I would like to prove that the VC dimension of a set system $(X,\mathcal{R})$ never takes values in $(0,1).$
For the sake of completeness, I'll define some basic ideas in this context.
Definition: A set system $(X,\mathcal{R})$ is a set $X$ (possibly infinite) and a collection of subsets of $X$, $\mathcal{R}\subseteq 2^X$.
A subset $Y\subseteq X$ is shattered if every subset of $Y$ can be obtained by intersection with a member of $\mathcal{R}$: $\mathcal{R}|_Y=\{R \cap Y \mid R\in \mathcal{R}\}=2^Y$.
The VC dimension of a set system $(X,\mathcal{R})$ is the smallest integer $d$ such that there exists no shattered subset of size $d+1$.
It is known that set system with bounded VC dimension grow polynomially: For a set system $(X,\mathcal{R})$ with VC dimension $d$ and every $m$-point subset $Y\subseteq X$, the size of the system is $\mathcal{R}|_Y =\mathscr{O}(m^d)$ (intuitively, one can only include all subsets of $Y$ of size at most $d$).
That brings me to the next definition:
Definition(VC-exponent): The VC-exponent of a set system $(X,\mathcal{R})$ is the infimum over all numbers $s\in \mathbb{R}$ such that for every $m$-point set $Y\subseteq X$, $|\mathcal{R}|_Y| = \mathscr{O}(m^s)$.
I've been able to prove that the VC dimension $d$ is finite if and only if the VC-exponent $s$ is. But I've run into the claim that $s$ never takes values in $(0,1)$.
That makes sense, since any sublinear amount of sets in $\mathcal{R}|_Y$ means the exponent is $\mathcal{R}|_Y = \mathscr{O} (|Y|^\varepsilon) $ for every $\varepsilon >0$). And it also makes sense that the 'next step' would be a VC exponent of 1. But I haven't been able to prove this.
My question is how do you prove that the VC-exponent of a set system $(X,\mathcal{R})$ never takes values in $(0,1)$. That is, $s\notin (0,1)$.
I'd appreciate any help understanding this better and anything leading up to a proof.
