Talk:Order Isomorphism between Wosets is Unique

Hello, I am reading this book. I have a question. Is the order isomorphism from a woset to itself unique?

My thoughts (though not rigorously presented) are as follows:

Let $(S,\preceq)$ be a woset and let $h:S \to S$ be an order isomorphism.

Let $x_0$ be the smallest element of $S$

As $\forall x \in S: x \preceq h(x)$. Suppose $x_0 \prec h(x_0)$.

(and this is the bit where I may have made a mistake):

$h$ is not surjective as nothing gets mapped to $x_0$. So then $h(x_0)=x_0$.

Then I was going to use the woset version of induction to prove that $h(x)=x$ for the remaining elements.

Am I making a mistake here?

--Jshflynn 12:41, 6 July 2012 (UTC)


 * I think not. You may want to use the actual definition of an order iso that $x \preceq_1 y \implies f(x) \preceq_2 f(y)$ from which the questioned part of your reasoning directly follows. Induction then may indeed be employed to prove this (as usual, take care to distinguish the successor and limit element cases). Alternatively, just invoke the proof on this page; it works. --Lord_Farin 13:17, 6 July 2012 (UTC)


 * I think I worded my question wrong. Perhaps this is better "Does there exist a woset $(S,\preceq)$ for which there are two distinct order automorphisms?"


 * Again, just invoke the theorem this page is the talk of; there is no imposition that $(S_1,\preceq_1)$ and $(S_2, \preceq_2)$ be distinct. Therefore, no. --Lord_Farin 14:07, 6 July 2012 (UTC)


 * Thanks LF. I was trying to prove this special case (without referencing this theorem) so that I could provide an alternative proof for this theorem. Using: $f \circ g = id \implies f^{-1}=g$ (if $f$ and $g$ are bijections). But the proof on this page seems more economical anyway. --Jshflynn 14:37, 6 July 2012 (UTC)


 * In that case, read my first answer again, I think it answered your question alright after all. --Lord_Farin 21:39, 6 July 2012 (UTC)