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Let $X$ be a nonempty convex subet of $A$. I need to Show that $z$ is an extreme point of $X$ if and only if the set $X − \{z\}$ is a convex set.

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You probably mean $X - \{z\}$ at the end. Anyway: what have you tried and where are you stuck? Which direction is causing you difficulties? (and what's $A$?) –  t.b. Dec 7 '11 at 8:51
    
We do not vandalize questions here, Michael. Please don't do that again. –  J. M. Dec 7 '11 at 10:41
    
You could explain the reason why you deleted (again) the paragraph: I first tried to show that if $z$ is an extreme point then $A − \{z\}$ us a convex set. I used the definition of a convex set, $\lambda x+(1-\lambda)y\;$ is in set $S$. I also attempted to go on the idea that $z=\lambda x_1+(1-\lambda)x_2$, so $x_1=x_2=z$. –  Did Dec 7 '11 at 19:03
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2 Answers 2

If $z$ is not an extreme point of $X$, there are $x,y\in X\setminus \{z\}$ such that $z=tx+(1-t)y$ for some $t\in(0,1)$. This easily implies that $X\setminus \{z\}$ is not convex; why?

Now suppose that $z$ is an extreme point of $X$, and let $x,y\in X\setminus\{z\}$; you need to show that $tx+(1-t)y\in X\setminus\{z\}$ for each $t\in[0,1]$. $X$ is convex, so $tx+(1-t)y\in X$ for each $t\in[0,1]$. How could $tx+(1-t)y$ fail to be in $X\setminus\{z\}$? There’s only one way; why can’t it actually happen?

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im not sure how that exactly answers my question –  Michael Buck Dec 7 '11 at 9:04
    
@Michael: It doesn’t answer it completely; it provides a pretty big hint for each of the two things that you have to prove. –  Brian M. Scott Dec 7 '11 at 9:09
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This is just Brian's answer, somewhat elaborated:

The proof should be transparent, it just uses the definitions.

A set $X$ is convex if and only if every line segment between two points of $X$ is contained in $X$.

A point $z\in X$ is an extreme point of the convex set $X$ if and only if it is not an interior point of any line segment in X (it can be an endpoint of a line segment contained in $X$).

Let $z$ be an extreme point of the convex set $X$. Since $X$ is convex, every line segment between two points in $X\setminus\{z\}$ is contained in $X$. Now, since $z$ is an extreme point of $X$ and since $z$ is not in $X\setminus\{z\}$, none of these line segments contain $z$. So, in fact, every line segment between two points in $X\setminus\{z\}$ is contained in $X\setminus\{z\}$. Thus, $X\setminus\{z\}$ is convex.

If $X\setminus\{z\}$ is convex, then, every line segment between two points in $X\setminus\{z\}$ is contained in $X\setminus\{z\}$. Thus, $z$ cannot be on any line segment between two points in $X\setminus\{z\}$. Of course, then, $z$ cannot be on the interior of any line segment in $X$; which shows that $z$ is an extreme point of $X$.

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