If we can $\epsilon-$cover one set with $\frac{K}{2}$ elements and another with 2 elements why isn't their set product $2^{\frac{K}{2}}$ cover? I was reading the following paper on Robustness and Generalization (and covering numbers)and I was reading their example 1:

which didn't quite make sense to me. The reason it didn't make sense is that intuitively it seems to me that each region $X_i$ could have the label either +1 or -1, so it seems to me that the partition of Z should be exponential:
$$2^{\frac{K}{2}}$$
because it region can have either label +1 or -1. However, instead it seems to be that instead if 2 times the epsilon cover of the other set X, which seems puzzling. Any ideas why?

Ok for background definition:
Epsilon cover means the smallest set size that can $\epsilon$ cover some target set T. i.e.:
$$ N(\epsilon,T,\rho) = \min \{ |\hat T | \mid T \subset \cup_{\hat t \in \hat T} B(\hat t; \epsilon, \rho ) \}$$
where $B(\hat t; \epsilon, \rho ) = \{ x \mid \rho(\hat t, x) \leq \epsilon \}$ is the epsilon ball around $\hat t$.
Ab algorithm $A_s(x)$ has a margin $\gamma$ if every point in the input space around it would be classified the same way. i.e. as the paper wrote:
$$ A_{s}(x) = A_s(s_{j | x}) ; \forall x: \| x - s_{j|x}\|_2 &lt; \gamma $$
for more definition the paper has them pretty much on the page and the page before Example 1 I am referencing.
Given a cover for $\mathcal{X}$, you seek to "cover" $\mathcal{X}\times\{-1,1\}$.
To do so, you take any set $A$ from the cover of $\mathcal{X}$, and consider the sets $A\times\{1\}$, $A\times\{-1\}$. If your original cover was of size $K$, then this process results in $2K$ sets (the reason you don't end up with $2^K$ sets is that fixing the label for one set does not affect the labeling of another, so there is no multiplication going on).

In a more straightforward approach, let $\mathcal{C}$ be your original $\frac{\gamma}{2}$-cover for $\mathcal{X}$, then a partition $\mathcal{P}$ of $\mathcal{X}\times\{-1,1\}$ such that for all $ p\in\mathcal{P}$ and $(x_1,y_1),(x_2,y_2)\in p$ it holds that $y_1=y_2$ and $||x_1-x_2||\le\gamma$, is given by $\mathcal{C}\times\{-1,1\}$, which is of size $2|\mathcal{C}|$.