A set of edges in a graph $G$ is a matching if no two of the edges have a vertex in common. For an even integer $n$, an $n$-vertex graph $G$ has a perfect matching if there are $\frac{n}{2}$ edges $x_1 y_1 ,x_2 y_2, \ldots ,x_{\frac{n}{2}}y_{\frac{n}{2}}$ involving all of the vertices of $G$. If $n$ is odd, $G$ has an odd-perfect matching if there are $\frac{(n-1)}{2}$ such edges involving all but one the vertices of $G$. A set $F$ of edges of $G$ is called a preclusion set if $G-F$ has no perfect and odd-perfect matching. The matching preclusion number of $G$, $mp(G)$, is the size of a minimum preclusion set.
My question is that:
If $G$ a graph with even order $n$, then why $mp(G)\leq \delta(G)$?
What I tried: Let $v$ be a vertex with $\deg (v)=\delta(G)$. Let $F$ be the set of all edges incident with vertex $v$. I want to show that $G-F$ has no perfect (or odd-perfect matching). If $G-F$ has a matching, then $v(G-F)=even$ hence $v(G)=odd$ which is a contradiction. But I don't know how to show $G-F$ hos no odd-perfect matching.
Thanks in advance.
I think when $n$ is even, we need to take $F$ the set of all edges incident with vertex $v$. Then there is no edge in $G-F$ incident to $v$ hence there is no perfect matching. Therefore $mp(G)\leq \delta(G)$. But when $n$ is odd, that's not the case at all because by the definition of $F$, $G-F$ can have an odd-perfect matching (no need to have an edge incident to $v$ by definition of odd-perfect matching). For example, we can consider a path with 5 vertices with start $v=v_1$. If we remove the edge $v_1 v_2$, then remaining graph is an odd-perfect matching (i.e. $mp(G)\not \leq \delta(G)$). I think in this case, we need to take $F$ the set of all edges incident with vertex $v$ or $w$, where $w$ is a vertex with $\delta(G)\leq \deg (w)$ and $\deg(w)\leq$ any other degrees. Then $mp(G)\leq \delta(G)+deg(w)$.