Let $A\subseteq\mathbb{R}^d$. Do we have $\bar A\subseteq \text{conv}(A)$?Counterexample?
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No. If $A$ is open and convex, then $\textbf{conv}(A) = A$. For a specific example, suppose $A$ is the open unit ball. Then $\textbf{conv}(A) = A \nsupseteq \bar{A}$. |
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