Arrangements of graphs of linear functions

Fix $n>0$ and $X\subseteq\mathbb{R}^n$. Assume $X$ is unbounded.

A function $f:X\longrightarrow\mathbb{R}$ is linear if it is of the form $$f(\bar{x})=a_1x_1+\ldots+a_nx_n+b$$ for some $a_i,b\in\mathbb{R}$.

Suppose we have linear functions $f_1,\ldots,f_t$ and $g_1,\ldots,g_t$ with the following property:

For any $\bar{x}\in X$ there are permutations $\sigma$ and $\tau$ of $\{1,\ldots,t\}$ such that $$f_{\sigma(1)}(\bar{x})\leq g_{\tau(1)}(\bar{x})< f_{\sigma(2)}(\bar{x})\leq g_{\tau(2)}(\bar{x})<\ldots< f_{\sigma(t)}(\bar{x})\leq g_{\tau(t)}(\bar{x}).$$

Is it true that there are $i,j\in\{1,\ldots,t\}$ and some bounded subset $A\subseteq X$ such that $f_i(\bar{x})\leq g_j(\bar{x})$ for all $\bar{x}\in X\backslash A$?

Notes:

1. This is true for $n=1$. In this case it is enough to assume that for all $\bar{x}\in X$ and $i\in\{1,\ldots,t\}$ there are $j,k\in\{1,\ldots,t\}$ such that $f_i(\bar{x})\leq g_j(\bar{x})$ and $f_k(\bar{x})\leq g_i(\bar{x})$. We can also take $A=\emptyset$.

2. I am also interested to know if it makes a difference to also assume that the interior of $X$ is unbounded.

-
Are you sure you want strict inequalities between g and f? In this case this would be quite easy to prove as $f_i=f_j+c$ and $g_i=g_j+c$ would hold for any i and j with some appropriate constants. Otherwise, I don't think this problem is that hard, I would try induction on $n$. –  domotorp May 29 '13 at 6:51