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

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The discussion confirms that any convex combination of elements from a convex set S remains within that set. Specifically, if S is a convex set and x1, x2, x3, ..., xn are elements in S, then their convex combination, expressed as x = α1x1 + α2x2 + α3x3 with αi ≥ 0 and Σiαi = 1, is also in S. The proof utilizes a transformation of the coefficients to demonstrate that the resulting combination adheres to the properties of convexity, ensuring that x is an element of S.

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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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