I am working with an integer linear optimization problem of the form
Find $\vec{x}$ such that $\mathbf{A}\vec{x} < 0$ and such that the sum of the entries of $\vec{x}$ is as small as possible. The entries of $\vec{x}$ must furthermore be nonnegative integers.
Now the entries of $\mathbf{A}$ are all in the set $\{-1, 0, +1\}$, except for a few which have value $-\infty$. In low-dimensional terms, a constraint like $[a,b,-\infty]\cdot [x,y,z] < 0$ and $x,y,z\geq 0$ means that either ($ax + by < 0$ and $z=0$ and $x,y\geq 0$) or ($z>0$ and $x,y\geq 0$).
I know that the solution space for a linear inequality is usually convex. But although I have been experimenting with low-dimensional test cases, I am not confident about whether or when the solution space remains convex if the matrix $\mathbf{A}$ is allowed to have negative infinities in it. (note that the infinite entries are always negative infinities, and the constraint is always $\mathbf{A}\vec{x}< 0$).
The reason I am concerned is because these negative infinities sort of split the problem into a union of solutions to separate optimization problems, and I know that in general the union of convex spaces is not necessarily convex.
So:
In linear optimization problems like the one noted above, is the solution space generally convex even when negative infinities are allowed as coefficients in $\mathbf{A}$?