Well ordered Isomorphic sets (ZF) Let A be a infinite set. Prove that if there is a well order
≤1 in A, then there is a well order ≤2 in A, for which <A, ≤1> and <A, ≤2>
are not isomorphic.
My attempt:
I start with the use of the definition of well ordered set, hence for every non empty B in A there exists a minimal element to ≤:
(∀ B)[B != ∅ & B ⊆ A → (∃ y)[y ∈ A & (∀ x )[x ∈ A → y ≤ x ]]].\
Also we say that <A, ≤1> is isomorphic to <A, ≤2> if there is a bijection f:A1-> A2 such that if for all x and all y, <x,y> is in R1, then <f(x1),f(x2)> is in R2.
If I asume that A1 and A2 are well ordered, then they have a minimal element x0,y0, respectively.Then if A1,A2 are isomorphic there is bijection f such that f(x0)=z1, z1 in A2. Since f preserves the order=> z1 != y0 is a minimal element-contradiction. Hence A1 and A2 are not isomorphic.
I beg someone to check my attemp. I will be thankful to anyone who share their ideas.
 A: There is no actual argument here. You start by assuming that there is an isomorphism $f$ from $\langle A,\le_1\rangle$ to $\langle A,\le_2\rangle$, and then you simply assert that this isomorphism is in fact not an isomorphism, because $f(\min_{\le_1}A)\ne\min_{\le_2}A$. Assuming something and then asserting its negation is not a proof by contradiction: it is simply asserting two contrary things independently.
Moreover, the result holds only for infinite sets, and nowhere have you used the fact that $A$ is infinite.
What you must do is assume that $\le_1$ well-orders $A$ and use $\le_1$ to produce a new well-order $\le_2$ that is not isomorphic to $\le_1$. I will tell you one way to do this, but I will leave it to you to prove that it works.
Let $a=\min_{\le_1}A$, the least element of $A$ with respect to the order $\le_1$. To form the new order $\le_2$, move $a$ to the other end. That is, define $x\le_2y$ if and only if $x\le_1y$ and $x\ne a$, or $y=a$. Pictorially, if $\langle A,\le_1\rangle$ looks like this:
$$a\;\bullet\;\bullet\;\bullet\;\cdots$$
then $\langle A,\le_2\rangle$ looks like this:
$$\bullet\;\bullet\;\bullet\;\cdots\;a$$
