proof of Krein-Milman theorem
The proof is consist of three steps for good understanding.
We will show initially that the set of extreme points of , is non-empty, .
We consider that extreme.
The family set ordered by has a minimal element, in other words there exist such as we have that .
We consider . The ordering relation is a partially relation on . We must show that is maximal element for . We apply Zorn’s lemma.We suppose that
is a chain of .Witout loss of
generality we take and then . has the property of finite intersections
and it is consist of closed sets. So we have that . It is easy to see that .
Also , for any , so we have that , for any .
Every minimal element of is a set which has only one point.
We suppose that there exist a minimal element of which has at least two points, . There exist such as , witout loss of generality we have that . is compact set (closed subset of the compact ). Also there exist such that and . It is obvious that is an extreme subset of , is an extreme subset of ,. since and that contradicts to the fact that is minimal extreme subset of .
From the above two steps we have that .
where denotes the closed convex hull of extreme points of .
Let . Then is closed subset of , therefore it is compact, and convex clearly by the definition. We suppose that . Then there exist . Let use Hahn-Banach theorem(geometric form). There exist such as . Let , . Similar to Step2 is extreme subset of . is compact and from step1 and step2 we have that . It is true that . Now let then and if . That is a contradiction.
|Title||proof of Krein-Milman theorem|
|Date of creation||2013-03-22 15:24:40|
|Last modified on||2013-03-22 15:24:40|
|Last modified by||georgiosl (7242)|