Does there exist such a convex hull of infinite points?

For example, consider infinite number of points of which form a circle in $\mathbb{R}^2$. Is this considered as a convex hull?

  • 4
    $\begingroup$ There exists the convex hull of every set of a vector space. en.wikipedia.org/wiki/Convex_hull $\endgroup$ Nov 30, 2013 at 19:15
  • $\begingroup$ Why the convex optimization tag? $\endgroup$
    – hhsaffar
    Nov 30, 2013 at 19:33
  • $\begingroup$ I corrected to convex-analysis. Sorry $\endgroup$
    – dresden
    Nov 30, 2013 at 19:44

2 Answers 2


For any subset $S$ of $\mathbb R^N$, the convex hull of $S$ is the intersection of all convex subsets of $\mathbb R^N$ which contain $S$.

A circle in $\mathbb R^2$ isn't convex; the convex hull of a circle in $\mathbb R^2$ is a closed disk.


Let $X$ be a vector space. A set $C\subset X$ is convex if $[x,y]\subset C$ for all $x,y\in C$, where $[x,y]:=\{(1-\alpha)x+\alpha y: \alpha\in[0,1]\}$ (the closed line segment from $x$ to $y$).

The convex hull of $A\subset X$ is the smallest convex set containing $A$.

Lemma. Let $A\subset X$. Then the convex hull of $A$ equals the set of all sums $$ t_1 a_1 + \cdots + t_n a_n $$ where $n\in\{1,2,\ldots\}$, $t_i\ge0$, $a_i\in A$ for all $i$, and $t_1+\cdots+t_n=1$.

(Clearly the lemma is true even if we require $t_i>0$ for all $i$. It holds for both real and complex vector spaces.)

Proof: The proof of Theorem 2.5 here applies without changes (although it is meant for $R^n$ only). https://www.ti.inf.ethz.ch/ew/lehre/CG10/lecture/ln2.pdf

Some of this and more can be found in https://en.wikipedia.org/wiki/Convex_hull


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