Initial segment of well ordered set

Question

An initial segment (B) of an well-ordered set $(A,<)$ is a subset $X⊊A$ such that, for all $x \in X$ and for all $y∈A$ such that $y<x, y∈X$.

I want to prove that this set is equal to $A_z =$ { $a \in A: a<z $}

Where $z$ is the minimum element of $A-B$.

I have shown that $A_z \subseteq B$.

Answer 1

For the other direction we notice that $\forall b \in B: b < z$, because if this were not the case, we have an elment $b \in B$ s.t. $b >z$ and by definition $z$ would have to be an element of $B$, this is a contradiction. So we now know that $z$ is an upper bound of $B$, so clearly $B \subseteq A_z$.