Set Theory Ordinal Problem I have this problem from Enderton's Elements of Set Theory which says, "Assume that $\lt$A, < $\gt$ is a well-ordered structure with ordinal number $\alpha$ and that B$\subseteq$A. Let $\beta$ be the ordinal number of the well-ordered structure $\lt$B, <$^\circ$$\gt$ and show that $\beta \in \alpha$ or $\beta=\alpha$". 
My first approach was to let $a$ be the least element of A-B and say that E(seg($a$))=$\beta$ (where E is the epsilon-image function) and continue from there, but now I realize this won't work, as B isn't necessarily an initial segment of A; we could have A={a,b,c,d,e} (where a< b< c< d< e) and B={b,c,d,e}. Then B$\subseteq$A, but E(seg($a$))$\neq\beta$. So now I'm not sure what to try. It seems obvious that it's true, but I'm having trouble turning that into an actual proof. 
I'm thinking now that, using trichotomy of ordinals, I really just have to show that it's NOT true that $\alpha\in\beta$, so I might be able to do a proof by contradiction. If anyone might be able to give me a hint as to where I should look next, I'd appreciate it.
Edit: Could I argue something like this?
Suppose $\alpha\in\beta$. Then $\lt\alpha, \in_\alpha\gt$ $\simeq$ $\lt A$, <$\gt$ $\simeq$ $\lt$ seg(a), <$^\circ\gt$ for some a$\in$B, but B$\subseteq$A so a$\in$A, and we have A is isomorphic to a segment of itself, which is a contradiction?
 A: Hint: Can there be an order-preserving embedding $\beta \to \alpha$ when $\alpha \in \beta$ ?
A: Let us say $A=\alpha$ just for convenience. Then we have a map $f:\beta 
\to \alpha$ which sends $\beta$ in an order preserving fashion onto $B\subseteq \alpha$. Now every order preserving map has the property that $\mu\leq f(\mu)$ for all $\mu$. Thus we have the following, 
if $$\mu \in \beta$$ then
$$\mu\leq f(\mu)$$ and 
$$f(\mu)\in \alpha$$ and thus 
$$\mu \in \alpha.$$ This shows that $\beta \subseteq \alpha$ and this is equivalent to $\beta \leq \alpha$.
A: I found problem in Rene's answer. I don't think it can be proved that every order preserving map has the property $\mu \le f(\mu)$. Maxime's hint is the right direction and helped me finish my proof:
Prove by contradiction. Assume $\alpha \in \beta$. Then there exists $t \in B$ that $\langle A, \lt\rangle \cong \langle seg_B(t), \lt^t\rangle$, which means we have an order preserving map $f : A \to seg_B(t)$.
Since $B \subseteq A$ and then $t \in A$, we have $f(t) \in seg_B(t)$, which means $f(t) \lt t$. Let $\{x \in B|f(x) \lt x\}$ be C. Then C is nonempty as we have shown that t fits in. So by well order principle there is minimum m in C. But that means $f(m) \lt m$ which will violate the minimality of m if we can show that $f(m) \in C$.
To show $f(m) \in C$, first notice that $f(m) \in seg_B(t)$ so $f(m) \in B$. And the last thing we have to show is $f(f(m)) \lt f(m)$. It can be obtained from assumption $f(m) \lt m$ and the order preserving property of f.
