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I have found the following Lindenbaum theorem (here, page7)

It states that:

Suppose $X$, $Y$ are linear orders. If $X$ is isomorphic to an initial segment of $Y$ (with $f$) , and $Y$ is isomorphic to a final segment of $X$ (with $g$), then X $\simeq$ Y.

I can't seem to prove it. I have the bijection in the constructed out of $f$, $g$ as in the classical proof of the Schroeder-Bernstein theorem, but I can't show that it's order embedding. Any tips ?

Edit: In the article it's written " the hypotheses(f, g are isomorphisms) guarantee that this bijection is order-preserving"

But how does it guarantee it ?

Paul
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1 Answers1

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Disclaimer: I'm not an expert, just want to try solving this problem.

Let $X_f$ be final segment of $X$ and $X_i = X \setminus X_f$, i.e. $X = X_i \sqcup X_f$. Let $Y_i$ be initial segment of $Y$ and $Y_f = Y \setminus Y_i$, i.e. $Y = Y_i \sqcup Y_f$. Let's construct a chain $$ \cdots \rightarrow f^{-1}(g^{-1}(a)) \rightarrow g^{-1}(a) \rightarrow a \rightarrow f(a) \rightarrow g(f(a)) \rightarrow \cdots $$ This sequence may terminate on the left either in set $X$, or in set $Y$, or it may never terminate. Thus we can split all elements into $X$-stoppers, $Y$-stoppers, and never-stoppers.

  • We use $f$ as bijection for $X$-stoppers and never-stoppers, obviously we can compare any two of them because $f$ preserves order.
  • We use $g$ as bijection for $Y$-stoppers, obviously we can compare any two of them because $g$ preserves order.
  • We have to prove that $X$-stopper can be compared to $Y$-stopper.

Statement 1 All $X$-stopper chains originate in $X_i$. Proof. Consider the last $f^{-1}$ in the chain above that yielded an element $x \in X$. By the definition of $X$-stopper there is no such $y \in Y$ that $g(y)=x$. But $Y$ is isomorphic to $X_f$, thus $x \notin X_f \rightarrow x \in X_i$.

Statement 2 All $Y$-stopper chains originate in $Y_f$.

Final Statement All $X$-stoppers are smaller than $Y$-stoppers.

Proof Consider two elements $x_X$ and $x_Y$ from $X$-stopper and $Y$-stopper chains. Let me apply $f^{-1}$ and $g^{-1}$ until exhaustion. Note that sign $<$ or $>$ between $x_X$ and $x_Y$ remains unchanged at every step since we are applying the same function to both elements simultaneously. Suppose, the $X$-chain was the first one to exhaust, which means we can draw our chains as $$ \begin{aligned} &x'_X \xrightarrow[]{f} \dots \xrightarrow[]{g} x_X \xrightarrow[]{f} \dots \\ y''_Y \xrightarrow[]{g} \dots \xrightarrow[]{g} &x'_Y \xrightarrow[]{f} \dots \xrightarrow[]{g} x_Y \xrightarrow[]{f} \dots \\ \end{aligned} $$ By the statement 1 $x'_X \in X_i$. The other chain did not stop since it is a $Y$-chain, thus there exists $g^{-1}(x'_Y)$ and $x'_Y \in X_f$. Thus $x'_X < x'_Y$ and as a result we get $x_X < x_Y$. If we stop in $Y$, similar reasoning applies.

guest
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  • I am confused at "All X-stopper chains originate in X". I am not sure the proof is full? We have an element x which is part of X-STOPPER . Can't we have x -> g^-1(x) -> f^-1(g^-1(x)) -> NONE. X is part of a X-stopper but X has y \in Y such that (g(y) = x) – Paul Feb 20 '21 at 11:30
  • @PaulKeseru , yes, $x$ may be part of $X$-stopper chain, but not its origin. By "originate" I mean that the first element that has no predecessor should be in $X_i$. That's $x'_X$ in my designations. The trick is to iterate backwards until we hit the origin, at the origin we gain enough information to make comparison between $x'_X$ and $x'_Y$, then we propagate this sign forward until we hit $x_X$ and $x_Y$ (itś possible since $f$ and $g$ are order-preserving) – guest Feb 20 '21 at 12:10