Every member of an ordinal is an ordinal 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?
 A: 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. $\Box$
A: Suppose $\beta\in\alpha$ and $\alpha$ is an ordinal. Let $\gamma\in\beta$. We need to prove that $\gamma\subseteq\beta$. Accordingly, let $\delta\in\gamma$. We need to show that $\delta\in\beta$. 
Since $\alpha$ is an ordinal, and $\beta\in\alpha$, then $\beta\subseteq\alpha$. Therefore, $\gamma\in\alpha$. Again, this gives us that $\gamma\subseteq\alpha$. Thus $\delta\in\alpha$.
Now consider the set $\{\beta,\delta\}$. Since $\alpha$ is well-ordered by $\in$, we have that one of the following holds: $\beta=\delta$, $\beta\in\delta$, $\delta\in\beta$. We need to rule out the first two options.
If $\beta=\delta$, consider the set $\{\beta,\gamma\}$ and note that it contradicts well-foundedness: $\gamma\in\beta$ and $\beta=\delta\in\gamma$.
If $\beta\in\delta$, consider the set $\{\beta,\gamma,\delta\}$ and note that $\beta\in\delta\in\gamma\in\beta$, again contradicting well-foundedness.
The only option we have left is that $\delta\in\beta$, as we wanted.
