What are the maximum number of solutions to a system of non-linear inequalities? In particular:
Let $A, B, C$ be real nonzero numbers. Consider the set $S = \{\frac 1 A, \frac 1 B, \frac 1 C, \frac {1 -A} B, \frac {1 - B} A, \frac {1 - A } C, \frac {1 -C } A, \frac {1 - B} C, \frac {1 - C} B\}$.
Let $T = \{ s \in S | 0 \leq s \leq 1 \}.$ What is the cardinality of $T$?
I conjecture it can be any of $0, 1, 2, 3, 4, 5, 6$, but no greater.
Motivation: I believe $\#T$ is the number of vertices formed by intersections between an (almost) arbitrary plane ($Ax + By + Cz = 1$) and the unit cube.
What is $\#T$? Does it vary if we make the inequality strict?
Update
To address @nickalh's questions: The question intends to ask: Given set $S$, how many elements $s$ of $S$ can satisfy $0 \leq s \leq 1$ (or $0 < s < 1$)?
The question does not ask "How many values of $A, B, C$ can result in such an $s$?", which would clearly be infinite, as nickalh points out.
This is analogous to asking "How many intersections can two distinct real algebraic curves of degree 2 have (in the real plane)?" where the answer is "Either 0, 1, 2, 3 or 4". There are infinite curves and infinite coefficients which result in two intersections, infinite places in the plane where they may intersect, but the maximum number of intersections is 4.
If there is a preferred way to express this question, or better terms to use, I'd appreciate learning them.
Finally, nickalh asks about the tag "calculus". This is because I attempted to solve this using calculus (by finding extrema or tangents to identify "turning points".)