1
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Related: math.stackexchange.com/questions/75781 $\endgroup$
    – Watson
    Jul 25 '16 at 16:11
  • $\begingroup$ 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 . . ." $\endgroup$
    – bof
    Jul 26 '16 at 1:01
  • $\begingroup$ 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?? $\endgroup$
    – bof
    Jul 26 '16 at 1:05
2
$\begingroup$

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.

Other than that, I'd say this proof has the right idea.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!

$\endgroup$
2
  • 1
    $\begingroup$ 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.) $\endgroup$ Jul 25 '16 at 16:19
  • 1
    $\begingroup$ @Mitchell Spector: Thanks for the input! $\endgroup$ Jul 26 '16 at 0:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.