I am currently learning about polytopes and I have reached the stage of convex polytopes. Wiki says that a
convex polytope is a special case of a polytope, having the additional property that it is also a convex set contained in the $n$-dimensional Euclidean space $\mathbb{R}^{n}$.
then we go to the definition of a convex set:
In geometry, a subset of a Euclidean space, or more generally an affine space over the reals, is convex if, given any two points in the subset, the subset contains the whole line segment that joins them
Now as the title of this question says, I am wondering if we can define a convex polytope on the mixed-integer space $\mathbb{Z}^m \times \mathbb{R}^n$.
I believe the answer is 'no' because the integer-set does not have the convex set property required of a convex polytope. But I am still very much a novice in this area so wanted to reach out to more knowledgeable people to see if there is in fact a version (or some other object) over the mixed space.
Thanks
EDIT:
Following the current fad with LLMs, this is what ChatGPT gives as answer to this question.
Mathematically, a convex polytope in $\mathbb{Z}^m \times \mathbb{R}^n$ can be defined as follows:
Let $A \in \mathbb{R}^{(m+n) \times (m+n)}$ be a matrix, and $\mathbf{b} \in \mathbb{R}^{m+n}$ be a vector. Then the convex polytope can be defined as:
$$P = \{(x,y) \in \mathbb{Z}^m \times \mathbb{R}^n \mid Ax + By \leq \mathbf{b}\},$$
where $x \in \mathbb{Z}^m$ represents the integer coordinates, $y \in \mathbb{R}^n$ represents the real coordinates, and $A$, $B$, and $\mathbf{b}$ define the linear inequalities that the points must satisfy to belong to the polytope.
But then I wonder, do you need linear inequalities do define this type of convex polytope? That seems odd.