Union of 2 non-disjoint open intervals is open interval. I will not use word open later.

I want to obtain formal proof of this fact that is different from mine.

My attemp:

We know that union of two open sets is open set. Assume it is not an interval. Then it must be a union of intervals (because any open set is union of intervals). But union of more than one interval is not connected - we can just apply defintion of connectedness. And we know that union of nondisjoint connected sets is connected. Contradiction shows it must be an interval.

I am very unhappy with my attemp because I am trying to prove that any open set is union of disjoint intervals and in order to do I must prove that that union of non-dijsoint open intervals is open interval. So it is just cheating from my side, and in fact I do not understand two proofs instead of one proof. In fact I am not even sure that my attemp could be saved as a proof.

Is it possible to prove the fact using only properties of real numbers, including completeness and archimedean property without using connectedness?

Or maybe there is another way to deal with these 2 propositions at once without logic flaws?


2 Answers 2


Say $(a, b)$ and $(c, d)$ are your intervals. If there's an $x$ in both intervals, then that x is strictly between $c$ and $b$, so $c<b$, and so the union of these intervals is $(a, d)$ or $(a, b)$, depending on which of $b, d$ is larger.

  • $\begingroup$ $(a,b)=(1, \infty)$ and $(c,d)= (2,3)$, their union is not $(1,3)$. $\endgroup$
    – Hedgehog
    Oct 28, 2014 at 18:17
  • $\begingroup$ Yeah, let me try to fix it. $\endgroup$
    – Nishant
    Oct 28, 2014 at 18:18

Intervals in $\mathbb R$ are precisely the convex subsets (possibly without the empty set). So you just need to prove that every union of such nondisjoint sets is convex. Can you do that?

It follows that every union of nondisjoint intervals is an interval, which is even more than what you asked because the openness is trivial if they are both open.

  • $\begingroup$ I like you idea, but to be honest I do not want to use ideas of convexity for this particular proof. $\endgroup$
    – Hedgehog
    Oct 28, 2014 at 18:03
  • $\begingroup$ @Hedgehog Well, it's just another way to say that you only need to prove that for every $a<b$ in the union, the whole interval $[a,b]$ is in the union. If $x$ is a common point of both intervals and $a,b$ are not both in one of the intervals, then $a < x < b$ which implies $[a,b]$ is in the union. $\endgroup$ Oct 28, 2014 at 18:06
  • $\begingroup$ @Hedgehog And you can drop openness and this generalizes to the fact that every union of 2 nondisjoint intervals is an interval. I'll reformulate the answer to separate these concepts. $\endgroup$ Oct 28, 2014 at 18:08
  • $\begingroup$ Ok, with your approach I need to be able to prove that union of 2 disjoint intevals is not convex, that intervals are the only convex subsets and that union of 2 nondisjoint intervals is convex. Have I missed something? $\endgroup$
    – Hedgehog
    Oct 28, 2014 at 18:27
  • $\begingroup$ @Hedgehog No! You only need that a nonempty convex subset of $\mathbb R$ is an interval because the idea is to show that the union of nondisjoint intervals is convex. $\endgroup$ Oct 28, 2014 at 21:31

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