I am doing Exercise 8.10 (Submodular function minimization) of Introduction to Linear Optimization by Bertsimas and Tsitsiklis and may have found a mistake in (b).
Exercise 8.10* (Submodular function minimization) We are given a finite set $N=\{1,...,n\}$, and a function $f(S)$ defined for all subsets $S$ of $N$. We assume that $f(S)$ is submodular, that is $$ f(S)+f(T)\ge f(S\cap T)+f(S\cup T),\quad\forall\,S,T\subset N. $$ In this exercise, we show how to use the equivalence of separation and optimization to solve the problem of minimizing $f(S)$ over all $S\subset N$.
(a) Suppose first that $f(S)$ is submodular and nondecreasing, i.e. $f(S)\le f(T)$ if $S\subset T$, and $f(\emptyset)=0$. Consider the polyhedron $$ P(f)=\left\{\mathbf{x\ge 0}:\sum_{j\in S}x_j\le f(S),\ \forall\,S\subset N\right\}, $$ ...
(b) Consider a submodular. but not necessarily nondecreasing, set function $f(S)$. Let $$ f_{\text{mon}}(S)=\min\{f(T)\mid S\subset T\subset N\}-\min\{f(T)\mid T\subset N\}. $$ Prove that $f_{\text{mon}}(S)$ is submodular and nondecreasing, and that $P(f)=P(f_{\text{mon}})$.
The part I am having doubts about is $P(f)=P(f_{\text{mon}})$. Clearly $P(f)\supset P(f_{\text{mon}})$. But I do not think $P(f)\subset P(f_{\text{mon}})$. For example, let $N=\{1\}$ and let $f(S)=1$ for all $S\subset N$. Then $f_{\text{mon}}(S)=0$ for all $S\subset N$. In this case, $$ P(f)=[0,1]\neq\{0\}=P(f_{\text{mon}}). $$
However, if I use $\tilde{f}_{\text{mon}}=\min\{f(T)\mid S\subset T\subset N\}$ (that is, without the constant term $\min\{f(T)\mid T\subset N\}$), then I can show that $P(f)=P(\tilde{f}_{\text{mon}})$.
Did I miss something? Or is this question wrong?