Let $X$ be a finite set of ($n$) points in $\mathbb R^2$ of diameter $1$, i.e. any two points in $X$ have distance at most $1$. We need to prove that $X$ can be included in a disk of radius at most $1/\sqrt 3$.
I proceeded with the inductive proof - I can prove the base case of $n = 3$: by considering an equilateral triangle of side $1$.
Next, I assume the statement true for all $3 \le i \le n$, and I am trying to prove for $n+1$.
Consider set $X_n$ of $n$ points for which it is true by inductive hypothesis. "Insert" the $n+1$st point and call the resulting set $X_{n+1}$ There are 2 cases:
- Convex hull of $X_{n+1}$ is defined by $\le n$ points. Then by convexity we can consider only the points on the boundary, and the statement it true using inductive hypothesis.
- Convex hull of $X_{n+1}$ is defined by $n+1$ points. Here, I am not sure how to use the inductive hypothesis.
Any hints or lemmas that I need to prove for proving the second case?
This is a known exercise on Helly's theorem.
Suppose that $n\ge3$, and let the points be $P_1,P_2,\ldots,P_n$. The following statements are equivalent:
$\quad$ (1) There exists some point $C$ such that $P_k\in\overline{B}(C,\tfrac1{\sqrt3})$ for $k=1,2,\ldots,n$.
$\quad$ (2) There exists some point $C$ such that $\overline{P_k C}\le \tfrac1{\sqrt3}$ for $k=1,2,\ldots,n$.
$\quad$ (3a) There exists some point $C$ such that $C\in \overline{B}(P_k,\tfrac1{\sqrt3})$ for $k=1,2,\ldots,n$.
$\quad$ (3b) $\bigcap\limits_{k=1}^n \overline{B}(P_k,\tfrac1{\sqrt3})\ne\emptyset$.
The disks $\overline{B}(P_k,\tfrac1{\sqrt3})\ne\emptyset$ are convex, so for proving (3), by Helly's theorem, so it is sufficient if every three of them have a common point.