Prove that if $(A,<)$ is a well ordering, then $(A,<)\nless(A,<)$

Question

Prove that if $(A,<)$ is a well ordering, then $(A,<)\nless(A,<)$

I'm trying to teach myself set theory for a course I am taking and am struggling a bit here. I need to suppose for contradiction that there is an isomorphism $f ∶ (A, <) → (pred(b, A, <), <)$ for some $b \in A$. Let $a\in A$ be the least such that $f(a) ≠ a$ and derive a contradiction from here.

Also the notation in my book is a bit hard to read. When I search online for help I wonder if perhaps the text only appears to be writing $<$ when it means $\prec$?

Excuse my ignorance, but I should also just make sure that I am reading this notation correctly. $(A,<)$ simply means some $A$ that is well-ordered yes? If not please feel free to correct me.

As I understand, the starting point I'm provided with says that there exists a function such that for some well ordered set, the predecessors of $b$ in set $A$ also belongs to a well ordered set. From here I need to assume $a$ to be my least valued element of my well ordered set? Moving from this is the suggestion to find a contradiction in $f(a)\neq a$. Here the logic loses me completely. If someone could guide me through this problem I'd greatly appreciate it! Thank you!

Answer 1

If $f\colon (A, <) \to (A,<)$ is 1-1, then $a\le f(a)$ for all $a\in A$. Otherwise, there is a least $a_0\in A$ such that $f(a_0) < a_0$. Let $b = f(a_0)$. We have $b<a_0$, so $f(b) < f(a_0) = b$. But $b \le f(b)$, because $a\le f(a)$ for every $a<a_0$. Contradiction.

In particular we can't have $image(f)\subseteq pred(b)$ for any $b\in A$, because $f(b)\in (A\setminus pred(b))$.

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By the way, here's a different proof, not by contradiction: we show $a\le f(a)$ by well-founded induction on $a$ along $<$. Suppose $a\in A$ and $(\forall x<a)\, x\le f(x)$. Then for all $x<a$, $x\le f(x) < f(a)$, so $$ \sup_{x<a} x \le \sup_{x<a} f(x) \le f(a).\tag{*} $$ There are two cases:

If $a$ has no immediate predecessor, then $a = \sup_{x<a} x $, so $a\le f(a)$ by (*).

Otherwise, $a = b^+$ for some $b\in A$. From $b<a$ we get $f(b) < f(a)$. By induction hypothesis, $b\le f(b)$; thus $a = b^+ \le f(b)^+ \le f(a)$.

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A corollary of this result, as @user254665 points out in a comment:

Well-orderings are rigid with respect to order: a well-ordering has no nontrivial order isomorphisms.

Unlike for example the rationals with the usual ordering $(\Bbb Q, <)$, which has many self-similarities i.e. order automorphisms, a well-ordering has only the identity function as an order automorphism.