Countable Intervals of Reals

Let us consider $\{I_{\Lambda}\}_{\lambda \in \Lambda}$, the collection of non-empty pairwise-disjoint intervals on $\mathbb{R}$. We assume every interval contains at least two distinct elements. Prove the collection of these intervals are countable.

To prove $\Lambda$ is countable, we construct an injective function $f: \Lambda \rightarrow \mathbb{Q}$.

A function $f : A \rightarrow B$ is injective if $a_1 \neq a_2 \implies f(a_1) \neq f(a_2)$ for all $a_1,a_2 \in A$.

Since $\mathbb{Q}$ is dense in $\mathbb{R}$, between any two real numbers, there is a rational number. So for each $\lambda \in \Lambda$, a rational exists within each interval $I_\lambda$.

Let the function $f : \Lambda \rightarrow \mathbb{Q}$ be defined by letting $f(\lambda)$ be a rational number in $I_\lambda$.

We assume the Axiom Of Choice which asserts given a collection of nonempty sets $A_i$ for $i \in I$, there exists a choice function $f$ such that $f(i) \in A_i$ for all $i \in I$.

Suppose $f(\lambda) = f(\lambda’) = r$. We know $r \in I_\lambda$ and $r \in I_{\lambda’}$. By definition of intersection, $r \in I_\lambda \cap I_{\lambda’}$.

If $\lambda \neq \lambda’,$ then $I_\lambda \cap I_{\lambda’} = \emptyset$, but we said $r \in I_\lambda \cap I_{\lambda’}$, so it must be that $\lambda = \lambda’$. Therefore f must be injective. Since we have found an injection with a countable set $\mathbb{Q}$, we have proven that $\Lambda$ is countable.