Arrangements of affine hyperplanes Fix $n>0$ and $X\subseteq\mathbb{R}^n$. Call a function $f:X\longrightarrow \mathbb{R}$ linear if it is of the form
$$
f(\bar{x})=a_1x_1+\ldots+a_nx_n+b
$$
for some $a_i,b\in\mathbb{R}$.
Now suppose we have linear functions $f_1,\ldots,f_t$ and $g_1,\ldots,g_t$ with the following property:
For all $i\in\{1,\ldots,t\}$ and for all $\bar{x}\in X$,
$$
\textrm{there are }j,k\in\{1,\ldots,t\}\textrm{ such that }f_i(\bar{x})\leq g_j(\bar{x})\textrm{ and }f_k(\bar{x})\leq g_i(\bar{x}).
$$
Is it true that there must be some $i,j\in\{1,\ldots,t\}$ such that $f_i(\bar{x})\leq g_j(\bar{x})$ for all $\bar{x}\in X$?

Let's treat the case $n=1$. 
Assume that it is not true that there exist $i,j$ such that $f_i \leq g_j$ on all of $X$.
Then for every pair $i,j$ we can choose $x_{ij} \in X$ such that $f_i(x_{ij}) > g_j(x_{ij})$. Let $a=\min_{i,j}(x_{ij})$ and $b=\max_{i,j}(x_{ij})$.
Upon renumbering the $g_i$ if necessary, we can assume that $g_1(a)$ is the smallest value one of the $g_i$ attains at $a$. By assumption there exists $k$ such that $f_k(a) \leq g_1(a)$, and our assumption on $[a,b]$ above assures that $f_k(b) > g_1(b)$ (as $f_k$ and $g_1$ intersect in $[a,b]$, and lines can intersect at most once). Hence there must exist $j$ such that $f_k(b) \leq g_j(b)$. Again by the construction of $[a,b]$ there exists $\xi \in [a,b)$ such that $f_k(\xi) > g_j(\xi)$. In particular we must have $g_j(a) < f_k(a) \leq g_1(a)$, which contradicts minimality of $g_1(a)$.
