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 $\bar{x}\in X$?
 A: 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 $g_1(a)$.
A: I had an idea to approach this differently, and after fighting it, took a simpler case where each $f$ or $g$ has only one element that is non zero, in the hope that this is still useful and suggests a path to the general case. 
For the case when each $f$ or $g$ has only one element that is non zero (i.e. the planes are 'axis' aligned):
Given an index $p$, up to $n$, we can construct a vector $\bar v$ in $X$ with a 1 in the $p$ position and otherwise $0$s, and following the direction of $\bar v$ there is some point beyond which a function $f$` having the largest negative value in position p becomes (and remains) the minimum function $f$.
This gives us a lower bound for components in the $p$ position for all functions in $g$ - since any $g$ that has a lower value in the $p$ position will eventually fall below and remain below $f$` in the direction of $\bar v$, and thus break an assumption.
Heading in the reversed direction of $\bar v$, we can find a point at which the same $f \prime$ (or some parallel $f\prime\prime$ above it with equal largest negative component in position $p$) becomes and remains the maximum $f$ function. Since $f \prime$ and $f \prime\prime$ share the component in position p we are interested in, we will take the case where $f \prime$ is the upper function, without loss of generality.
Since we know there must be at least one function $g \prime$ which is eventually above $f \prime$ and remains above $f \prime$ in this direction by assumption (since there are a finite number of $g$, eventually at least one $g \prime$ must remain above $f \prime$ to infinity), we know this $g \prime$ must have a value in the p position of not more than that of that of $f \prime$, since otherwise it will eventually fall below $f \prime$. 
So now we have a maximum bound for the position p for $g \prime$, and since it is equal to the minimum bound we have above for all g in this position, we have an equality for this position p in (at least one) $g \prime$.
The conclusions then follows since for each position $p$, we determine that the function $f \prime$ which has the minimum value of all $f$ in this position, has some $g \prime$ parallel and $\ge$ to it (possibly also parallel and $\ge$ to some parallel $f \prime\prime$ above it).
Note: I think it can be argued similarly for each position $p$ from both the maximum positive and minimum negative values in $f$, since both will force this situation for some $g$.
I think the underlying idea also works for the general case, but requires taking combinations of multiple positions and allowing for $g$ that are parallel to intersections of planes, and I've not managed to find a clean (and reasonably short) way to attack that yet.
