Showing that there's no matching of G that matches all v's of V1

Question

I need to show that there's no matching of $G$ that matches all of the vertices of $V_1$, using Hall's Theorem. So Hall's Theorem is when $|X|$ is less than or equal to $|N(X)|$ (neighbours of the vertices in $X$)

so if $X$ is $V_1$ then $X = \{w,x,y,z\}$ and then $N(X)$ would be $\{a,b,c,d\}$ which means that's a complete matching, but the question says there shouldn't be one so I'm wondering where I've gone wrong?

Answer 1

You are missing the “for all $X$” part of Hall’s Theorem. The $X$ you exhibit satisfies the condition, but consider $X=\{y,z\}$.