# A collection of pairwise disjoint open intervals must be countable

Let $U$ be a collection of pairwise disjoint open intervals. That is, members of $U$ are open intervals in $\mathbb{R}$ and any two distinct members of $U$ are disjoint. Show that $U$ is countable.

My attempted proof:

Let $U \subseteq \mathbb{R}$ be a collection of pairwise disjoint opern intervals. Let $(a,b) \subseteq U$ be one such interval. Since $\mathbb{Q}$ is dense in $\mathbb{R}$, there exists a rational number $q_{a,b} \in (a,b)$. Define $f: U \to \mathbb{Q}$ by $f((a,b))=q_{a,b}$. As $f(a_1,b_1)=f(a_2,b_2)$, or $q_{a_1,b_1}=q_{a_2,b_2}$, implies $(a_1,b_1)=(a_2,b_2)$, we deduce that $f$ is injective. It was proved already in my lecture class that $\mathbb{Q}$ is countable, and that union of countable sets is countable. Thus, we deduce that $U$ is countable. (End of attempted proof)

I feel confident about my proof except the injectivity part. I was looking for help there, assuming the rest of my proof is cogent. If not, please let me know how I can improve on it.

• Jul 25 '16 at 16:11
• Where did you use disjointness? I think the fact that the intervals are disjoint must be important, yet nowhere in the proof did you say "since the intervals are disjoint . . ."
– bof
Jul 26 '16 at 1:01
• It's not any union of countable sets that is countable, it's countable union of countable sets that's countable. But I don't understand why you bring this up. How do you use this fact in your argument? What "union of countable sets" are you talking about??
– bof
Jul 26 '16 at 1:05

Once you have set up a map $f:U \to \Bbb Q$ such that $f(I) \in \Bbb Q$ and $f(I) \in I$ for each $I \in U$ (and I think you might need the Axiom of Choice to do this; but I'm no expert on set theory so if someone more knowledgeable wants to chime in, I for one would welcome it . . .), then injectivity follows from the fact that members of $U$ are disjoint, for if $I_1, I_2 \in U$ with $f(I_1) = f(I_2) = q \in \Bbb Q$, we have $q \in I_1$ and $q \in I_2$, contradicting $I_1 \cap I_2 = \emptyset$. Since $\text{Image}(f) \subset \Bbb Q$, it is countable by virtue of being a subset of a countable set. It seems to me that that the OP's argument for injectivity is basically the same as mine, but I don't quite see where one needs the fact that a (countable) union of countable sets is countable.
• The Axiom of Choice is not needed for this. You can set $f(I) =$ the first rational in Cantor's enumeration of $\Bbb Q$ which happens to belongs to $I.$ (By the way, just as you mentioned, the OP uses the fact that a subset of a countable set is countable, not the fact that a countable union of countable sets is countable.) Jul 25 '16 at 16:19