5 to the best of my knowledge this particular proof

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5 To the best of my knowledge, this particular proof is due to Jonathon Letwin ( Amer- ican Mathematical Monthly , Volume 98, 1991 , pp. 353–4). If you know about transfinite induction, there are more direct proofs. 13
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Exercise 37. (i) Show that if f : R R is continuous and satisfies the equation f ( x + y ) = f ( x ) + f ( y ) for all x, y R then there exists a c such that f ( x ) = cx for all x R . (ii) Show that there exists a discontinuous function f : R R and satis- fying the equation f ( x + y ) = f ( x ) + f ( y ) for all x, y R . [Hint. Consider R as a vector space over Q .] The rest of this section is devoted to a proof of Tychonov’s theorem. Theorem 38 (Tychonov). The product of compact spaces is itself compact. We follow the presentation in [1]. (The method of proof is due to Bour- baki.) The following result should be familiar to almost all of my readers. Lemma 39 (Finite intersection property). (i) If a topological space is compact then, whenever a non-empty collection of closed sets F has the property that T n j =1 F j 6 = , for any F 1 , F 2 , . . . , F n ∈ F it follows that T F ∈F F 6 = . (ii) A topological space is compact if whenever a non-empty collection of sets A has the property that T n j =1 A j 6 = for any A 1 , A 2 , . . . , A n ∈ A it follows that T A ∈A ¯ A 6 = . Definition 40. A system F of subsets of a given set S is said to be of finite character if whenever every finite subset of a set A S belongs to F it follows that A ∈ F . Lemma 41 (Tukey’s lemma). If a system F of subsets of a given set S has finite character and F ∈ F then F has a maximal (with respect to inclusion) element containing F . We now prove Tychonov’s theorem. The reason why Tychonov’s theorem demands the axiom of choice is made clear by the final result of this section. Lemma 42. Tychonov’s theorem implies the axiom of choice. 14
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6 The Hahn-Banach theorem A good example of the use of Zorn’s lemma occurs when we ask if given a Banach space ( U, k k ) (over C , say) there exist any non-trivial continuous linear maps T : U C . For any space that we can think of, the answer is obviously yes, but to show that the result is always yes we need Zorn’s lemma 6 . Our proof uses the theorem of Hahn-Banach. One form of this theorem is the following. Theorem 43. (Hahn-Banach) Let U be a real vector space. Suppose p : U R is such that p ( u + v ) p ( u ) + p ( v ) and p ( au ) = ap ( u ) for all u, v U and all real and positive a . If E is a subspace of U and there exists a linear map T : E R with Tx p ( x ) for all x E then there exists a linear map ˜ T : U R with Tx p ( x ) for all x U and ˜ T ( x ) = Tx for all x E . [Note that we do not assume that the vector space U is normed but we do assume that the vector space is real.] We have the following important corollary Theorem 44. Let ( U, k k ) be a real normed vector space. If E is a subspace of U and there exists a continuous linear map T : E R , then there exists a continuous linear map ˜ T : U R with k ˜ T k = k T k .
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