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This is Exercise III.2.20 of Bourbaki's Set Theory.

(Von Neumann ordinals are actually called "pseudo-ordinals" by Bourbaki, but I simply call them ordinals here)

Let $X$ be a transitive set, and suppose that each $x\in X$ is an ordinal. Then $X$ is an ordinal (Hint: for each $x\in X$, $x\cup\{x\}$ is an ordinal contained in $X$).

This statement is proven in many textbooks. The problem is that none of them uses Bourbaki's definition of ordinal:

For Bourbaki, a set $X$ is an ordinal if every proper transitive subset of $X$ is an element of $X$. [Edit: I'm not sure if this implies e.g. that $X$ is well-ordered by $\in$, since Bourbaki has no Axiom of Foundation.]

A proof that this implies one of the usual definitions (e.g. $X$ is a transitive set whose members are transitive) would be enough, too. [Sorry!]

This should be easy, but I don't know where to start. I'm glad for any help.

Second Edit: The statement I want to show in a hopefully clearer form:

Let $X$ be a transitive set such that any $x\in X$ has the property that any proper transitive subset of $x$ is an element of $x$. Then $X$ has the same property.

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I'm really not sure what your question is after the edit. –  Asaf Karagila Mar 26 '11 at 15:23
    
@Asaf: Thanks for your comment. I hope it's clearer now. –  Stefan Walter Mar 26 '11 at 15:46

1 Answer 1

up vote 2 down vote accepted

Remark: I see in your edit that Bourbaki doesn't have an axiom of foundation. This doesn't affects my proof in any substantial way, since my contradiction is of the form $B\in B$. Thus a slight modification by letting $B=\bigcup\{A\subset X : A \textrm{ transitive and well-founded}\}$ in the first part and analogously letting $B=\bigcup\{A\subset X\setminus\{a\} : A \text{ transitive and well-founded}\}$ in the second part is enough to show that the ordinals as defined by Bourbaki are the standard well-founded ordinals. I also added another way to show this in my edit that again doesn't use the axiom of foundation.

You want to show that if every transitive proper subset of $X$ is an element of $X$ then $X$ is a transitive set whose elements are transitive:

  • If there exists an element $a\in X$ such that $a\nsubseteq X$ then take the set $$B=\bigcup\{A\subset X : A \text{ transitive and well-founded}\}$$ This set is transitive and well-founded as the union of transitive well-founded sets and we have that $a\notin B$ since otherwise we would have $a\subset B\subset X$. Thus $B$ is a proper subset of $X$ and so $B\in X$. We also have that $B\neq a$ since $B\subset X$ while $a\nsubseteq X$. Now the set $B\cup\{B\}$ is a transitive subset of $X$, which means that $B\cup\{B\}\subset B$ (by the definition of $B$). Thus $B\in B$, which is impossible since $B$ is well founded. We arrived at a contradiction because we assumed that $a\nsubseteq X$, therefore $a\subset X$.

  • A similar argument shows that $a\in X$ means that $a$ is transitive: In this case take the set $$B=\bigcup\{A\subset X\setminus\{a\} : A \text{ transitive and well-founded}\}$$ If $a$ is not transitive then $B\neq a$ since $a$ is not transitive while $B$ is. We have that $B\in X$ and therefore $B\cup\{B\}\subset X\setminus\{a\}$. Again this gives us $B\in B$ which is a contradiction.


Edit: The converse it easy. Given an ordinal $\alpha$ let $X\subsetneq\alpha$ be transitive. Then since every element of $\alpha$ is an ordinal and thus transitive, we have that $X$ is a transitive set whose elements are transitive sets, i.e. an ordinal. Since it's a proper subset of $\alpha$ it cannot be $\alpha$ nor can it be longer than $\alpha$ since then we would have $\alpha\in X$ which would give us $\alpha\in\alpha$ a contradiction. Thus $X\in\alpha$. Therefore the two definitions are equivalent. This is enough (since you know that a transitive set of ordinals is an ordinal) to show exactly what you want.

Also here's a quicker way to show the first part of my answer namely that the property implies the fact that $X$ is an ordinal: Let $\alpha$ be the greatest ordinal such that $\alpha\subset X$. Observe that this ordinal indeed exists, since if there is an increasing sequence of ordinals such that each of them is a subset of $X$ then their limit will be a subset of $X$ (because a limit ordinal is the union of a sequence that approaches it and the union of subsets of a set is a subset of that set). Now it's easy to see that $X=\alpha$. If $\alpha\subsetneq X$ then $\alpha\in X$ and thus $\alpha\cup\{\alpha\}\subset X$, which is a contradiction by the definition of $\alpha$.

P.S.: Sorry for the delayed edit but I was out of town for the weekend.

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Thank you! I asked about this, but now I realize that I actually need the converse. If you can prove that as well, it would be very nice of you to extend your answer. –  Stefan Walter Mar 26 '11 at 13:39
    
I'm really sorry. I didn't notice your edit until today. I've become used to getting a notification about anything new and didn't realize that an edit would not cause such a notification. –  Stefan Walter Apr 23 '11 at 17:37
    
If you feel like taking the question up again: The problem is that without the axiom of foundation a transitive set of transitive sets is not necessarily well-ordered by the $\in$ relation. (At least it says so on Wikipedia) I think that in the first paragraph of the edit, the conclusion $X\in \alpha$ requires that $\in$ is a total order. –  Stefan Walter Apr 23 '11 at 18:06
    
@Stefan: You can as well show that the set with that property is in the well-founded universe. If it isn't it contains an element that is not well founded. Then by transfinite induction every ordinal is an element of that set (since it's always going to be a proper subset) which is a contradiction. But furthermore, I already provided with a different way to prove what you want. As for the first paragraph of my edit the $\in$ relation is a total order for ordinals. –  Apostolos Apr 25 '11 at 14:17
    
We're talking past each other. In the edit and in your last comment, what is your definition of ordinal? –  Stefan Walter Apr 25 '11 at 15:54

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