Let $X$ be a Hilbet space of finite dimension. Consider the functions $g_1,\cdots,g_p:X\longrightarrow\mathbb{R}$ , $h_1,\cdots,h_q:X\longrightarrow\mathbb{R}$ and the set $I(x)=\{i\in\{1,\cdots,p\}: g_i(x)=0\}$
Let $\bar{x}\in\mathbb{R}$ .
how can i prove that the sets $$A=\left\{ \sum_{j=1}^q\lambda_j\nabla h_j(\bar{x}) : \lambda_1,\cdots,\lambda_q\in\mathbb{R}\right\}$$ $$B=\left\{ \sum_{i\in I(\bar{x})}\mu_i\nabla g_i(\bar{x}) : \sum_{i\in I(\bar{x})}\mu_i=1,\mu_i\ge0,i\in I(\bar{x})\right\}$$
are closed sets? even more, B is compact?
$A$ is closed, as it is a subspace of the finite-dimensional space $X^*$; consider coordinate vectors of a sequence in $A$, and use the Heine-Borel theorem to find a subsequence so that each coordinate converges.
$B$ is a polytope; it's equal to the convex hull of the finite family of points $\{\nabla g_i(\overline{x}) : i \in I(\overline{x})\}$ (remember $I(\overline{x}) \subseteq \{1, \ldots, p\}$). As such, it is, convex, closed, and bounded (by the maximum norm of all the extreme points), which makes it compact in a finite-dimensional space.