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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 $f(w)<w$, a contradiction.

I have a problem understanding the proof. I've found that the proof makes sense if $w \in X$ since
(i) $z\in X \rightarrow f(z)<z$ 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?

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  • $\begingroup$ $f(z) < z$ implies $f(f(z)) < f(z)$ since $f$ is increasing, hence $f(z) \in X$. $\endgroup$
    – Tan
    Commented Jan 22, 2021 at 15:51
  • $\begingroup$ Ahhh thx,, I was stupid as hell. thank you. $\endgroup$
    – 이승우
    Commented Jan 22, 2021 at 15:58
  • $\begingroup$ 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 $\frac x2\leqslant x$. $\endgroup$
    – Mr Pie
    Commented Jan 22, 2021 at 16:23
  • $\begingroup$ @MrPie $\mathbb R$ is not well-ordered by the usual order. $\endgroup$
    – Tan
    Commented Jan 22, 2021 at 16:27
  • 1
    $\begingroup$ @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. $\endgroup$ Commented Jan 22, 2021 at 17:37

1 Answer 1

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Since $f$ is increasing and $w<z$, $f(w)<f(z)=w$.

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  • $\begingroup$ Thank you for your answer. I was confused with such a small thing lol $\endgroup$
    – 이승우
    Commented Jan 22, 2021 at 15:59

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