MHB Is Every Convex Combination of Elements in a Convex Set Also in the Set?

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If S is a convex set and x1, x2, x3, ..., xn are elements in S, then their convex combination also belongs to S. The proof involves showing that if x is defined as a convex combination of these elements, it can be expressed in terms of other convex combinations that maintain the conditions of convexity. By adjusting the coefficients of the combination while ensuring they remain non-negative and sum to one, it can be demonstrated that the resulting element is still within the set S. This reasoning can be generalized for any number of elements in the convex set. Thus, every convex combination of elements in a convex set is indeed contained within that set.
Fernando Revilla
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I quote a question from Yahoo! Answers

I need to Show the Following
if S is a convex set and x1, x2, x3, . . . xn are n elements in S then Their convex combination is also in S .
Please help me .
Thanxs

I have given a link to the topic there so the OP can see my response.
 
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Firstly, we prove that if $x=\alpha_1x_1+\alpha_2x_2+\alpha_3x_3$ for $\alpha_i\ge 0$ and $\sum_i\alpha_i=1$, then $x\in S$. Consider $$\alpha'_i=\frac{\alpha_i}{\alpha_1+\alpha_2} \quad (i=1,2)$$ Note that $\sum_i\alpha'_i=1$ and $\alpha'_i\ge 0.$ We have $$x=\alpha_1x_1+\alpha_2x_2+\alpha_3x_3=(\alpha_1+\alpha_2)(\alpha'_1x_1+\alpha'_2x_2)+\alpha_3x_3$$ By hypothesis $\alpha'_1x_1+\alpha'_2x_2\in S.$ Denote $x_4=\alpha'_1x_1+\alpha'_2x_2$, then $$x=(\alpha_1+\alpha_2)x_4+\alpha_3$$ But $x$ is a convex combination of two elements of $S$, so $x\in S.$ This argument can be generalized for any $i$.
 

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