# Definition of ordinal

I am confused with the definition of ordinals: "A set $a$ is an ordinal if it is transitive and totally ordered by $\in$." Then an ordinal is a set but we also know that all elements of ordinals are ordinals themselves.

If we take $\alpha=\{x,\{x\}\}$ then $\alpha$ is a transitive set and totally ordered by $\in$ so it is an ordinal. But $x\in\alpha$ so $x$ is an ordinal but $x$ is not a set and ordinals are sets.

Can someone please fill in this gap in my understanding?

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Elements of sets are sets. If $x$ is not a set, then $\{x,\{x\}\}$ is not a set. – Chris Eagle Dec 22 '11 at 18:11
It is not exactly as Chris says, but that is the idea. More precisely, the subset relation is only defined between sets. To be transitive means that every element is a subset, so in particular, any element of an ordinal is a set. – Andrés E. Caicedo Dec 22 '11 at 18:15
@AndresCaicedo Thank you. Crystal clear now. – user18096 Dec 22 '11 at 18:21
@usr18096 : In English, "every" is singular. Thus one may write "Every element of a set is a set". "All" is plural, so one can write "All elements of sets are sets." (I changed "every" to "all" in your question.) – Michael Hardy Dec 22 '11 at 18:21
Thank you. (English is not my first language) – user18096 Dec 22 '11 at 18:34

## 1 Answer

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

Well-ordered means that every non-empty subset has a least element. Transitive means that every element is also a subset.

What this means for your example is that $\alpha = \{ x, \{ x \}\}$ is transitive if $x \subset \alpha$. The only case where $\alpha$ is transitive is if $x = \emptyset$, in all other cases $\alpha$ is not actually an ordinal because then you don't have $x \subset \alpha$.

Edit

The ordinals are an extension of the natural numbers, see here, and the natural numbers start at $0$ which is the empty set $\emptyset$. So every ordinal has to contain $0$, that is, the empty set.

Hope this helps.

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Can you please give me an example of ordinal other than involving emptyset? – user18096 Dec 22 '11 at 18:32
@user18096, every ordinal that is not $\varnothing$ itself contains $\varnothing$ as an element. – Henning Makholm Dec 22 '11 at 18:41
@Matt, if we have the Axiom of Foundation, a transitive set totally ordered by $\in$ will automatically be well-ordered. – Henning Makholm Dec 22 '11 at 18:43
@user18096 I added an answer to your comment into my answer. – Rudy the Reindeer Dec 22 '11 at 18:45
@HenningMakholm Thanks for pointing that out Henning! – Rudy the Reindeer Dec 22 '11 at 18:46