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Let $(A_i: i \in I)$ be a family of sets in a topological vector space such that for all $i,j \in I$ there exists $k \in I$ for which $A_i \cap A_j=A_k$. Denote by $\overline{co}(X)$ the closure of the convex hull of $X$. Is it true that $$ \bigcap_{i \in I}\overline{co}(A_i)=\overline{co}\left(\bigcap_{i \in I}\overline{A_i}\right)\,\,\,? $$

One inclusion: Since $\overline{A_i} \subseteq \overline{co}(A_i)$ for all $i$ then $\bigcap_i \overline{A_i} \subseteq \bigcap_i \overline{co}(A_i)$. Therefore $$ \overline{co}\left(\bigcap_i \overline{A_i}\right) \subseteq \overline{co}\left(\bigcap_i \overline{co}(A_i)\right)=\bigcap_i \overline{co}(A_i). $$

What about the other?


Comment 1. The hypothesis on the family $\{A_i:i \in I\}$ is necessary, see Kavi's answer.

Comment 2. The claim is easily seen to be verified if $\{A_i: i \in I\}$ has a minimum, which holds, e.g., if $I$ is finite.

Comment 3. The claim $\bigcap_{i}\overline{co}(A_i)=\overline{co}\left(\bigcap_{i}A_i\right)$ (i.e., replacing $\overline{A_i}$ with $A_i$ in the right side) is false. As a counterexample, let $A_i=(0,2^{-i})\cup\{1\}$ for all integers $i\ge 1$. Then the left side is $[0,1]$ and the right side is $\{1\}$.

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This answer is due to Salvo Tringali.

The claim is false. For each positive integer $n$ let $A_n$ be the set of integers with absolute value greater than $n$. Then the left hand side is $\mathbf{R}$ and the right hand side is $\emptyset$.

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This answer is for an earlier version of the question.

Let $A=\{1,4\}, B=\{2,3\}$. The RHS is empty. But $co(A)=[1,4],co(B)=[2,3]$ so $2 \in LHS$.

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