Polar set of $S=\{s\in\mathbb{R}^n:1^Ts≤1,\; s≥0\}$

Question

The definition of a polar set, that I am working with, is: $M^{pol}=\{x\in\mathbb{R}^n:m^Tx≤1, \forall m\in M\} $
Now I want to find the polar $S^{pol}$ of $S$, where $S=\{s\in\mathbb{R}^n:$1$^Ts≤1,\; s≥0\}$
S is the set of vectors in $\mathbb{R}^n$ for which each $s_i$ is bigger than $0$ and for which the sum of all $s_i$ is smaller than $1$. Would then $S^{pol}$ be all vectors $x\in\mathbb{R}^n$ such that 1$^Tx≤1$? Therefore: $S^{pol}=\{x\in\mathbb{R}^n:$1$^Tx≤1\}$
I don't think my answer is wrong but I feel it may be incomplete and I am unsure how to prove that for $x$ with 1$^Tx>1:$ $s^Tx>1$, therefore I would appreciate any help with this problem.

Answer 1

$S^{pol}=U:=\{u\in \mathbb R^n \ | \ \max_{u_i\ge 0} u_i\le 1\}$ for $S=\{s\in \mathbb R^n \ | \ 1^Ts \le 1\text{ and } s\ge 0\}$:

(1) $U\subseteq S^{pol}:\quad$ let $u\in U$ and $s\in S$, then by defining $u_+:=\max(u,0)$ and $u_-:=\max(-u,0)$; both componentwise, we have: $$u^Ts=(u_+-u_-)^Ts=u_+^Ts-u_-^Ts\le u_+^Ts=\sum_{u_i\ge 0}u_is_i\le \Big(\max_{u_i\ge 0} u_i\Big)\sum_i s_i\le \max_{u_i\ge 0}u_i\le 1.$$ Thus, $u\in S^{pol}$.

(2) $S^{pol}\subseteq U: \quad$ let $p\in S^{pol}$, and $i^*\in argmax_{p_i\ge 0} \ p_i$, then choose $s=\textbf{e}_{i^*}\in \mathbb R^n.$ We see that $s\in S$ and $s^Tp=\max_{p_i\ge 0} p_i\le 1$. Hence, $p\in U$.