$C$ is a set consisting of points in $\mathbb{R}^n$, $x\in int(conv(C))$ . Question is: Does there exist a closed subset of $conv(C)$, namely $A$, s.t. $x\in int(A)$, and $A$ is a polytope, and the vertexes of $A$ are in $C$?
I know that if $A$ is not restricted to have an interior point there exists a polytope(simplex) $A$ such that $x\in A$ and the vertexes of $A$ are in $C$. This is because $x$ can be represented as a convex combination of points in $C$, $x=\sum_{i=1}^mw_ix_i,\ x_i\in C$. If there exists $\alpha_1\cdots \alpha_m$ s.t. $\sum_{i=2}^m\alpha_i(x_i-x_1)=0$, then choose $b=\min\{\dfrac{w_i}{\alpha_i}\ |\ \alpha_i>0,i=2\cdots m\}$, then $x=(w_1+\sum_{i=2}^mb\alpha_i)x_1+\sum_{i=2}^m(w_i-b\alpha_i)x_i$ is a convex combination with less points. In this way, we can find a subset of $\{x_1\cdots x_m\}$ such that $x$ is a convex combination of these points and they are vertexes of a simplex. However, I want $A$ to have interior but the $A$ we find above may not have one. For example, if we choose $C=\{(a,b)\in\mathbb{R}^2\ |\ a,b\in \mathbb{Z}\}$, and $x=(1,1)$. Following the above method, it is possible that we will find $A=\{(a,1)\ |\ 0\leq a\leq 2\}$, which has no interior. Instead, I want $A$ to be a set like $\{(a,b)\ |\ 0\leq a\leq 2,0\leq b\leq 2\}$.
I don't need to construct an $A$, just wonder whether there exists one. Could someone give me some help?
Yes. First of all, all polytopes (convex hulls of finitely many points in $\mathbb{R}^n$) are closed, because they are continuous images of simplices (which are compact). So we just have to show that, whenever $x \in \operatorname{int}(\operatorname{conv}(C))$, then there exist $x_1, \dots, x_k \in C$ such that $x \in \operatorname{int}(\operatorname{conv}(\{x_1, \dots, x_k\}))$. Here is a proof sketch, see if you can fill in the details!
Step 1: Since $x \in \operatorname{int}(\operatorname{conv}(C))$, there is some $\varepsilon > 0$ such that $B_{\varepsilon}(x) \subseteq \operatorname{conv}(C)$.
Step 2: $x \in \operatorname{int}(\operatorname{conv}(\{x + s \frac{\varepsilon}{2} e_i : 1 \leq i \leq n, s \in \{-1,1\}\}))$, where $e_1, \dots, e_n$ are the standard basis vectors of $\mathbb{R}^n$.
Step 3: Each $x + s \frac{\varepsilon}{2} e_i \in \operatorname{conv}(C)$, so there is a finite set $X_{s,i} \subseteq C$ such that $x + s \frac{\varepsilon}{2} e_i \in \operatorname{conv}(X_{s,i})$.
Step 4: $$x \in \operatorname{int}\left(\operatorname{conv}\left(\bigcup_{s \in \{-1,1\}} \bigcup_{i=1}^n X_{s,i}\right)\right).$$