I'm curious about this. Say we have X Banach space and Y is a totally bounded subset (a more mathematically mature friend told me this was synonymous with Y being precompact). We can prove the closure of Y is compact but does that imply the convex hull of Y is also compact? How is the closure related to the conv. hull?
Closure and convex hull of a set
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functional-analysis
banach-spaces
compactness
1 Answers
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No. Simple example is an open ball in finite dimensional space. It is totally bounded, the closure is compact. But it is convex, so the convex hull is the ball itself.
I'm not aware of any significant relationship between closures and convex hulls.