Thread: Advanced Geometry - Convex sets and Extreme points proof

1. 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.

2. Re: Advanced Geometry - Convex sets and Extreme points proof

I know it's hard but anything written down could be good.

3. Re: Advanced Geometry - Convex sets and Extreme points proof

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.

4. 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?

5. Re: Advanced Geometry - Convex sets and Extreme points proof

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.