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If we are working over a general topological vector space V (i.e. not necessarily $R^d$) and we consider a subset $A \subset V$ which itself it not necessarily convex and a convex set $G$ which contains $A$, why is it that any convex combination

$$ t_1x_1 + ... + t_nx_n $$

of elements $x_1, ..., x_n \in A$ and $t_i \in (0,1)$ with the sum of the $t_i's$ being equal to $1$ is an element of $G$? Intuitively it makes enough sense, since we are sort of 'filling out' the convex set $G$ by lines drawn between finitely many points $x_k$ in $A$, but is there a simple way of proving that any convex combination like above will be an element of $G$?

Billy Bob
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  • Can you prove it for $n = 2$? for $n = 3$? for $n = k+1$ if it is true for $n = k$? It might help to write $t_n$ as $1 - (t_1 + \cdots + t_{n-1})$. Also, are you familiar with the definition of "convex hull"? – Eric Towers Apr 25 '21 at 03:15
  • The easiest way is by induction and you can choose the $t_i \in [0,1]$ as long as they sum to $1$. – CyclotomicField Apr 25 '21 at 03:22

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Any convex combination of elements of $G$ will be inside $G$.

Since $x_i \in A$ and $A$ is a subset of $G$, $x_i \in G$.

Hence the convex combination is in $G$.

Siong Thye Goh
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