you! Shir

Answer 1

Your proof starts with "Take $b_1\in A$". You can't do that in general. And the case where you can't gives you an ordinal not having $\emptyset$ as element ...

Apart from that: $\emptyset$ is an ordinal and the class of ordinals is ordered by $\in$. Hence for any ordinal $A\ne\emptyset$ we have either $A\in\emptyset$ (which is absurd) or indeed $\emptyset \in A$.

Answer 2

This is false, but only for trivial reasons: The empty set itself is an ordinal.

However, if the ordinal is non-empty, then the empty set is indeed one of its elements. Your proof is correct, but it may be better to use well-foundedness directly: Your ordinal $\alpha$ is well-ordered by $\in$, so it has a least element (here we use that $\alpha$ is non-empty). Call it $b$. The point is that ordinals are transitive, so $b$ is a subset of $\alpha$. If $c$ is any element of $b$, then $c\in b\cap\alpha$, contradicting minimality. So $b$ is empty.