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Is the convex hull of a finite set of points in $\mathbb R^2$ closed? Intuitively, yes. But not sure how to show that. Thanks!

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  • $\begingroup$ What have you tried? How have you used the description of the convex hull, or maybe of an arbitrary element of the hull? $\endgroup$
    – Lubin
    Apr 11 '14 at 15:50
  • $\begingroup$ You can show by induction that the convex hull of $n$ points is compact, so it is also closed. Hint: Write that hull as the continuous image of $K\times[0,1]$ for a compact convex $K$. $\endgroup$ Apr 11 '14 at 15:55
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Generally, the convex hull $\{\lambda_1x_1+\dots+\lambda_nx_n\colon0\le\lambda_1,\dots,\lambda_n\le1,\lambda_1+\dotsb+\lambda_n=1\}$ of $x_1,\dotsc,x_n\in\mathbb R^d$ is compact, hence closed, since it's the continuous image of the closed simplex $\{(\lambda_1,\dotsc,\lambda_n)\colon0\le\lambda_1,\dots,\lambda_n\le1,\lambda_1+\dotsb+\lambda_n=1\}\subseteq\mathbb R^n$.

To fill in the details, you can prove by induction that the closed simplex is really compact as Hamcke mentioned in the comment, but you can do it simpler: as a subspace of $\mathbb R^n$, it's obviously bounded and it could be written as an intersection of a (in fact finite) collection of closed subsets of $\mathbb R^n$.

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  • $\begingroup$ what finite intersection of closed sets then? $\endgroup$
    – Britta
    Jun 26 '20 at 9:52
  • $\begingroup$ @Britta Given a topological space $E$ and a continuous function $f\colon E\to\mathbb R$, the subset $f^{-1}([0,1])\subseteq E$ is closed, so is the subset $f^{-1}(0)\subseteq E$. Then choose proper $f$'s to get $n+1$ closed subsets of $\mathbb R^n$ of which the intersection is the closed simplex. $\endgroup$
    – Yai0Phah
    Jun 26 '20 at 14:53
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I'd like to generalize this to the statement that if $K$ is a convex compact subset of a topological vector space $V$ and $z$ is some vector, then the convex hull $H$ of $K$ and $z$ is compact.
Proof: Each convex combination $x=\sum_{i=1}^n t_i x_i+sz$ of points in $K\cup\{z\}$, where $x_i\in K$, $t_i,s\ge 0,\ s+t_1+\cdots +t_n=1$, can be written as $$x=sz+(1-s)k,\qquad k=\sum \frac{t_i}{\sum t_i}x_i$$ where $k\in K$. If we define $$h:K\times[0,1]\to V,\qquad (k,s)\mapsto sz+(1-s)k$$ then $h$ is continuous and its image is $H$. Since the domain is compact, $H$ must be compact, too. Hence $H$ is closed.

Now assume by induction that $K$, the convex hull of $n-1$ points is compact, and it follows that the hull of all $n$ points, which is just $H$, is compact.

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Although the fact needs a proof, the hull of a finite set is also the intersection of all closed half-spaces containing the set. So closed.

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