Tutte's Theorem: Why can we assume a maximum counter example?

Question

Tutte's Theorem: A graph G has a 1-factor if and only if $o(G-S) \leq |S|$ for all $S \subseteq V(G)$.

Every proof I've seen of this theorem makes an assumption that I don't understand. Assuming there is a graph $G$ such that $o(G-S) \leq |S|$ for all $S \subseteq V(G)$ and trying to prove that G must have a 1-factor, almost every proof I've seen is by contradiction and says something to the effect of "We may assume a maximum counter example where G satisfies Tutte's condition (that $o(G-S) \leq |S|$ for all S) but does not have a 1-factor and has the additional property that adding any edge to G creates a one factor."

I understand the strategy behind using a maximum counter-example, but it seems like we're assuming that if there is any counter-example, then there must be one that has this arbitrary property that we just made up. I don't see why that's true. Anyone have any thoughts?

Answer 1

The additional property follows from $G$ being a maximal counterexample (you didn't say with respect to what, but let's assume number of edges).

Suppose $G'$ is obtained from $G$ by adding an edge. Since $G$ was assumed to be a maximal counterexample and $G'$ has more edges than $G$, the theorem must hold for $G'$. But $o(G'-S)\leq o(G-S)\leq|S|$, so $G'$ has a 1-factor.