# Advanced Geometry - Convex sets and Extreme points proof

• Nov 28th 2011, 09:23 AM
maxgunn555
Advanced Geometry - Convex sets and Extreme points proof
Let K subset of R^n be a convex set. We call x1 element of K an extreme point of K if K/{x1} is convex too.

Prove that x1 element of K is an extreme point iff cx + (1-c)y = x1 for 0<c<1, c element of R (real numbers as above) implies x = y = x1.

any answer or help will be greatly appreciated.
• Nov 30th 2011, 05:38 PM
maxgunn555
Re: Advanced Geometry - Convex sets and Extreme points proof
I know it's hard but anything written down could be good.
• Dec 1st 2011, 05:24 AM
Opalg
Re: Advanced Geometry - Convex sets and Extreme points proof
Quote:

Originally Posted by maxgunn555
Let K subset of R^n be a convex set. We call x1 element of K an extreme point of K if K/{x1} is convex too.

Prove that x1 element of K is an extreme point iff cx + (1-c)y = x1 for 0<c<1, c element of R (real numbers as above) implies x = y = x1.

any answer or help will be greatly appreciated.

For $\displaystyle x\in X$, write C(x) for the following condition:

$\displaystyle C(x)\qquad \text{If } x = cy+(1-c)z \text{ (where }y,z\in X, 0<c<1 \text{), then }y=z=x.$

Notice that, in that condition, if $\displaystyle y\ne x$ then $\displaystyle z\ne x$. So the condition can be written in the modified (but equivalent) form

$\displaystyle C(x)\qquad \text{If }y,z\in X\setminus\{x\} \text{ and } 0<c<1 \text{ then }x \ne cy+(1-c)z.$

Now suppose that $\displaystyle x_1$ satisfies the condition $\displaystyle C(x_1)$. Then $\displaystyle X\setminus\{x_1\}$ satisfies the definition of convexity. The reason is that if $\displaystyle y,z\in X\setminus\{x_1\}$ then $\displaystyle cy+(1-c)z\in X$ (because X is convex), but $\displaystyle cy+(1-c)z\ne x_1$ (because of the modified form of the condition $\displaystyle C(x_1)$). Therefore $\displaystyle cy+(1-c)z\in X\setminus \{x_1\}.$

The converse implication comes straight from the definition. If $\displaystyle X\setminus \{x_1\}$ is convex, and $\displaystyle y,z\in X\setminus \{x_1\}$, then $\displaystyle cy+(1-c)z\in X\setminus \{x_1\}.$ Thus $\displaystyle cy+(1-c)z$ cannot be equal to $\displaystyle x_1$. That shows that the modified form of condition $\displaystyle C(x_1)$ holds.
• Dec 1st 2011, 04:22 PM
maxgunn555
Re: Advanced Geometry - Convex sets and Extreme points proof
thank you very much. i understand now. btw where do you get all those symbols and mathematical notation to make posts with?
• Dec 2nd 2011, 04:14 AM
Opalg
Re: Advanced Geometry - Convex sets and Extreme points proof
Quote:

Originally Posted by maxgunn555
where do you get all those symbols and mathematical notation to make posts with?

Look at the "sticky" posts in the LaTeX help forum.