I could use a hand understanding a proof from Jech's Set Theory.
Firstly, note that Jech defines that an increasing function $f : P \rightarrow Q$ is a function that preserves strict inequalities (where $P$ and $Q$ are partially ordered sets). We then have
Lemma 2.4. If $(W,<)$ is a well-ordered set and $f : W \rightarrow W$ is an increasing function, then $f(x) ≥ x$ for each $x \in W$.
I'm having difficulty understanding the proof. It goes like so.
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 get the first sentence - we're doing a proof by contradiction, which is why $X$ is non-empty, and since $W$ is well-ordered, we can conclude that $X$ has a minimum element, and we call it $z$.
What's going on in the second sentence?
In the proof we pick $z$ to be the least, then we set $w = f(z)$, but we know from the definition of the set that $f(z) < z$, so $w < z$. However, $f$ is an (strictly) increasing function, that is $w < z$ implies $f(w) < f(z) = w$. This in turn implies that $w$ should be a part of $X$, but $w < z$ and contradicts that we have picked the least element (which we could because $W$ is well-ordered).
Alternative (and I think that a bit more intuitive) view on this proof is that an existence of element $z$ such that $$f(z) < z$$ implies that $$z > f(z) > f(f(z)) > f^{(3)}(z) > f^{(4)} > \ldots$$ is an infinite strictly decreasing sequence, which is a contradiction due to the fact that $W$ is well-founded.
I hope this helps ;-)