TITLE: Every member of an ordinal is an ordinal
QUESTION [13 upvotes]: How to prove that, if a is an ordinal and b is in a, then b is an ordinal?
Here are the definitions I'm using.
A set is an ordinal number if it is transitive and well-ordered by ∈.
A set T is transitive if every element of T is a subset of T.
My difficulty is mainly that I can't prove b is transitive. What's the magic?

REPLY [5 votes]: Simpler proof than the one given:
Suppose $\beta\in\alpha$ and $\alpha$ is an ordinal. Suppose that $\gamma\in\beta$ and that $\delta\in\gamma$.
Since $\alpha$ is an ordinal, $\alpha$ is a transitive set. Therefore, since $\beta\in\alpha$ and $\gamma\in\beta$, we must have $\gamma\in\alpha$. Again, since $\gamma\in\alpha$ and $\delta\in\gamma$, $\delta\in\alpha$.
Now we have all the facts we need:

$\in$ is a well-order on $\alpha$, which means that $\in$ is transitive on $\alpha$.
$\beta,\gamma,\delta\in\alpha$
$\gamma\in\beta$ and $\delta\in\gamma$

Since $\in$ is transitive on $\alpha$ and $\delta\in\gamma$ and $\gamma\in\beta$ and $\beta,\gamma,\delta\in\alpha$, we must have $\delta\in\beta$. Thus, $\beta$ is a transitive set. $\Box$