Well-ordered set and countable set Question 1: Let $P$ is a well-ordered set of real numbers with the usual order. Is $P$ always countable? I think it is true, but I can't prove it.
Question 2: Let $P$ is a well-ordered set. Is $P$ always countable? 
Thanks in advance.
 A: Suppose $A \subseteq \mathbb{R}$ is well-ordered in the order $<$ from $\mathbb{R}$. Then it is at most countable.
Suppose $A$ is well-ordered in $<$.
Then for each $a \in A$, there is some $r_a>0$ such that
$A \cap [a, a+r_a) = \{a\}$ (it's a right-sided limit point).
For, if this were not the case:
$$\forall r>0: \exists a'(r) \in A: a'(r) \neq a \land a'(r) \in [a, a+r)$$
And then take $a_1 = a'(1)$, and having defined $a_n$ set $a_{n+1} = a'(a_n -a)$. This defines a decreasing sequence $a_n$ in $A$, and so $\{a_n : n \in \mathbb{N}\}$ does not have a minimum, contradicting well-orderedness. So the claim holds, and we have the required $r_a$. Note that all intervals $[a, a+r_a)$ are pairwise disjoint by construction, and the set of open intervals
$(a, a+r_a)_{a \in A}$ is a family of pairwise disjoint non-empty open intervals of $\mathbb{R}$ of size $|A|$ and each interval contains a (necessarily different) rational, and $\mathbb{Q}$ is countable. 
So $A$ is at most countable.
As to 2): In ZF we can develop countable ordinals and take the union of all of them. This is $\omega_1$ : the first uncountable ordinal, which is well-ordered (as all ordinals). So there are uncountable well-ordered sets.
Using AC (the axiom of choice) we can prove that all sets regardless of size, can have a well-order defined on them. 
A: Q1: For every $x\in P$, the set $\{\,y\in P\mid y>x\,\}\cup \{x+1\}$ has a minimal element $y_x$. Pick a rational number $q_x\in [x,y_x)$. This gives us an injective  map $P\to\Bbb Q$.
Q2: No. In set-theory, well-order manifests best in ordinals. Since I do not know if aou are acquainted with them, these are, in short,  sets $\alpha$ with the properties that a) $x\in \alpha$ implies $x\subseteq \alpha$ and b) the $\in$ relation is a well-order on $\alpha$.
Some elementary facts: 


*

*It is vacuously true that $\emptyset$ is an ordinal.

*If $\alpha$ is an ordinal, then $\alpha\cup\{\alpha\}$ is an ordinal.

*If $S$ is a set of ordinals, then $\bigcup S$ is an ordinal

*Every well-ordered set is order-isomorphic to a unique ordinal


Let $B$ the set of well-order relations on $\Bbb N$ (that need not be related to the usual order on $\Bbb N$), and for each $b\in B$, let $\alpha_b$ be the unique ordinal order-isomorphic to $b$. Let $S=\{\,\alpha_b\mid b\in B\,\}$ and $\beta=\bigcup S$. Then $\beta\notin S$, i.e., $\beta$ is not order-isomorphic to any well-ordering of $A$. Hence there cannot exist a bijection $\Bbb N\to\beta$. On the other hand, the standard order $<$ of $\Bbb N$ is a well-order, hence  then there does exist a bijection $\Bbb N\to \alpha_<\subseteq \beta$. It follows that $\beta$ is uncountable. 
(Remark: If we replace $\Bbb N$ with any other set, the same proof shows that if a set is well-orderable, then there exists a well-orderable set of larger cardinality)
