How can I prove that for 2 ordinals $a,b \ s.t. a \ne b$ that $a\in b\vee b\in a$?
The definition in my course for an ordinal is: A set $A$ is an ordinal if and only if:
- $A$ is a transitive set. ($\forall a,b\ s.t.\ a\in b\in A,\ a\in A$).
- $\forall x\in A, x\notin x$.
- $\forall x,y,z\in A\ s.t.\ x\in y\in z,\ x\in z$.
- $\forall x,y\in A\ s.t.\ x\ne y, x\in y \vee y \in x$
- $\forall \varnothing\ne X\subseteq A$, there exists a minimal element in $X$ according to the membership relation.
This is a small part of my h.w. and this part is the only part I don't know where to start from. Thnaks.
If you're allowed to use transfinite recursion at this stage in your course, we may use a proof by contradiction. If a counterexample exists, choose the least $a$ for which such a $b$ exists, then choose $b$ to be minimal for that $a$. Since each $c\in b$ satisfies $c=a\lor c\in a\lor a\in c$, we have $a\in b$ unless each $c\in b$ satisfies $c\in a$. In that case $b\subseteq a$. By the same logic, if $a\notin b\notin a$ then $a\subseteq b$. By extensionality, $a=b$.