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@Iosif Pinelis (i) It is immediately verified that it is equivalent for the case of $\phi(\sigma)$ being closed in $I$. The latter is of course implied by the closed graph property I stated in the question. (ii) The closed graph property $\phi$ shall satisfy is completely independent of the topology on $2^I$. So a topology on $2^I$ is not required. However, you could equip $2^I$ with the topology induced by the upper Hausdorff hemimetric . Then the required closed graph property should be equivalent to $\phi$ being continuous and $\phi(\sigma)$ being closed in $I$.
@PietroMajer Thank you for your comment. Unfortunately, I had the same idea and it does not work. The set you describe may not be convex. Problems may arise when the boundary of $O$ and $K$ (both boundaries w.r.t. the topology of $T$) share points in common.
