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Is the closure of the convex hull of some set $A\subseteq\mathbb R^d$ equal to the convex hull of the closure of $A$, i.e. $$\text{cl}(\text{conv}(A))=\text{conv}(\text{cl}(A))?$$

If not, what are the general relations between them?

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  • $\begingroup$ How do you distinguish the first and last sets? $\endgroup$
    – copper.hat
    Nov 5 '12 at 16:32
  • $\begingroup$ Well, $\text{conv}(A) \subset \overline{\text{conv}}(A)$, hence $\text{cl}(\text{conv}(A)) \subset \overline{\text{conv}}(A)$ and $\text{cl}(\text{conv}(A))$ is closed and convex, hence we must have $\text{cl}(\text{conv}(A)) = \overline{\text{conv}}(A)$. $\endgroup$
    – copper.hat
    Nov 5 '12 at 16:37
  • $\begingroup$ conv(cl(A)) is neither of the sets you mentioned, which was the original question $\endgroup$ Jan 25 '17 at 5:15
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No.

Let $A = \{(x,e^{-x})\}_{x\geq 0} \cup \{(x,-e^{-x})\}_{x\geq 0}$. Then $A$ is closed, and $\mathrm{co} A = (\{0\}\times [-1,1]) \cup ((0,\infty)\times (-1,1))$, which is not closed (take $(x_n,y_n) = (1, 1-\frac{1}{n})$).

Hence $\mathrm{co} A = \mathrm{co} \overline{A} $ is strictly contained in $\overline{\mathrm{co}} A = [0,\infty)\times [-1,1]$.

If $A$ is compact, the result is true (using, eg, Carathéodory's theorem).

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  • $\begingroup$ Your example is one, where $conv(cl(A))$ is stritly contained in $cl(conv(A))$. Do you know an example, where we have $cl(conv(A))$ is strictly contained in $conv(cl(A))$? $\endgroup$
    – Andy Teich
    Jun 11 '13 at 8:56
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    $\begingroup$ No. Since $A \subset \mathrm{co} A$, you have $\overline{A} \subset \overline{\mathrm{co}} A$. Since $\overline{\mathrm{co}} A$ is convex, we have $\mathrm{co} \overline{A} \subset \overline{\mathrm{co}} A$. $\endgroup$
    – copper.hat
    Jun 11 '13 at 15:52
  • $\begingroup$ What if $\overline{A}$ is compact? $\endgroup$ Apr 25 '16 at 8:51
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    $\begingroup$ @TigranSaluev: My previous comment shows that $\mathrm{co} \overline{A} \subset \overline{\mathrm{co}} A$ for any $A$. If $A$ is relatively compact, we have $\mathrm{co} \overline{A} = \overline{\mathrm{co}} \overline{A}$ and since $A \subset \overline{A}$ we have $\overline{\mathrm{co}} A \subset \overline{\mathrm{co}} \overline{A} = \mathrm{co} \overline{A}$ from which we have $\overline{\mathrm{co}} A = \mathrm{co} \overline{A}$. $\endgroup$
    – copper.hat
    Apr 25 '16 at 15:14
  • $\begingroup$ In an infinite dimensions Banach space, is this true for compact set? That means, convex hull of a compact set is closed. Is this true? $\endgroup$
    – A learner
    Nov 27 '21 at 15:36
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Not necessarily. Let $$A=\Bigl\{(x,y) : y\geq {1\over 1+x^2}\Bigr\}$$ Then the closure of the convex hull is the closed upper half plane $\{(x,y) : y\geq 0\}$, but the convex hull of the closure is the open upper half plane $\{(x,y) : y > 0\}$.

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  • $\begingroup$ Hi!, how can I prove that $B=\{(x,y): y>0\}$ is the convex hull of $A$? Indeed, $B$ is convex set, but how can I prove that $B$ is the convex hull of $A$? $\endgroup$
    – mathproof
    Nov 15 '20 at 6:49
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    $\begingroup$ @Ramanujan If $(x,y)$ is any point in $B$ but not in $A$, it will lie on the straight line between $(-N,y)$ and $(N,y)$, where $N$ is chosen large enough so that the two points lie in $A$. $\endgroup$
    – Per Manne
    Nov 15 '20 at 8:10
  • $\begingroup$ Oh, I understand. Thank you! $\endgroup$
    – mathproof
    Nov 15 '20 at 8:16

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