Ordinals trichotomy property > Given that $A,B$ are ordinals, prove that $A=B$ or $A \in B$ or $B \in A$ **My attempt** Assume $A \not = B$ and $B \not\in A$ and i want to prove that $A \in B$ Let $C = A \cap B$. I think if I prove $A = C$ I am done. $C \subset A$ because $ C = A \cap B$ , but I am stuck on the other side. Proving that $A \subset C$ ?! I think I need to use the fact that in any non-empty subset of ordinals there is a minimum)

Assume $A\
otin B$, $B\
otin A$ and $A\
eq B$. By symmetry we can assume that there is $C_1\in B\setminus A$.

Then $C_1\subset B$, $A\
otin C_1$, $C_1\
otin A$ and $C_1\
eq A$.

Repeat the same argument with $A$ and $C_1$ to get a $C_2\in$ either $A\setminus C_1$ or $C_1\setminus A$.

You find an infinite sequence of $C_i$ belonging to either $B$ or $A$. This is impossible due to the well-ordering. Therefore $A\
otin B$, $B\
otin A$ and $A\
eq B$ is false.