Unique solution of LP Hi I am working on the following question:

If $c \in int(N_P(x))$, then $x$ is a unique solution.

I have proven that this is true if $x$ is a vertex. Well I am wondering if the following is a counter example: Let $Q=[-1,1]^2$. Choose $x=(1,0)$. Then $N_Q(x)=pos((1,0))$ where $pos$ is the positive hull. Since $aff(x)=lin((1,0)^T)$ ($aff,lin$ the affine respectively linear space) $c=(1,0) \in int(N_Q(x))$, but $x'=(1,0.5)$ is also an optimal solution. 

Consider the LP max $c^Tx,x \in P$, where $c \in \mathbb{R^n}$  and $P= \{x \in \mathbb{R^n}: Ax \leq b \}$ is a fulldimensional polyhedron with $A\in \mathbb{R^{m\times n}}$ and $b\in \mathbb{R^m}$ Moreover let $x \in P$. 
  If $c \in int(N_P(x))$ then there exists a unique dual optimal solution. True or False?

Probably not if the primal has no unique solution the dual won't. 
Is it possible to conlcude the primal has a unique solution if and only if the dual has?


"Is it possible to conclude the primal has a unique solution if and
  only if the dual has?"

There is a way to verify whether LP has unique solution by running another LP. 
Appa, G. “On the Uniqueness of Solutions to Linear Programs.” The Journal of the Operational Research Society, vol. 53, no. 10, 2002, pp. 1127–1132., www.jstor.org/stable/822972.
Here is a counter example to the statement that primal unique and dual not.
$$max \ 0x_1+0x_2$$
s.t.
$$-x_1-x_2>=-2 $$
$$0x_1+x_2>=1$$
$$x_1-x_2>=0$$
This LP has unique solution $x_1=x_2=1$;
Dual is given as follows
$$min -2y_1+y_2+0y_3$$
s.t.
$$-y_1+0y_2+y_3=0$$
$$-y_1+y_2-y_3=0$$
$$y_1,y_2,y_3<=0$$
This LP has infinite many feasible solution.
