Logical Labyrinths

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180 III. Infinity x, if each successor of x has only finitely many descendants, so does x. Therefore, by the above theorem, every point of the tree has only finitely many descendants. In particular, the origin has only finitely many descendants, so the tree is finite. This proves the Fan Theorem. Ball Games Revisited In the solution to Problem 16.8 we proved Theorem 16.7 (Ball Game Theorem). Every ball game of the type described in Problem 16.8 must terminate in finitely many steps. This theorem is closely related to the Fan Theorem—indeed, it can be derived as a corollary of it, thus yielding an interesting alternate proof: Without loss of generality, we can assume that there is only one ball initially in the box, because if we start with a finite set S of balls in the box, we could instead start with just one ball x whose label is a greater number than any of those in S, and on the first move, replace x by the balls in S. So we shall assume that we initially have only one ball in the box. Then to any such ball game B, we associate the following tree, which we will denote tr(B) (“the tree of B”). We take the ball that is initially in the box as the origin, and for any ball that is ever in the box, we take its successors to be the balls that replace it (balls numbered 1 are of course end points). Since each ball is replaced by only finitely many balls, the tree is finitely generated. Since each ball is replaced by balls bearing lower numbers, each path must be finite. Then, by the Fan Theorem, the whole tree is finite—there are only finitely many points on the tree, which means that only finitely many balls ever enter the box, which wouldn’t be the case if the game never terminated. Therefore the game must terminate. We have now seen two different proofs of the Ball Game Theorem— the first was by mathematical induction on the greatest number assigned to any ball originally in the box, and the second, by the Fan Theorem. But there is a simpler proof yet, which uses neither the induction technique of the first proof nor the Fan Theorem. The proof is simply this: In the tree tr(B) associated with a ball game, not only are all paths finite, but no path can be of length greater than n, where n is the number assigned to the original ball! Thus only finitely many levels are hit, and since in a finitely generated tree, each level can have only finitely many points (Problem 16.10), the tree must be finite. You see, a ball game is really more than just a finitely generated tree whose paths are all finite; it is that together with an assignment of positive integers to all points of the tree such that each end point is assigned the number 1 and each other point x is assigned a greater number than
16. Generalized Induction, König’s Lemma, Compactness 181 any of its successors. Such a tree, we will say, is well-numbered. Can every finite tree be well-numbered? (If so, then every finite tree is the tree of some ball game.) As a matter of fact, even an infinite tree can be wellnumbered, provided that there is an upper bound to the lengths of all its branches. Problem 16.13. How can such a tree be well-numbered? Compactness König’s Lemma will be seen to have interesting applications to propositional and first-order logic. We shall shortly consider another principle that has equally important such applications, but we first turn to some related problems. We consider a universe V with denumerably many inhabitants. They have formed various clubs. A club C is called maximal if it is not a proper subset of any other club. Thus if C is a maximal club, then for any set S that contains all the people in C, if S contains so much as one person who is not in C, then S fails to be a club. Problem 16.14. (a) In a denumerable universe V, assuming that there is at least one club, is there necessarily a maximal club? (b) What about if V is finite instead of denumerable? If there is at least one club, is there necessarily a maximal club? Now for the key problem! Problem 16.15. Again we consider a universe V of denumerably many people, and we assume there is at least one club. We have seen that it does not follow that there must then be a maximal club, but now, suppose we are given the additional information that for any set S of inhabitants of V, S is a club if and only if every finite subset of S is a club. Then it does follow that there must be a maximal club! Better yet, it follows that every club C is a subset of some maximal club. The problem is to prove this. Here are the key steps in the proof. (1) Show that, under the given conditions, every subset of a club must be a club. (2) Since V is denumerable, its inhabitants can be arranged in a denumerable sequence x 0 , x 1 , x 2 , . . . , x n , . . .. Let C be any club. Now
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Logical Labyrinths
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Editorial, Sales, and Customer Serv
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vi Contents III Infinity 133 14 The
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viii Preface In Part III, we journe
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- Chapter 1 - The Logic of Lying an
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1. The Logic of Lying and Truth-Tel
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6. A Unification 39 To find out if
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44 II. Be Wise, Symbolize! Negation
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11. The Tableau Method 95 Some Vita
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322 References [17] R. M. Smullyan.
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