Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have an issue with a definition in Rudin's Functional Analysis in the paragraph regarding the Krein-Milman Theorem.

"Let $K$ be a subset of a vector space $X$. A nonempty set $S$ in $K$ is called an extreme set if no point of $S$ is an internal point of a line interval whose end points are in $K$ but not in $S$. Analytically, the condition can be expressed as follows: if $x$ and $y$ are in $K$, if $t$ is in $(0, 1)$, and if $tx + (1 - t)y$ is in $S$, then $x$ and $y$ are in $S$. The extreme points of $K$ are the extreme sets that consist of just one point."

For this condition to be equivalent to the definition, one should replace the conclusion by: "$\dots$ then $x$ is in $S$ or $y$ is in $S$." It turns out that this is indeed equivalent when $S$ consists of a single point, but not in general.

So my question is: what is the good definition for an extreme set?

share|improve this question
    
    
@t.b.: thank you. Rudin's proof does not assume its extreme sets to be convex, though. Anyway, it looks like the condition should be taken to be the definition here. –  user26770 Mar 12 '12 at 19:43
add comment

1 Answer 1

The second sentence is convoluted and, to me, hard to understand. The third sentence conforms to several other texts. If you want a fancier definition you can use the following:

Let $\mathsf{con}$ denote the convex hull operator. For $S \subseteq K$, the set $S$ is an extreme subset of $K$ if and only if for all $D \subseteq K$ we have $S \cap \mathsf{con}(D) = S \cap \mathsf{con}(S \cap D)$. If you want to, you only have to consider finite subsets $D$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.