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Let $\mathscr{F}$ be a finite family of open or closed intervals in the line $\mathbb{R}^1$. Show by an elementary proof (without referring to Helly's theorem), that if any $2$ of the intervals intersect, then all of them intersect. (Note: any $2$ of them does not mean only $2$ is sufficient, but that any $2$ chosen from the family must intersect.)

Here is my attempt:

Since $\mathscr{F}$ is a finite family, let it have cardinality $n$, and let's list the intervals in the family as $F_1,F_2, \dots , F_n$ such that $\inf F_i \le \inf F_{i+1}$ (note that in the case when some number of infimums of some intervals are equal, then just arbitrarily assign them continuing values of the sequence). Now, from the hypothesis we can suppose that $F_1 \cap F_i \not = \emptyset$ for $i=2, \dots , n$. Then $\sup F_1 \ge \inf F_2$ in order for $F_1$ to intersect $F_2$. Since $F_1$ intersects all other intervals as well, we must have that $\sup F_1 \ge \inf F_i$, and in particular, $\sup F_1 \ge \inf F_n$. Similarly, $\sup F_i \ge \inf F_n$ for $i=2,3,\dots , n-1$.

Now, the region (point) between (shared by) $\inf F_n$ and $\min \{\sup F_i$ : $i=1,2,\dots , n-1$} is a region (point) of intersection for all $F_i$ so that $\bigcap \mathscr{F} \not = \emptyset$


Is this proof correct? Also, would it be possible to do a proof by induction?

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    $\begingroup$ Could you state what title means more clearly in question itself? E.g. does it mean any two intervals in some collection of them intersect in a nonempty set? $\endgroup$ – coffeemath Feb 24 at 11:07
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    $\begingroup$ Why can't you have $\inf F_i=\inf F_{i+1}$? $\endgroup$ – Gerry Myerson Feb 24 at 11:56
  • $\begingroup$ @Myerson I am first trying to prove when assuming that all the sets are open, since it's easier $\endgroup$ – Wesley Feb 24 at 12:46
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    $\begingroup$ Your proof breaks down in the last line. The idea is good, but you've stated in a way that fails to guarantee the result. If the intervals are all open, you can have $\inf F_n = \sup _i$ for all $i < n$, (which satisfies all you've argued), but intersection would be empty since that common point is not actually in any of the intervals. Instead of infimums and supremums, you should work with actual points of intersection, since you know that those will be in the intervals. $\endgroup$ – Paul Sinclair Feb 24 at 19:53
  • $\begingroup$ @Sinclair Here imgur.com/a/UlXhLRB there is a proof that does what you say to do, use the points of intersection. However, I find it kind of hard to process and is not intuitive for me. Is it at all clear there would be an easier way to write up the proof, or is the way presented the "best" way? $\endgroup$ – Wesley Feb 24 at 21:32

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