TITLE: A well-order on a uncountable set
QUESTION [8 upvotes]: I can't find an example of a well-order on an uncountable set. Is possible to prove that exists with the Axiom of Choice? How can I give a pratical construction?
I try to define a well-order on Cantor Set, but I'm not able to do.

REPLY [5 votes]: You cannot define a well-order on the Cantor set in ZF without AC (Axiom of Choice) because we can show without AC that there is a bijection from the Cantor set to the reals, and it has been shown (by using inner models of Forcing extensions) that it is consistent with ZF  that the reals cannot be well-ordered.
Without AC, we still can show that the cardinal ordinal $\omega_1$ exists..... $\omega_1$ is uncountable, and well-ordered by $\epsilon.$
First, we develop enough of the theory of well-orders to show that any well-order $<$ on any set $S$ is isomorphic to the  $\epsilon $-order on a unique ordinal. 
Second, let $W$ be the set of subsets of $P(\omega \times \omega)$ that are (the graphs of) well-orders. Now for each $T\in W$ there is a unique ordinal $T^*$ to which $T$ is isomorphic, so   by Replacement and Comprehension,we obtain the set $U=\{T^*:T\in W\}.$
(1) : Any member of $U$ is a countable ordinal.
(2): If $d$ is a countable ordinal there is a bijection $f:d\to e$ for some $e\subset \omega,$  and $T_d=\{(\;f(x),f(y)\;):x\in y\in d\}$ belongs to $W,$ and we have $T_d^*=d.$ So every countable ordinal belongs to $U.$ 
(3).So we obtain the existence of the set of all countable ordinals (namely, $U$). It is easily seen that $U$ is, in fact, the least uncountable ordinal: $U=\omega_1.$ 
Remark on notation: $P(B)$ means the Power-Set of $B$, which is the set of all subsets of $B$.  We cannot get uncountable sets without the Power-Set Axiom, because the class of all hereditarily countable sets (which does not satisfy Power) satisfies all the other axioms of ZF, but also satisfies "All sets are countable."