A well-order on a uncountable set

Question

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.

Answer 1

You can, without AC, prove that the first uncountable well-order exists. Without going too much into every detail, it may be done like this:

Using the normal $<$ ordering of $\Bbb Q$, we get that some elements of $\mathcal P(\Bbb Q)$ are well-ordered, and, more importantly, every countable well-order appears as such a set. For any countable ordinal $\alpha$, let $X_\alpha \subseteq \mathcal P(\Bbb Q)$ be the set of all subsets of $\Bbb Q$ that are order isomorphic to $\alpha$. Also note that we can impose an order of those sets in a very natural way: $X_\alpha < X_\beta$ iff $\alpha < \beta$. Then $$ \{X_\alpha \mid \alpha \text{ is a countable ordinal}\} \subseteq \mathcal P(\mathcal P(\Bbb Q)) $$ is actually well-ordered, and order isomorphic to the first uncountable ordinal.

Answer 2

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."

Answer 3

$\operatorname{ZF}$ proves the equivalency of the axiom of choice and the so called well order principle. The well order principle simply states: For every nonempty set $x$ there is a relation $R \subseteq x \times x$ such that $(x; R)$ is a well-order. Hence, in $\operatorname{ZFC}$, there exists a well order on the Cantor set. However, one cannot hope to explicitly construct such a set. So there is no 'practical' solution, if I interpret the meaning of this correctly.

To prove the well order principle in $\operatorname{ZFC}$, the following suffices: Let $x$ be a nonempty set and let $F \colon \mathcal P(x) \setminus \emptyset \to x$ be a choice function. Recursively construct a sequence $(x_{i} \mid i < \kappa)$, $x_i \in x$, such that given $(x_i \mid i < \kappa)$, for some $\kappa$, we stop the construction if $\{ x_{i} \mid i < \kappa \} = x$ or otherwise let $x_{\kappa} = F(x \setminus \{x_{i} \mid i < \kappa\})$. (Intuitively, $F$ picks the next element of $x \setminus \{x_i \mid i < \kappa \}$.)

This construction must eventually stop, simply because $x$ has (as a set) bounded rank. We hence get an enumeration $x = \{ x_i \mid i < \kappa \}$ for some ordinal $\kappa$. We may then define a well order $R \subseteq x \times x$ by letting $x_i R x_j$ iff $i \le j$ in the enumeration above. It's easy to verify that this is indeed a well order on $x$.