# Prove that Open Sets in $\mathbb{R}$ are The Disjoint Union of Open Intervals Without the Axioms of Choice

There are several proofs I have seen of this, but they all seem to use choice subtely at some point. Is there any way to prove this without choice, or is it possibly unproveable?

• As far as I know, none of the proofs I've seen use the axiom of choice. The use of choice must be too subtle for me. Could you please give an example of such a proof and point out where choice is used? Considering that the decomposition is unique, it seems unlikely that choice is involved. Is choice needed to show that a topological space is the disjoint union of its connected components? Or that the connected components of an open set in $\mathbb R$ are open intervals? – bof Jan 3 '15 at 18:14
• @bof: You can always force the axiom of choice in there. (E.g. start by enumerating the real numbers) :-) – Asaf Karagila Jan 3 '15 at 18:15
• @bof: The answer given uses the axiom of choice if you just say that a rational exists for each interval, but you can get around that. Choice is used in the proofs a lot in claiming existance, when existance can be shown without choice anyway. – Tim Jan 3 '15 at 18:17
• @Asaf Karagila: I know you didn't use choice. I didn't say that you did in the above comment. I commented on your answer below saying that it made sense. Not sure what the confusion is here. – Tim Jan 3 '15 at 18:23
• I misread that comment, sorry. I'll remove my comment now. – Asaf Karagila Jan 3 '15 at 18:26

The usual proof uses nothing of the axiom of choice.

Given a non-empty open set $U$, define an equivalence relation on $U$ by: $$x\sim y\iff\exists I\text{ an interval}, I\subseteq U, x,y\in I.$$

Now show that each equivalence class is open, simply by definition. Any point inside has an open interval around it contained in the equivalence class.

Therefore $U$ is the disjoint union of open intervals. And we can do more. We can use the fact that the rational numbers are countable to enumerate them, and choose a unique one from each interval, thus proving that the number of intervals is finite or $\aleph_0$ as well.

• At least countable choice appears in the choice of one rational per interval. – Unit Jan 3 '15 at 18:08
• Haven't you used choice to pick a rational from each interval? – Tim Jan 3 '15 at 18:09
• @Unit: No, I have not. The rational numbers can be enumerated (they are countable, without using the axiom of choice). Now choose the least rational in the enumeration from each interval. – Asaf Karagila Jan 3 '15 at 18:10
• @Tim: Ping for the previous comment. – Asaf Karagila Jan 3 '15 at 18:10
• Ah, that makes sense. – Tim Jan 3 '15 at 18:12