On convex hulls and intersections of chains of compact sets Let $V$ be a topological vector space, let $\{ C_i \}_{i \in I}$ be a set of compact subsets of $V$ which forms a chain with respect to inclusion. For now, assume the following stronger properties:


*

*$V$ is a Euclidean space.

*$I$ is countable.


Question 1: Is it true that conv($\bigcap_{i \in I} C_i$) = $\bigcap_{i \in I} \text{conv}(C_i)$, where conv( $\cdot$ ) denotes the convex hull?
Question 2: If the answer to Question 1 is yes, can the assumptions 1 and 2 be relaxed such that conv($\bigcap_{i \in I} C_i$) = $\bigcap_{i \in I} \text{conv}(C_i)$ (or an appropriately altered statement) is still true?
 A: The Answer to question 1 is yes:
Let $V = \mathbb R^n$ and $I = \mathbb N$. It is clear that $\operatorname{conv}(\bigcap_{i \in \mathbb N} C_i) \subset \operatorname{conv}(C_j)$ for every $j \in \mathbb N$ since $\bigcap_{i \in \mathbb N} C_i \subset C_j$ for every $j \in \mathbb N$. Thus $\operatorname{conv}(\bigcap_{i \in \mathbb N} C_i) \subset \bigcap_{i \in \mathbb N}\operatorname{conv}(C_i)$.
Now let $x \in \bigcap_{i \in \mathbb N}\operatorname{conv}(C_i)$, so $x \in \operatorname{conv}(C_j)$ for all $j \in \mathbb N$. We use the following fact: 
Every point in $\operatorname{conv}(C_j)$ can be written as the convexcombination of at most $n + 1$ points of $C_j$. 
Thus for every $j \in I$ there exist $x_j^1, \dots ,x_j^{n + 1} \in C_j$ and $\lambda_j^1, \dots , \lambda_j^{n + 1} \in [0,1]$ such that 
$$x = \Sigma_k \lambda_j^kx_j^k.$$
Since $C_j \subset C_0$ for all $j$ and $C_0$ is compact we may assume, after taking subsequences, that all the sequences converge: $x_j^k \to x^k$ for $j \to \infty$ and $\lambda_j^k \to \lambda^k$ for $j \to \infty$ for all $k$. It is easy to see that $x^k \in \bigcap_{j \in \mathbb N}C_j$ for all $k \in \{1, \dots n + 1\}$. Thus 
$$x = \lim_{j \to \infty}\Sigma_k \lambda_j^kx_j^k = \Sigma_k \lambda^kx^k \in \operatorname{conv}(\bigcap_{i \in \mathbb N} C_i).$$
About question 2: Given that $V = \mathbb R^n$ i guess the second assumption can be dropped. For that notice that the argument for the first inclusion does not use this. To apply the argument for the other inclusion one may first take $J \subset I$ which is countable such that $\bigcup_J C_i = \bigcup_I C_i$.
