VC dimension of only the rim of a unit disk Suppose we have an origin centered circle, ie $x^2 + y^2 =1$, so it's in $\mathbb{R}^2$ (2D). It will be classified as 1 if it lies only on this arc, and will be labeled 0 otherwise. What is the VC dimension? 

I think the answer is 2. Any two points can be shattered $(++, -+, +-, --)$. But if we have three points $\{(x_1,y_1),\dots,(x_3,y_3)\}$ that all have the label 1, with radius $r_1 = r_2 = 1$, and $r_3 = 0$, it is impossible to shatter them. Is my intuition correct?

Or can I construct the sample points to my liking (to get the maximum VC dim) so that only the positive labels will be on the rim (have some combination of $x_i^2 + y_i^2 = 1$), and have 0 labels to not lie on the rim, therefore have infinite VC dimension?
Let us prove the following general result:


  Let $\mathcal F$ be a class of functions from $\mathcal X$ to $\{0,1\}$. If $\mathcal F$ has VC dimension $d$ then $|\mathcal F| \geq 2^d$.


Indeed, if $\mathcal F$ has VC dimension $d$ then $\mathcal F$ shatters some set $S \subseteq \mathcal X$ of size $d$. This means that for any function $\phi\colon S \to \{0,1\}$ we can find a function $f_\phi \in \mathcal{F}$ that agrees with $\phi$ on $S$. Note that if $\phi \neq \psi$ then $f_\phi \neq f_\psi$. Since there are $2^d$ different functions from $S$ to $\{0,1\}$, we conclude that $\mathcal F$ must contain at least $2^d$ functions. $\square$

In particular, the VC dimension of a class of functions containing a single function is zero.