Does it make sense to compare sets (polygons) with different dimensions? In the context of integer programming, I am considering 3 different linear models for a given problem. The goal is to determine which formulation is the tightest, that is, the one that gives the least fractional solution when omitting integrity constraints.
Let $X$, $Y$ and $Z$ be the polytopes defined by each of these formulations, without integrity constraints, and let $z_1, z_2, z_3$ denote the optimal values of the formulations over $X,Y,Z$, respectively. Assume we are dealing with a minimization problem. I have managed to prove that
$$
z_1 \ge z_2 \ge z_3
$$
in other words, the formulation over polytope $X$ is the tightest.
From there, do the following inclusions hold (or make any sense) ? :
$$
X\subseteq Y \subseteq Z
$$
My guess is that if the same number of variables is used for each formulation, then yes. But if these numbers are different, I am not sure the polytopes are "comparable" (my wording may be incorrect), and perhaps it makes no sense to use the symbol $\subseteq$.
 A: Afik, the question is still ill-posed, but here is what I can tell you:
Suppose, $X$ and $Y$ are (closed) convex polytopes in $R^n$ such that for every linear function $f: R^n\to R$, we have
$$
\min f|_Y\le \min f|_X.
$$
Then $X\subset Y$. Here $\min f|_Z= \min \{f(z): z\in Z\}$.
A proof of this is quite straightforward. I will prove the contrapositive: If $X$ is not contained in $Y$ then there exists a linear function function
$f: R^n\to R$, such that
$$
\min f|_X < \min f|_Y.
$$
Since $X$ is not contained in $Y$, one of the vertices $v$ of $X$ is not contained in $Y$. Let $C$ denote the convex hull of $X\cup Y$. Then $v$ is still a vertex of $C$.
Given any convex polytope $P\subset R^n$ and its vertex $p$, there exists a linear function $h: R^n\to R$ such that $v$ is the unique minimum point of $h$ on $P$: $h(p)=a$ and $h(q)>a$ for all $q\in P, q\ne p$. Applying this to the polytope $C$ and its vertex $v$, we obtain a linear function $f: R^n\to R$ such that $f(v)=0$ and $f(c)>a$ for all $c\in C$, except $c=v$. Since $v\notin Y$, we get
$$
a=\min f|_X < \min f|_Y. 
$$
qed
It is, however, possible that the question that you had in mind is a bit different:
Suppose that $f_1,...,f_k$ are given defining linear functions of the polytope $X\subset R^n$  (i.e. $X$ is given by a system of linear inequalities $f_i\ge a_i, i=1,...,k$). Suppose that $Y$ is another polytope in $R^n$ such that for each $i=1,...,k$,
$$
\min f_i|_Y\le \min f_i|_X.
$$
Does it follow that $X\subset Y$? This is indeed false even when $n=2$.
The difference between the two questions is that in the first formulation, I am allowed to enlarge the given defining set of functions for $X$.
