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By Marcus (Du ) Sautoy

Titre unique: The song of the Primes
Traduit de l'anglais par Raymond Clarinard

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4). Johnson) shows that Res (L) a lattice does not follow from the fact that L is a lattice. 3. Construct a lattice L as follows: take the four element lattice which has the following Hasse diagram: π o bo^ \ O a FOUNDATIONS 39 and insert three copies of the field of real numbers as suggested by the following diagram: - \ \ \ v\ \ v x remove-·- b o ^ \ \ \Χν^ν\Νο3 Remove the element b and enlarge the ordering as suggested by the dotted lines. To be more specific, for any r e R3 let r* denote the same real number in Rx and for r,seR3 define r < s* if and only if r < s in the usual ordering of the reals.

It follows that f(y) = y a n d so yeF. This shows that (a) => (6). If now (b) holds, then / o r each xeE we have g(x)e(7 e F and so /[g(x)] = g(x), giving f o g = g which is (c). If now (c) holds, then from g > ìdE we have g = g ° g = g o / o g > g o / o id E = g o / and from / > id E we have g °f> g- Thus g = g o/which is (d). Finally, if (d) holds then for each x e E, we have/(x) = (id £ of) (x) < (g of) (x) = g(x) and (a) follows. 5. Let E be an ordered set, let K(E) denote its set of closure subsets and let C{E) be its set of closure mappings.

The following conditions are equivalent: (1) there exists a closure mapping f: E -► E such that the set off-closed elements is F; (2) for each xe E the set [ x , - > ] n F admits a minimum element. Proof Suppose that (1) holds. Then for any xe E the set [x, ->] n F is not empty, for it clearly contains the element/^). Moreover, if z e [x,-*] n F, then x < z and/(x) < f(z) = z. Consequently [x,-+] nF admits a minimum element, namely/(x). Conversely, suppose that (2) holds. Let x^ denote the minimum element of [x, ->] n F and define a mapping fiE^Eby the prescription/(x) = x^.

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