Let G be a graph without a perfect matching, thus, Tutte’s conditions fail for G which implies that there exists a set $S ⊂ V (G)$ such that the number of odd components in $G − S$ exceeds $|S|$. Now let us assume that G is bipartite with bipartition $V (G) = V' \sqcup V''$. Show that Hall’s conditions fail for at least one of the sets $S ∩ V'$and $S ∩ V''$
I tried to draw a graph that fails Tutte's condition as shown below:
But when I verified it for hall's theorem it did not fail:
Since Hall's condition states that for a bipartite graph with parts $V'$ and $V''$ the condition $ \forall A \subseteq V', \lvert N(A) \rvert \geqslant \lvert A \rvert$ is necessary and sufficient for the existence of a matching that covers all vertices in $V'$
Am I doing something wrong? Could you please help me to solve this. Thanks.