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?
Tell me more
×
Mathematics Stack Exchange is a question and answer site for
people studying math at any level and professionals in related fields. It's 100% free, no registration required.
|
|
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). |
||||
|
|
|
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 $\{(x,y) : y > 0\}$. |
|||
|
|
