I want to prove that if "there are not well-ordered sets then there are incomparable sets. In fact if $a$ is not a well-ordened set then there is an ordinal $\alpha $ such that neither $\alpha \preceq a$ nor $a \preceq \alpha$"
idea: suppose $a\notin BO$, I've already know that doesn't exist $\gamma [a\preceq \gamma ]$ i.e. for all $\gamma [a \not\preceq \gamma ]$. Then it´s sufficient to prove that you can find $\alpha $ such that $\alpha \not \preceq a$, so, let $B_{a}= \left \{\beta /\exists u,v[u\subseteq a \wedge v\subseteq u\times u \left \langle u,v \right \rangle\in COBO \wedge \tau (u,v)=\beta ] \right \}$ then $B_{a} \in V$ then $B_{a}$ is the set that works, but i can´t prove that $B_{a}$ is a set!, I thought it was easy but I'm stuck; I want to follow this idea; I just have to prove that detail. Please help, thanks!
HINT: Show that there is an $n\in\omega$ such that $\langle u,v\rangle\in\wp^n(a)$ whenever $u\subseteq a$ and $v$ is a well-order on $u$. Then you can use the comprehension schema to define
$$W=\{\langle u,v\rangle\in\wp^n(a):u\subseteq a\land\langle u,v\rangle\text{ is a well-order}\}$$
and the replacement schema to get your set $B_a$.