By my definition a similarity is a bijective function $f:X \rightarrow Y$ between well ordered sets such that $x\leqslant y \iff f(x) \leqslant f(y)$. I understand some people refer to this as an order isomorphism.
What I am struggling to grasp is the following theorem in my notes:
“Let $X$ be a well-ordered set and let $f:X \rightarrow X$ be a mapping such that $x<y \implies f(x)<f(y) \;\;\forall x,y \in X$. Then $x\leqslant f(x) \;\; \forall x\in X$.”
This may seem like a silly question but what is this theorem actually telling us? Why is a mapping from a set onto itself and why has the order changed to strict order?
I have done some working on this and I believe $x<y \implies f(x)<f(y) \;\;\forall x,y \in X$ is equivalent to the similarity condition so I’m confused why this is a theorem because this isn’t an additional special property (if I’m right by my working that they are equivalent?
The point of the theorem is that if $X$ is well-ordered, and $f:X\to X$ is strictly order-preserving, then it is not possible for $f$ to send some $x\in X$ to a smaller member of $X$: there is no $x\in X$ such that $f(x)<x$. This turns out to be a significant and useful fact about well-orders that is probably being proved now so that it can be used soon.
Note that it does not follow if we assume only that $f$ is weakly order-preserving, i.e., that $x\le y$ implies $f(x)\le f(y)$, as in that case $f$ could even be a constant function sending everything to the smallest member of $X$. Note also that there is no requirement here that $f$ be a surjection: it might, for instance, send each $x\in X$ to its immediate successor.