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 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 $\displaystyle tz_1+ (1-t)z_2$? (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.) Tags convex, prove Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post masterofmath Algebra 1 May 21st, 2013 11:54 AM masterofmath Algebra 1 May 18th, 2013 11:49 AM octaveous Number Theory 13 September 23rd, 2010 04:36 AM qweiop90 Algebra 1 July 31st, 2008 06:27 AM qweiop90 New Users 1 December 31st, 1969 04:00 PM

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