1
$\begingroup$

I came across with the following claim: If three intervals $I, J$ and $K$ intersect, $I\cap J\cap K\neq\emptyset$, then necessarily one of them is contained in the union of the two other. From picture this seems obvious, but is it true in general? Is there an easy way to prove it?

$\endgroup$
  • $\begingroup$ By "intersect" you mean $I\cap J \cap K\ne \phi$, right? $\endgroup$ – kimchi lover Oct 29 '17 at 14:43
  • $\begingroup$ @kimchilover yes, edited that now $\endgroup$ – peastick Oct 29 '17 at 14:46
  • $\begingroup$ Assume the intervals are finite and closed. Let $a$ be in their common intersection, let $b$ be a minimal element of $I\cup J\cup K$, let $c$ be a maximal element; the two intervals you want are the intervals containing $b$ and $c$. $\endgroup$ – kimchi lover Oct 29 '17 at 14:50
2
$\begingroup$

You can suppose, without loss of generality, that $\inf I\leqslant\inf J$ and that $\inf I\leqslant \inf K$. There are three cases then:

  • If $\sup I\geqslant\sup J$ and $\sup I\geqslant\sup K$, then $J,K\subset I$.
  • If $\sup J\geqslant\sup I$, then $K\subset I\cup J$.
  • If $\sup K\geqslant\sup I$, then $J\subset I\cup K$.
$\endgroup$
2
$\begingroup$

There is a point $a$ in the intersection of the three intervals $J_k$. Therefore the union $U$ of the $J_k$ is again an interval. Let $\inf U=\inf J_{k_1}$ and $\sup U=\sup J_{k_2}$. Since $a\in J_{k_1}\cap J_{k_2}$ it follows that $U$ is the union of $J_{k_1}$ and $J_{k_2}$.

$\endgroup$
1
$\begingroup$

Let $a\in I\cap J\cap K$. Consider $I^+=[a,\infty)\cap I$ etc. One of the three intervals $I^+$, $J^+$, $K^+$ contains the other two. Likewise if $I^-=(-\infty,a]\cap I$ then one of the three intervals $I^-$, $J^-$, $K^-$ contains the other two. If, say, they are $I^+$ and $J^-$ then $K\subseteq I\cup J$ etc.

$\endgroup$
  • $\begingroup$ thank you for this answer $\endgroup$ – peastick Oct 29 '17 at 14:57

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.