For $i,j \in N$ let the indicator vector $e^{\{i\}} \in \mathbb{R}^n$ be such that $e^{\{i\}}_j = 1$ if $i=j$ and $0$ if $i \neq j$. Let $\operatorname{veto}(v)$ be the set of veto players, defined by $\operatorname{veto}(v) = \bigcap \{S: \> v(S) = 1 \}$.
Let $v \in TU^N$ be a simple game and prove that $$C(v) = \text{Conv}(\{e^{\{i\}} | i \in \operatorname{veto}(v)\}) $$ where $C(v)$ is the core of $v$ and $\text{Conv}$ denotes a convex hull.
I already know that $C(v) \neq \emptyset \iff \operatorname{veto}(v) \neq 0.$ However I do not know how to prove this. Intuitively it is clear, as I've done examples by hand with 3 players, wherein I solved a system of inequalities to show the core is empty if there are no veto players. But I don't know how to go about extending this result to all $N$.
I checked the proof of Them 16.12(ii) in the Peters' Game Theory book, which is wrong. The proof runs through contraposition and not by contradiction.
A proof by contraposition establishes $(\neg B \Rightarrow \neg A)$ which is equivalent to $(A \Rightarrow B)$. We need to establish that whenever $\mathbf{x} \not\in conv(\{\mathbf{e}^{i} \in \mathbb{R}^{n}\,\big\arrowvert\,i \in veto(v) \})$, then $\mathbf{x} \not\in C(v)$, which is equivalent to $\mathbf{x} \in C(v)$, then $\mathbf{x} \in conv(\{\mathbf{e}^{i} \in \mathbb{R}^{n}\,\big\arrowvert\,i \in veto(v) \})$ Hence, there is no need to assume that $\mathbf{x}$ is in the core.
This means, we suppose that $i \not\in veto(v)$ with $x_{i} > 0$ for some non-veto player $i$ is satisfied, i.e., $\neg B$ holds. Take $S$ with $v(S) = 1$ and $i \not\in S$, then $x(S)=x(N)-x(N\backslash S) \le 1 - x_{i} < 1 = v(S)$. Thus, it holds $\mathbf{x} \not\in C(v)$. Therefore, $\neg A$ holds. This concludes the inclusion $C(v) \subseteq conv(\{\mathbf{e}^{i} \in \mathbb{R}^{n}\,\big\arrowvert\,i \in veto(v) \})$.