Prove a theorem about a vector space and convex sets

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SUMMARY

The discussion centers on proving that if a set X of vectors {x1,...,xn} belongs to a vector space E and is convex, then all convex combinations of X also belong to X. A convex combination is defined as t1*x1 + t2*x2 + ... + tn*xn, where t1,...,tn are non-negative and sum to 1. The challenge lies in demonstrating that these combinations remain within the bounds of X, particularly when considering the ordering of the vectors. Induction is suggested as a method for proof, starting with simple cases such as n=2 and n=3.

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Summary:: Be the set X of vectors {x1,...,xn} belong to the vector space E. If this set X is convex, prove that all the convex combination of X yet belong to X. Where convex combination are the expression t1*x1 + t2*x2 + ... + tn*xn where t1,...,tn >= 0 and t1 + ... + tn = 1

I tried to suppose xn > xn-1 > ... > x1, so in this way we have two limits, and the convex requires that all elements of E [v1,vn] belongs to X.
Now here i smell a rat: I suppose that xn > t1*x1 + t2*x2 + ... + tn*xn > x1, in such way that it will automatically belongs to E. The problem is how to prove my statement...

[Moderator's note: Moved from a technical forum and thus no template.]
 
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xn > t1*x1 + t2*x2 + ... + tn*xn > x1

Remember these are vectors, you don't have a natural way to compare them. Convexity of E only says this sum is in E if it happens to lie on the line between x1 and xn, which is very unlikely for random coefficients.

I would suggest trying to do induction. n=2 is as easy as you think. What about n=3? This is just a triangle, so a picture might help you think about it.
 

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