# We shall look at well ordered chains that is to say

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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

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(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 .
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