Good evening, I'm reading about the proof of Helly's Intersection Theorem:
The author applies the inductive hypothesis to $D_{1}, \ldots, D_{n-1}$ where $D_{i}=C_{i} \cap H$ for all $i=1, \ldots, n-1$. To apply the inductive hypothesis, we need $D_{i}=C_{i} \cap H \neq \emptyset$ for all $i=1, \ldots, n-1$.
My question: How does it follow that $D_{i}=C_{i} \cap H \neq \emptyset$ for all $i=1, \ldots, n-1$?
Thank you so much for your help!
Best Answer
For each $i=1, \ldots, n-1$ the set $C_i$ intersects both the $>a$ and the $<a$ sides of $H$, and hence $C_i$, being convex, must also intersect $H$.
Note that $C_i$ intersects the $>a$ side of $H$ because $C_i$ intersects $C_{-n}$.
(Clearly $C_{-n}=\cap_{j\neq n}C_j\subseteq C_i$ so $C_i\cap C_{-n}=C_{-n}$, and $\emptyset\not=C_{-n}$ by hypothesis.)
On the other hand, $C_i$ intersects the $<a$ side of $H$ because $C_i$ intersects $C_n$, because $i\not=n$ and $|\{i,n\}|=2<n\ge3$.
(Pick $k<n$ with $k\neq i$, then $\emptyset\neq\cap_{j\neq k}C_j\subseteq C_i\cap C_n$.)