Helly's theorem.

Question

Let F be a finite family of segments in R such that among any n of them there are two intersecting. Prove that it is possible to divide F into n−1 families such that any two segments in one family are intersecting.

Answer 1

(Fill in the gaps as needed. If you're stuck, write out your working and thought process to demonstrate where you're at.)

Rephrase the statement to

Let F be a finite family of segments in R such that among any $n$ of them there are two intersecting. Prove that there are $n-1$ values, such that each segment contains at least 1 of these values.

You can do a proof by induction on $n$.

The case of $n=2$ is Helly's theorem (or you can prove it directly by considering the left most right endpoint).

Suppose the statement is true for some $k$. Consider $k+1$.
Given any $m \geq k+1$ segments $R_i = [a_i, b_i]$ that satisfy the condition that any $k+1$ segments have 2 segments what intersect.
WLOG, let $R_1$ be the segment with the left-most right endpoint (IE $b_1 = \min(b_i)$)
Any set which intersects $R_1$ must contain $b_1$.
Consider any $k$ sets which do not intersect $R_1$. By the assumption applied on these $k$ sets along with $R_1$, there exists 2 which intersect. $R_1$ cannot be involved, so it's 2 of these $k$ sets. Apply the induction hypothesis, and there are $k-1$ points such that these sets contain at least 1 of these values. Take the union of these $k-1$ points along with $b_1$, and we are done.