Proving $a\le f(a)$ for a monotonically increasing function on a well-ordered set

Question

Let $(x,\le)$ be well-ordered set and let $f: \ x \rightarrow x$ be monotonically increasing function.

Prove that $\forall a \in x$ $$a \le f(a)$$

Find an example of set $x$ linearly ordered such that the statement doesn't hold.

My try:

Assume $X=\{a \in x \ : \ a \ge f(a) \}$ is non-empty set with a minimal element S (we can assume that there is such element, because the set is well ordered).

According to the definition of set $X$ we know that $S\ge f(S)$, but $f(S)$ is in X, thus we found an element smaller than S and that is contradiction.

It seems too easy and I made no use of monotone characteristic of the function.

I have no idea of how to find proper example and what I need to do to complete the proof.

Please help, thank you!

Answer 1

Your attempt is almost fine.

1. $X$ should be $\{a\mid a>f(a)\}$, rather than $\geq$.
2. Monotonity comes into play where you claim that $f(S)$ is $X$ and therefore larger than $S$. You can use it in a clearer manner by pointing out that $S>f(S)>f(f(S))$ and therefore $f(S)$ should be the minimum of $X$ rather than $S$.

Finally, as an example, consider the integers, and something which shifts everything down. Any other linear order which is unbounded from below will do as well (although you can find examples which do have a minimum too).