June 28th, 2016, 10:39 PM  #1 
Newbie Joined: May 2016 From: Denmark Posts: 2 Thanks: 0  I need help to prove
Given a convex subset K is a subset of R^3. I(K)={z there exists x, y in R such that (x,y,z)^t is in K} $\displaystyle \subseteq$ R. I need to prove that I(K) is a convex subset. Can anybody help me? I don't know where to start. 
June 29th, 2016, 08:05 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
I would think that a good place to start proving that a set was convex would be with the definition of "convex"! A set, X, is convex if and only if, for any two points, p and q, in X, every point "between" them, tp+ (1 t)q, for 0< t< 1, is also in X. Here, suppose that $\displaystyle z_1$ and $\displaystyle z_2$ such "there exist $\displaystyle x_1$, $\displaystyle y_1$ and $\displaystyle x_2$ and $\displaystyle y_2$ such that $\displaystyle (x_1, y_1, z_1)$ is in K and $\displaystyle (x_2, y_2, z_2)$ is in K. What can we say about $\displaystyle tz_1+ (1t)z_2$? (Use the fact that K is convex: for any t between 0 and 1, $\displaystyle t(x_1, y_1, z_1)+ (1 t)(x_2, y_2, z_2)$ is in K.) 

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