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If $C$ is a set in a topological vector space (or in particular a metric space), can we say that $\text{cl}(\text{conv}(C)) = \text{conv}(\text{cl}(C))$, where cl$(\cdot)$ represents closure and conv$(\cdot)$ represents the convex hull.

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2 Answers 2

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No, because a convex hull of a closed set may be open.

Let $C=\{(x,y): y\geq\frac1{1+x^2}\}$. $C$ is closed, by its convex hull is $\mathbb{R}\times(0,\infty)$.

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Another example:

Let $C=\{e_n\} \subset l_2$. $C$ is isolated, hence closed (also bounded). If $x_n = { 1\over n}(e_1+\cdots+e_n) $, we see that $x_n$ is in the convex hull of $C$ and $\|x_n\|= {1 \over \sqrt{n}}$, hence $x_n \to 0$, however $0$ is not in the convex hull of $C$.

Of course, since $C \subset \operatorname{co} C$, we have $\overline{C} \subset \overline{\operatorname{co}} C$, and since the latter set is convex (not immediate), we have $\operatorname{co} \overline{C} \subset \overline{\operatorname{co}} C$.

In finite dimensions, one can use Caratheory's theorem to show that the convex hull of a compact set is compact.

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