Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

1 Answer

up vote 1 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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.