Injection from a Well-Ordered Set to $\mathbb{Q}$

Question

Call a set $W \subset \mathbb{R}$ well-ordered if every non-empty $A \subset W$ has a minimum element. Prove that if $W \subset \mathbb{R}$ is well-ordered, then there is an injection from $W$ to $\mathbb{Q}$.

This problem was presented in my Topology class independent of the textbook we are using which is Principles of Topology by Fred H. Croom.

I understand that a function is injective, or one-to-one, if $f(x_1) = f(x_2) \Rightarrow x_1=x_2$. How does this along with the definition of well-ordered tie into a proof of the above problem? Any assistance will be greatly appreciated, thanks in advance.

Answer 1

HINT: If $A\subseteq\Bbb R$, and there is not injection from $A$ into $\Bbb Q$, then $A$ must be an uncountable set. Thus, you need only show that if $A\subseteq\Bbb R$ is uncountable, then $A$ is not well-ordered. One way to do this is to show that $A$ contains an infinite strictly decreasing sequence.

Suppose not; then in particular no point of $A$ is the limit of a decreasing sequence in $A$, so for each $a\in A$ there is an $\epsilon_a>0$ such that $[a,a+\epsilon_a)\cap A=\{a\}$. (Why?)

• Show that the intervals $[a,a+\epsilon_a)$ for $a\in A$ are pairwise disjoint.
• Show that each of these intervals contains a rational number.
• Conclude that there can be only countably many of these intervals.