Linear order and well-order The question is:
A linear(or total) order (L,=<) is a well-order if and only if (1) for every infinite decreasing sequence 
x0 =< x1 =< x2 =< ...
in L, there is an n such that for all m>n, xn =xm.
My answer (sketch) so far:
Assume the linear order (L, =<) and (1). 
For every infinite sequence, we have (1). This gives us an xn ( with n the smallest number such that for every m>n, xn =xm.)* for every infinite sequence. These 'xn' will give us another infinite sequence, and with (1) will give us an element xn (with n again the smallest like before) such that for every x in L: xn =< x .
*= I got the hint that I had to use the axiom of choice, but I dont think I need it when I can just pick the smallest n for which this is true?
 A: You have to use the axiom of choice in order to prove that well-ordered are exactly those linear ordered without infinite decreasing sequences.
The reason is that it is consistent to have linearly ordered sets, which are not well-ordered, that have no countably infinite subsets. In fact it is even consistent that such set is a subset of the real numbers.
Your proof seems somewhat unclear to me. If $(L,\leq)$ is a linear order without an infinite decreasing sequence, take $A\subseteq L$ which is non-empty. Pick some $a_0\in A$, if it is was minimal, you are done. Otherwise there is some $a_1\in A$ such that $a_1<a_0$. Continue by induction and define a sequence $a_n$, and show that it is strictly decreasing, which is a contradiction to the assumption that no such sequence exists.
We use the axiom of choice in order to generate the sequence. By induction we only chose one next element, but the axiom of choice assures us that we can make all these choices uniformly and put them into one sequence.
