# Let $W$ be well-ordered and $f$ increasing. If $z$ is the smallest $x$ such that $f(x)<x$, why does $f(f(z))<f(z)$?

I am currently working on Set Theory of Thomas Jech, and I have a question about Lemma 2.4 which is :

Lemma
If $$(W, <)$$ is a well-ordered set and $$f:W \rightarrow W$$ is an increasing function, then $$f(x)\geq x$$ for each $$x \in W$$.

Proof
Assume that the set $$X=\{x \in W : f(x) < x\}$$ is nonempty and let $$z$$ be the least element of $$X$$. If $$w=f(z)$$, then $$f(w), a contradiction.

I have a problem understanding the proof. I've found that the proof makes sense if $$w \in X$$ since
(i) $$z\in X \rightarrow f(z) and
(ii) $$z$$ is least element of $$X \rightarrow (\forall x \in X)z \leq x)$$
(iii) $$f$$ is increasing $$\rightarrow (\forall x \in X)f(z) \leq f(x)$$
(iv) $$w \in X \rightarrow f(w) < w$$
But by (iii) and $$w \in X$$, $$f(z) \leq f(w) \iff w \leq f(w)$$. This contradicts with (iv).

However, I can't figure out why $$w \in X$$. Why is it? Or is there any mistake with my proof?

• $f(z) < z$ implies $f(f(z)) < f(z)$ since $f$ is increasing, hence $f(z) \in X$.
– Tan
Commented Jan 22, 2021 at 15:51
• Ahhh thx,, I was stupid as hell. thank you.
– 이승우
Commented Jan 22, 2021 at 15:58
• Isn't the lemma false, or am I misunderstanding it? $\mathbb R$ is a well-ordered set, and $f(x)=\frac x2$ is increasing. But $\frac x2\leqslant x$. Commented Jan 22, 2021 at 16:23
• @MrPie $\mathbb R$ is not well-ordered by the usual order.
– Tan
Commented Jan 22, 2021 at 16:27
• @HanulJeon My impression is that there is no serious alternative to Jech for the student aiming toward research in set theory. But I am not a set theorist, maybe that’s wrong. Commented Jan 22, 2021 at 17:37

Since $$f$$ is increasing and $$w, $$f(w).