It say that for any $x\in V(G)$,
show that $C-x$ satisties Tutte's condition:
For $S\subseteq V(C-x)$, $o((C-x)-S)\leq |S|$.

If $x\notin V(C)$ for some odd component $C$,
and when we check the case of $S=\emptyset$,
then $o((C-x)-S)=o(C-S)=o(C)=1>0=|S|$ and hence it fails.
Even if $x\in V(C)$ for some even component $C$,
and also when we check the case of $S=\emptyset$,
then $o((C-x)-S)=o(C-x)\geq 1>0=|S|$ and hence it fails, too.

In my opinion, it should be $x\in V(C)$ for odd component $C$,
and it still can be used to prove Tutte's theorem.

p.s. Here I use latex lanquage:
        \in := belongs to,
        \notin := not belongs to
        \subseteq :=contained in
        \leq :less or equal
        \emptyset := empty set

※ 編輯: killyou         來自: 140.112.231.114      (12/04 23:37)