We shall look at well ordered chains that is to say

Info icon This preview shows pages 12–14. Sign up to view the full content.

View Full Document Right Arrow Icon
We shall look at well ordered chains, that is to say, chains for which every non-empty subset has a minimum. (Formally, if S C is non-empty we can find an s 0 C such that s s 0 for all s S . We write min C = s 0 .) By the previous paragraph A C = { x : x ´ c for all c C } 6 = . Thus, if we write W for the set of all well ordered chains, the axiom of choice, tells us that there is a function κ : W → X such that κ ( C ) ´ c for all c C . We now consider ‘special chains’ defined to be well ordered chains C such that κ ( C x ) = x for all x C. 12
Image of page 12

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
(Note that ‘well ordering’ is an important general idea, but ‘special chains’ are an ad hoc notion for this particular proof. Note also that if C is a special chain and x C then C x is a special chain.) The key point is that, if K and L are special chains, then either K = L or K = L x for some x L or L = K x for some x K . Subproof If K = L , we are done. If not, at least one of K \ L and L \ K is non-empty. Suppose, without loss of generality, that K \ L 6 = . Since K is well ordered, x = min K \ L exists. We observe that K x L . If K x = L , we are done. We show that the remaining possibility K x 6 = L leads to contradiction. In this case, L \ K x 6 = so y = min L \ K x exists. By definition of y and the fact that K x L , we have K y = L y . But K and L are special chains so y = κ ( K y ) K contradicting the definition of y . End subproof We now take S to be the union of all special chains. Using the key observation, it is routine to see that: (i) S is a chain. (If a, b S , then a L and b K for some special chains. By our key observation, either L K of K L . Without loss of generality, K L so a, b K and a b or b a .) (ii) If a S , then S a is a special chain. (We must have a K for some special chain K . Since K S , we have K a S a . On the other hand, if b S a then b L for some special chain L and each of the three possible relationships given in our key observation imply b K a . Thus S a K a , so S a = K a and S a is a special chain.) (iii) S is well ordered. (If E is a non empty subset of S , pick an x E . If S x E = , then x is a minimum for E . If not, then S x E is a non-empty subset of the special, so well ordered chain S x , so min S x E exists and is a minimum for E .) (iv) S is a special chain. (If x S , we can find a special chain K such that x K . Let y = κ ( K ). Then L = K ∪ { y } is a special chain. As in (ii), S y = L y , so S x = L x and κ ( S x ) = κ ( L x ) = x .) We can now swiftly obtain a contradiction. Since S is well ordered κ ( S ) exists and does not lie in S . But S is special, so S κ ( S ) is, so S κ ( S ) S , so κ ( s ) lies in S . The required result follows by reductio ad absurdum 5 . Lemma 36 (Hammel basis theorem). (i) Every vector space has a basis. (ii) If U is an infinite dimensional normed space over F (with F = R or F = C ) then we can find a discontinuous linear map T : U F .
Image of page 13
Image of page 14
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern