Consider an arbitrary linear program:


$$\max \vec c \cdot \vec x$$


subject to:


$$\textbf{A}\cdot \vec x = 0, \quad \vec a \le \vec x \le \vec b$$


Assume that this program is feasible and bounded. Now suppose I perturb the bounds by small amounts:


$$\vec a' = \vec a + \delta \vec p, \quad \vec b' = \vec b + \delta \vec q$$


where $\vec p,\vec q$ are unit vectors and $\delta > 0$. Assume that the modified problem remains feasible and bounded.


**Prove or disprove:** For arbitrary $\epsilon > 0$, there exists a $\delta > 0$ such that there exist optimal solutions $\vec x, \vec x'$ to the original and perturbed linear programs (respectively) satisfying $|\vec x - \vec x'| < \epsilon$. 


Extra: Is there a way to define *uniform continuity* in this context, such that Linear Programming is, or isn't, uniformly continuous?



Linear programming is continuous
The answer is no. Consider the probem
$$x+y\to\max,$$
$$x\geq 0,\; y\geq 0,$$
$$x+y\leq 1.$$
It has infinitely many solutions. One of them is $(0,1)$.
Now change the last inequality to 
$$x+(1+\epsilon)y=1.$$
The new problem has a unique solution $(1,0)$ which is not close
to the solution of the first one.


One can also construct a counterexample with unique solution of the original problem: 


Consider this $x\to\max$, under the restrictions $x\geq 0,\; y\geq 0$,
$1\leq x+y\leq 1.$ The unique solution is $(1,0)$. Now you can perturb a little 
the last restriction, and you obtain a unique solution $(0,1)$. 


