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Suppose we have finitely many convex sets $c_1,\dots, c_m$ in $\mathbb{R}^n$ all include zero. For simplicity let $m=2$. Let $x\in c_1$ and $y \in c_2$. Suppose we create a vector $z$ by picking some coordinates from $x$ and some from $y$. Is $z \in c_1 \cup c_2$? If yes, how can I show it, if no, how I can refute it?

My thoughts

I do not know how to use $\lambda x+ (1-\lambda) y$ to show the above claim where $\lambda \in [0,1]$. I feel it cannot be true for an arbitrary convex set. Probably for certain class of convex sets, we can show it. More likely for sign symmetric sets like $\ell_p$ balls it would be possible.

Asaf Karagila
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Saeed
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1 Answers1

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Let $C_1$ be a horizontal $100$x$1$ box in $\mathbb{R}^2$ and $C_2$ be a $1$x$100$ vertical box. Boxes are clearly convex.

Vector $a=(100,0)=(x,y) \in C_1 $

Vector $b=(0,100)=(0,y) \in C_2$

But vector $(100,100)=(x,y) \notin C_1 \cup C_2 $

So your intuition is correct. We can't just pick coordinates arbitrarily and have the resulting tuple be a vector in our sets.

Draw it. There must be a $λ \in [0,1]$ for which the line segment $λx+(1−λ)y$ which is parameterized by lambda sits outside the union.

vallev
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