TITLE: Using Axiom of Choice To Find Decreasing Sequences in Linear Orders
QUESTION [6 upvotes]: Suppose I have a set along with a relation which is a total order. The set has no minimum element. Now, I want to create an infinite decreasing sequence of elements in this set. For this I have to use axiom of choice. But I'm not able to figure from which family of sets should I pick the elements so that I can use AoC.

REPLY [2 votes]: If you want to apply the axiom of choice directly, pay careful attention to what you're choosing:

First, you choose an element $x_0 \in X$
Second, you chose an element $x_1 \in \{ a \in X \mid a < x_0 \}$

Pay close attention to the second point -- we aren't really choosing an element from a set! Instead, we are choosing from a set-valued function of $x_0$, and so $x_1$ really should be a function of $x_0$!
Now how to apply choice is clear.

First, you choose $x_0 \in x$
Second, you choose $f_1$ from the subset $S_1$ of functions $X \to X$ satisfying $f(a) < a$
Third, you choose $f_2$ from the subset $S_2$ of functions $X \times S_1 \to X$ satisfying $g(a,f) < a$ and $g(a,f) < f(a)$
...

But to make things easier on ourselves, I will simplify:

First, you choose $x_0 \in X$
Second, you choose $f_1 : X \to X$ satisfying $f(a) < a$
Third, you choose $f_2 : X^2 \to X$ satisfying $f(a,b) < a$ and $f(a,b) < b$ 
...

Now, each of our sets is completely determined. We can show they are non-empty (by invoking induction and choice), and so we can apply choice to the collection.
Finally, our sequence is:

$x_0$
$x_1 = f_1(x_0)$
$x_2 = f_2(x_0, x_1)$
...

EDIT: just to make it explicit, the simplified version defines the sets $S_n$ to be the set of functions $g : X^n \to X$ satisfying $g(\vec{x}) < x_i$ for all $1 \leq i \leq n$.
(Depending on your tastes, you may want to special case $S_0 = X$)
For a choice $g \in S_n$, the value $g(\vec{x})$ is meant to be interpreted as "the choice I would make at at time $n$ if I had previously made the choices $x_0, x_1, \ldots, x_{n-1}$ at times $0, 1, \ldots, n-1$".
Now apply the axiom of choice to get a family of choices $f_n \in S_n$. Under the intended interpretation, our chosen decreasing sequence is
$$x_n = f_n(x_0, x_1, \ldots, x_{n-1})$$
(If you're special casing $0$, then $x_0 = f_0$ instead)
Of course, while this approach is systematic, it did not lead to the simplest result. At some point we might have recognized that our choice at time $n$ could be made to depend only on the choice made at time $n-1$ rather than the entire family of choices made at times $0$ through $n-1$. And if we do that, we might have then realized a single choice would work for all positive times. This would result in Arturo's answer.