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How do you show that, given two open intervals $I$ and $J$ in $\mathbb{R}$, $$(I\cap J\ne \emptyset)\Leftrightarrow (I\cup J\text{ is an open interval)}$$

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I was trying to avoid defining the "endpoints" of I and J and using max/min functions and casework. I wanted to use this definition of an open interval to prove it: given any two different points a and b in an open interval I, the open interval (a,b) is contained in I. I guess there is nothing wrong with defining the endpoints of I and J as in the two answers given so far... – angryavian Oct 9 '12 at 5:14
up vote 3 down vote accepted

Hint: I do not know whether you are supposed to consider also infinite intervals, such as $(5,\infty)$, $(-\infty, 17)$, and $(-\infty,\infty)$. If you are, they are not hard to deal with, it just lengthens the proof.

Let us start by considering finite intervals $(a,b)$ and $(c,d)$. Because of the symmetry, we can without loss of generality assume that $a\le c$.

Draw a picture. There are two cases to consider: (i) $c\lt b$ and (ii) $c\ge b$. In case (i), show that both sides of the "$\iff$" are true, and therefore the $\iff$ is true. In case (ii), show that both sides of the $\iff$ are false, and therefore the $\iff$ is true.

Remark: Actually, the "infinite interval" possibilities can be dealt with simultaneously with the "main" finite intervals case. But when one is a bit uncertain, attempts at compression can lead to loss of control.

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Hint for one direction: If $(a,b)\cap(c,d)\ne \emptyset$ then there is a real number $x$ with $a<x<b$ and $c<x<d$. Let $e=\min\{a,c\}$, $f=\max\{b,d\}$. Can you show that $(a,b)\cup(c,d)=(e,f)$? I.e. that $a<t<b\lor c<t<d$ implies $e<t<f$?

For the other direction assume $(a,b)\cup(c,d)=(e,f)$ and show that you cannot have $a<b<c<d$ nor $c<d<a<b$. Conclude that $(c,b)$ and $(a,d)$ are intervals (i.e. $a<d$ and $b<c$) and have nonempty intersection, which is also $(a,b)\cap(c,d)$.

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