I apologize for this naive question, but I'm new to set theory and I'm having a hard time figuring out properties of supremum of ordinals.
From here on out I will refer to $\alpha$ as a nonzero ordinal.
If I understand correctly the following are equivalent.
- $\alpha$ is a limit ordinal;
- $\bigcup \alpha = \alpha$.
Moreover, in general we have that $\bigcup \alpha$=$\text{sup }\alpha$.
Now I don't understand one thing, if $\bigcup \alpha$=$\text{sup }\alpha$, this means that whenever $x\in\alpha$ we have that $x\in\bigcup\alpha$ (because $\bigcup\alpha$ is the supremum of $\alpha$). However by transitivity $x\in\bigcup\alpha$ implies $x\in\alpha$. So it seems to me that property 2. hold for all ordinals, limit and successor.
Now, I know that there is some problem in my understanding, but where?
Also, consider $6= \{ 0,1,2,3,4,5 \}$, is it correct to say that $\bigcup 6 = \{ 0,1,2,3,4 \} $ ? If so, the previous point $x\in\alpha$ implies $x\in\bigcup\alpha$ is not true (because $5\in 6$, but $5\not\in\bigcup 6$), but then I must be missing something about the supremum properties. I thought the supremum was the least of the upperbounds, so as an upperbound, whenever $x\in\alpha$ we should have $x\in \text{sup } \alpha$.
The thing is that $x\in\alpha$ only means that $x\leq\sup\alpha$. So, knowing that $\sup\alpha=\bigcup\alpha$ we have that $x\in\bigcup\alpha$ OR $x=\bigcup\alpha$.
Remember, an ordinal is itself a set of ordinals, and the supremum is not "the least one not in the set", but rather "the least one that is *greater or equal to all the elements of the set". So, when a set of ordinal has a maximum, that maximum is the supremum, which is exactly what happens at successor ordinals.
In particular, $\bigcup 6=5$.