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Let $X$ a Banach space and $A \subset X$ a subset. I know the notation $\text{conv}(A)$ denotes the convex hull of $A$. What means the notation $\overline{\text{conv}}(A)$ used in comments in this discussion? My (naive) guess: one takes the intersection only over all closed convex subsets containing $A$. Is it correct?

I saw here that it is considered as the closure of the convex hull. But wouldn't in that case the notation $\overline{\text{conv}(A)}$ syntactically more meaningful? This lead me to conjecture that the notation $\overline{\text{conv}}(A)$ should mean something different by definition.

user267839
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    Yes, generally the overline means closure in this context. Both are used, it is typographically easier to use it over the co or conv. – copper.hat Oct 29 '22 at 02:08
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    Your proposed definition of $\overline {conv}(A)$ is equal to $\overline {conv(A)}$. – DanielWainfleet Oct 29 '22 at 02:13
  • Both definitions are equivalent, see the comments under hte linked question. – daw Oct 29 '22 at 17:04
  • @DanielWainfleet: yes, but even if my proposed (!) definition of $\overline {conv}(A)$ equals $ \overline {conv(A)}$, this not in general implies that this is the original definition of $\overline {conv}(A)$, just that it is equivalent to the other as daw noticed. It was just a guess. And my question was about the standard definition of $\overline {conv}(A)$ – user267839 Oct 31 '22 at 01:12
  • Can it be considered as a kind of convention if $\text{makesomething}(A)$ is an operation associating some boolean operation on certain subsets depending on $A$ (eg like taking union of all subset contained in $A$ or containing it, or intersecting them, or any other boolean combi allowed in the power set of $X$), then $\overline{makesomething}(A)$ should by convention be the restriction to do the operation but on only closed subsets related to $A$? – user267839 Oct 31 '22 at 01:19

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