Let us have $2n+1$ intervals on the real line such that for each $I_i$, $i = 1, 2, ... , 2n+1$, there are at least $n$ other intervals that intersect $I_i$. Prove there exists an interval such that it intersect all other intervals.
I have tried to use the probabilistic method but I am really not sure if I got it correct. I suspect there may be some errors related to dependency between events, so I'd be glad to hear the correct sketch of this proof. Is there a possibility that Lovász local lemma may come in handy here?
Let's have $X_{ij}$ the event that $I_i$ does not intersect $I_j$. Then $P[X_{ij}] \leq \dfrac{1}{2n+1} \times \dfrac{n}{2n} = \dfrac{1}{2(2n+1)}$.
$Y_i = \bigvee\limits_{j=1}^n X_{ij} \rightarrow P[Y_i] \leq \dfrac{n}{2(2n+1)} $ because at most there can be $n$ intervals not intersecting $I_i$.
Finally,
$Z = \bigwedge \limits_{i=1}^{2n+1} Y_i \rightarrow P[Z] \leq \Big(\dfrac{n}{2(2n+1)}\Big)^{(2n+1)} < 1$. Therefore, there exists an interval such that it intersects all other intervals.
I guess the probabilistic method is doomed to fail because the same claim with, say, convex sets in the plane is false.
We may assume that all our intervals are closed (why?). Wlog $I_1=[a_1,b_1]$ is an interval with leftmost right end and $I_2=[a_2,b_2]$ an interval with rightmost left end. (Note that $I_1=I_2$ would imply $I_1\subseteq I_i$ for all $i$). As there are only $2n-1<n+n$ intervals left, the pigeon-hole principle tells us that one interval ($I_3=[a_3,b_3]$, say) intersects both $I_1$ and $I_2$. For any other interval $I_k=[a_k,b_k]$, we conclude $$ b_k\ge b_1\ge a_3\qquad \text{and}\qquad a_<\le a_2\le b_3,$$ hence $I_k\cap I_3\ne \emptyset$.