I was reading this problema about ordinals. I'm following the Z.F theory.
Let $A$ be a nonempty set of ordinals. Prove that if $\cup A \notin A$ then $\cup A$ it's a limit ordinal, i.e $\cup A $ is nontempty but that is directly, and also is not a sucesor of any ordinal.
I proved that $ A\subset \cup A$. In the following way:
if $\alpha \in A$ , since $\cup A \notin A$ then $ \alpha \ne \cup A$ and since $\cup A$ is an ordinal, then $\alpha \in \cup A$ (since if $\cup A \in \alpha$ $\wedge$ $\alpha \in A$ $\Rightarrow$ $\cup A \in \cup A$ a contradiction if we consider the regularity axiom).Thus we have that $ A \subset \cup A$
Remark: I'm using the fact that $\cup A$ is an ordinal, and also that given two ordinals $\alpha$, $\beta$ are always comparable, in the sense that some of this occurs $\alpha \in \beta$, $\beta \in \alpha$ or $ \alpha=\beta$. And this is a theorem.
I don't know what can I do with the other inclusión $ A \subset \cup A$, it's easy to prove that a set A is $\in$-transitive ( $x\in A$ $\wedge y\in x$ $\Rightarrow y\in A$) if and only if $\cup A \subset A$ but it is always true that a set of ordinals is always $\in$-transitive? or this happens only in this case? Please I want a counterexample or the proof of that part )= , because I'm a Little confused
Note that transitive sets of ordinals are ordinals. So if $A$ is a transitive set then it is in fact an ordinal.
However what you can prove is that $\bigcup A=\sup A$, so whenever $\alpha\in A$ we have that $\alpha\leq\sup A=\bigcup A$. If $\bigcup A\notin A$ then $A$ does not have a maximal element, and therefore $\sup A$ must be a limit ordinal.