Odd cycle transversal Suppose we have a graph G. Say B a fundamental basis of the cycle space of G. Say LP a linear programming problem where there is a variable for each vertex of G, each variable can take value $\geq 0$, for each odd cycle of B we add to LP the constraint $x_{a} + x_{b} + x_{c} + ... + x_{i} \geq k$ where $x_{a},x_{b},x_{c},...,x_{i}$ are the verteces of the cycle and $k$ is the number of vertices of the cycle. The objective function of LP is $\min\sum\limits_{i=1}^{n}{x_{i}}$.
We say S an optimal solution of LP. Can we say that each vertex, whose variable takes a value $\gt 0$ in S, is a vertex of at least a minimum odd cycle transversal of G?
 A: No.  Let $G$ be the graph obtained by gluing a $3$-cycle $abc$ and a $5$-cycle $cdefg$ together at vertex $c$.  Then $(x_a, x_b, x_c, x_d, x_e, x_f, x_g)=(0,0,3,1,1,0,0)$ is an optimal solution of the LP.  However, neither $d$ nor $e$ are contained in a minimum odd cycle transversal of $G$, since $\{c\}$ is the unique minimum odd cycle transversal of $G$.  
The answer is also no to the updated question in the comment below.  Here is a counterexample for both versions.  Let $C_n$ be an odd cycle with $n \geq 5$.  Fix $e=ab \in E(C_n)$ and let $T=C_n - e$.  Let $G$ be the graph obtained from $C_n$ by adding all edges $f$ such that $T \cup f$ contains an even cycle.  Then, the fundamental basis of $G$ with respect to $T$ contains exactly one odd cycle, namely $C_n$.  Thus, $x=(\frac{1}{n}, \dots, \frac{1}{n})$ is an optimal solution to the revised LP (see comment below) and $x=(1, \dots, 1)$ is an optimal solution to the original LP.  However, it is not true that every vertex of $G$ is in a minimum odd cycle transversal, because the only minimum odd cycle transversals of $G$ are $\{a\}$ and $\{b\}$.  
