Shattering threshold functions in $\mathbb{R}$ (VC theory)

Question

The book "Understanding Machine Learning" has the following example in the section on VC dimension:

Let $\mathcal{H}$ be the class of threshold functions over $\mathbb{R}$ (real numbers). Take a set $C = \{c_1\}$. Now, if we take $a = c_1 + 1$, then we have $h_a(c_1) = 1$, and if we take $a = c_1 − 1$, then we have $h_a(c_1) = 0$. Therefore, $\mathcal{H}_C$ is the set of all functions from $C$ to $\{0,1\}$, and $\mathcal{H}$ shatters $C$. Now take a set $C = \{c_1,c_2\}$, where $c_1 \leq c_2$. No $h \in \mathcal{H}$ can account for the labeling $(0,1)$, because any threshold that assigns the label 0 to $c_1$ must assign the label 0 to $c_2$ as well. Therefore not all functions from $C$ to $\{0, 1\}$ are included in $\mathcal{H}_C$ ; hence C is not shattered by H.

I don't get why (0,1) labelling is not possible. You can choose a threshold to be between $c_1$ and $c_2$ that assigns a 0 to $c_1$ and a 1 to $c_2$.

I might be misunderstanding something here so any help is appreciated.

Answer 1

The quoted text is admittedly a little vague, but it sounds like the implicit definition of a single hypothesis is $$ h_a(x) = \mathbb{1}\{x \leq a\}, $$ where $\mathbb{1}$ is the indicator function. In this case, choosing a threshold $a$ such that $c_1 < a < c_2$ doesn't work, since this produces the labeling $(1,0)$ as opposed to $(0, 1)$.