Show that diagonal of a square is not a face

Question

Let C be a convex set in $R^n$. We say that F is a face of C if the following condition holds: if $x_1, x_2 \in C$ and $(1-\lambda)x_1+\lambda x_2 \in F$ for some $0\lt \lambda \lt 1$ then $x_1, x_2 \in F$

Based on this definition, how can we show that a diagonal of a square $F_1=conv({(0,0),(1,1)}) $ is not a face?

score 0 · Accepted Answer · answered Nov 27 '13 at 12:45

0

Take $x_1 = (0,1)$ and $x_2=(1,0)$. Then for $\lambda = .5$ we have $x_1, x_2 \not\in F$ but the combination is in $F$.

answered Nov 27 '13 at 12:45

Eric Auld

28,127

by the combination do you mean $(0.5, 0.5)$ ? – dresden Nov 27 '13 at 12:49
@dracc Yes.---- – Eric Auld Nov 27 '13 at 12:50

Show that diagonal of a square is not a face

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