Semi-infinite programming discretization theorem from the book ,,Theory, Methods, and Applications'

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I think this theorem from Kortanek and Hettich book ,,Semi-Infinite Programming: Theory, Methods, and Applications'' is false.

(P) is a semi-infinite programming problem in the following manner:

$$\max_z F(z)$$ $$s.t.\:\: g(z,t)\leq0\:\: \forall t\in B $$ $v(P)$ is the solution of this problem and $P(T_n)$ means that we take constraints $t\in T_n$, not from $B$ as in the original problem.

THEOREM 4.2. In Program $(P)$ assume that $B$ is compact, $F$ is concave, $g(z, t)$ is convex with respect to $z$ ($F$ and $g(\cdot, t)$ all finite over $\mathbb{R}^n$), and $v(P)$ is finite. Assume further that the following type of Slater condition holds. For every set of $(n + 1)$ points $t_o, ... , t_n\in B$, a $\overline{z}$ exists such that $g(\overline{z}, t_i) < 0$, $i = 0, ... , n$. Then there exists $T_n = \{t_1, ..., t_n\} \subset B$ such that $v(P) = v(P(T_n))$;

I believe this is false, because if assumptions were met for all sets consisting of $(n+1)$ points, then it would be true also for each set consisting of 1 point,

If for every set $$T=\{t_i\},\:\: i=\{1,2,\dots, n+1\}\:\:, T\subset B$$ consiting of $(n+1)$ points $$g(\overline{z},t_i)<0$$ is satisfied for some $\overline{z}$, then for every set consisting of one point $S=\{s\},\:\: s\in T$ the same condition holds, because for specific $s$ we can take the $\overline{z}$ we used to make sure the previous condition was satisfied. Therefore according to the theorem , there exists $T_1=\{\overline{t}\}\subset B$ such that $v(P)=v(P(T_1))$.

which would mean we could just maximize function $f$ with taking into account just one constraint. Correct me if I am wrong, thank you. The theorem has been copied from the book. As a solution I would accept either the explanation why this theorem is true or an improvement to the theorem, which would make it true (if such exists).

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Theorem 2.1 in the reference "Semi-Infinite Programming Duality: How Special is It?" contains a proof.

The part you seemingly missed is that $n$ in "$n+1$ points" refers back to the dimension of $x$.