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

https://proofwiki.org/wiki/Hausdorff_Maximal_Principle_implies_Well-Ordering_Theorem#:~:text=By%20the%20lemma%2C%20(B%E2%80%B2,a%20maximal%20subset%20of%20X.

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 $z \in S_y(C)$ ?

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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.

So yes, an initial segment of an initial segment, is an initial segment.

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  • $\begingroup$ I'm sorry I cannot follow you. Are you saying that $<_A = <_B$ when A is treated as an initial segment of B ? $\endgroup$ – Minto P Jun 11 '20 at 19:50
  • $\begingroup$ I am saying that $<_A$ is $<_B\cap A\times A$. $\endgroup$ – Asaf Karagila Jun 11 '20 at 20:01

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