Linear programming vs integer linear programming Given $A,b$, let $Ax \le b$ be an instance of linear programming on the variables $x=(x_1,\dots,x_n)$.  Assume that the constraints $0 \le x_i$ and $x_i \le 1$ are included in $A,b$.
Suppose that there is a feasible solution $x \in \mathbb{R}^n$ to the linear programming problem where $x_1 = 1$.  Also, suppose there exists a solution to the corresponding integer linear program, i.e., there exists $x' \in \{0,1\}^n$ such that $Ax' \le b$.  Are we guaranteed that there exists a solution with $x'_1=1$, i.e., there exists $x' \in \{0,1\}^n$ such that $Ax' \le b$ and $x'_1=1$?
To put it another way: if we solve the linear program associated with a zero-or-one ILP instance, and find that one of the variables gets assigned to 1 in some solution to the linear program, does it follow that there exists a solution to the ILP instance where we set that variable to 1?
I am skeptical but could not find either a proof or a counterexample.
The program $x_1 + 2x_2 = 2$ has the integer solution $(0,1)$ and the fractional solution $(1,1/2)$ but not an integer solution of the form $(1,\cdot)$.