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For any index set $I$, let $A_\iota\subseteq\mathbb{R}^d$ for $\iota\in I$ be closed sets. Do we have $\bigcap_{\iota\in I}\text{ri}(\text{conv}(A_\iota))\subseteq\text{ri}(\text{conv}(\bigcap_{\iota\in I}A_\iota)$? Here, $\text{ri}$ denotes the relative interior. If not, what is the relation between these sets? Any reference?

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up vote 2 down vote accepted

The rule of thumb is that you get a smaller set when you order operations so that intersection comes earlier.

For example, in one dimension $d=1$ let $A_i=[-1/i,1/i]$. Clearly, $0$ is in the interior of every $A_i$, and therefore $0\in \bigcap_{i\in I}\operatorname{ri}(\operatorname{conv}(A_i))$. On the other hand, $\bigcap_{i\in I}A_i$ is a single point, and therefore has empty interior. So the stated inclusion fails.

The reverse inclusion holds: $\operatorname{ri}(\operatorname{conv}(\bigcap_{i\in I}A_i)) \subseteq \bigcap_{i\in I}\operatorname{ri}(\operatorname{conv}(A_i))$. Indeed, for any $i$ we have $\operatorname{ri}(\operatorname{conv}(\bigcap_{i\in I}A_i)) \subseteq \operatorname{ri}(\operatorname{conv}(A_i))$ simply because the operations $\operatorname{ri}$ and $\operatorname{conv}$ preserve the order by inclusion. Then take intersection over $i$ on the right.

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