Hausdorff Maximal Principle implies Well-Ordering Theorem

While looking at a proof of "Maximum Principle implies Well-Ordering Theorem" in this web-page

I came across a Lemma inside this proof which includes a statement in it's own proof that I've found quite problematic. The Lemma states that:

Let $$\mathbf{X}$$ be a set. Let $$\mathcal{A}$$ be the set of all ordered pairs $$\left({A,<}\right)$$ such that $$A$$ is a subset of $$\mathbf{X}$$ and $$<$$ is a strict well-ordering of $$A$$.

Define $$\prec$$ as: \begin{align*} \left({A,<}\right) \prec \left({A',<'}\right) \end{align*} if and only if \begin{align*} \left({A,<}\right) \textrm{ equals an initial segment of } \left({A',<'}\right) \end{align*} Let $$\mathcal B$$ be a set of ordered pairs in $$\mathcal A$$ such that $$\mathcal B$$ is ordered by $$\prec$$.

Let $$B'$$ be the union of the sets $$B$$ for all $$\left({B,<}\right) \in \mathcal B$$.

Let $$<'$$ be the union of the relations $$<$$ for all $$\left({B,<}\right)$$.

Then $$\left({B',<'}\right)$$ is strictly well-ordered set.

This is where I've a problem in this proof here:

https://proofwiki.org/wiki/Equality_to_Initial_Segment_Imposes_Well-Ordering

"Suppose $$\left({A,<_A}\right)$$ equals an initial segment of $$\left({B,<_B}\right)$$ and $$\left({B,<_B}\right)$$ equals an initial segment of $$\left({C,<_C}\right)$$. Then $$\left({A,<_A}\right)$$ equals an initial segment of $$\left({C,<_C}\right)$$ from Equality is Transitive."

How can one make this bold assertion, when we don't know whether or not the well-ordered relation on $$A$$, $$B$$ and $$C$$ are same ? I ask this question because any initial segment of a well-ordered set $$\left({A,<_A}\right)$$ by an element of $$A$$ would depend upon the relation $$<_A$$.

If $$\left({A,<_A}\right)$$ equals an initial segment of $$\left({B,<_B}\right)$$ and $$\left({B,<_B}\right)$$ equals an initial segment of $$\left({C,<_C}\right)$$, it means $$\left({A,<_A}\right) = \left({S_x(B),<_B}\right), \textrm{ where } S_x(B) = \left\{ b \in B \textrm{ }|\textrm{ } b <_B x \right\}$$ and $$\left({B,<_B}\right) = \left({S_y(C),<_C}\right), \textrm{ where } S_y(C) = \left\{ c \in C \textrm{ }|\textrm{ } c <_C y \right\}$$ respectively.

How can we conclude $$S_x(B)$$ is an initial segment of $$\left({C,<_C}\right)$$ when we don't have any relationship between the relations $$<_B$$ and $$<_C$$ ? That is, how can we show that $$S_x(B) = \left\{ b \in B \textrm{ }|\textrm{ } b <_B x \right\} = S_z(C) = \left\{ c \in C \textrm{ }|\textrm{ } c <_C z \right\}$$, for some $$z \in S_y(C)$$ ?

Being an initial segment means, amongst other things, that $$A\subseteq B$$ and $$<_A=<_B\restriction A$$.
For example $$(\{0,1\},\{\langle 0,1\rangle\})$$ is an initial segment of $$\Bbb N$$ with the standard order, whereas $$(\{0,1\},\{\langle 1,0\rangle\})$$ is not.
• I'm sorry I cannot follow you. Are you saying that $<_A = <_B$ when A is treated as an initial segment of B ? – Minto P Jun 11 '20 at 19:50
• I am saying that $<_A$ is $<_B\cap A\times A$. – Asaf Karagila Jun 11 '20 at 20:01