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Smullyan Publisher: Courier Corporation ISBN: Category: Mathematics Page: View: Read Now » Written by a creative master of mathematical logic, this introductory text combines stories of great philosophers, quotations, and riddles with the fundamentals of mathematical logic. Author Raymond Smullyan offers clear, incremental presentations of difficult logic concepts. He highlights each subject with inventive explanations and unique problems.

Smullyan's accessible narrative provides memorable examples of concepts related to proofs, propositional logic and first-order logic, incompleteness theorems, and incompleteness proofs. Additional topics include undecidability, combinatoric logic, and recursion theory. Suitable for undergraduate and graduate courses, this book will also amuse and enlighten mathematically minded readers. Dover original publication. See every Dover book in print at www.

Featuring promising new voices alongside some of the foremost names in the field, The Best Writing on Mathematics makes available to a wide audience many articles not easily found anywhere else—and you don't need to be a mathematician to enjoy them.

These essays delve into the history, philosophy, teaching, and everyday aspects of math, offering surprising insights into its nature, meaning, and practice—and taking readers behind the scenes of today's hottest mathematical debates. In this volume, Moon Duchin explains how geometric-statistical methods can be used to combat gerrymandering, Jeremy Avigad illustrates the growing use of computation in making and verifying mathematical hypotheses, and Kokichi Sugihara describes how to construct geometrical objects with unusual visual properties.

In other essays, Neil Sloane presents some recent additions to the vast database of integer sequences he has catalogued, and Alessandro Di Bucchianico and his colleagues highlight how mathematical methods have been successfully applied to big-data problems. And there's much, much more. In addition to presenting the year's most memorable math writing, this must-have anthology includes an introduction by the editor and a bibliography of other notable writings on mathematics.

This is a must-read for anyone interested in where math has taken us—and where it is headed. It serves not only as a tribute to one of the great thinkers in logic, but also as a celebration of self-reference in general, to be enjoyed by all lovers of this field. Raymond Smullyan, mathematician, philosopher, musician and inventor of logic puzzles, made a lasting impact on the study of mathematical logic; accordingly, this book spans the many personalities through which Professor Smullyan operated, offering extensions and re-evaluations of his academic work on self-reference, applying self-referential logic to art and nature, and lastly, offering new puzzles designed to communicate otherwise esoteric concepts in mathematical logic, in the manner for which Professor Smullyan was so well known.

This book is suitable for students, scholars and logicians who are interested in learning more about Raymond Smullyan's work and life. Author : W. It begins with a survey of number systems and elementary set theory before moving on to treat data structures, counting, probability, relations and functions, graph theory, matrices, number theory and cryptography.

The end of each section contains problem sets with selected solutions, and good examples occur throughout the text. With each topic, a new challenge will be tackled to get a deeper knowledge of the Sparrow game framework and gain the skills to develop a complete mobile experience. This book is aimed at those who have always wanted to create their own games for iOS devices. Perhaps you've already dabbled in game development and want to know how to develop games for the Apple App Store, or maybe you have developed Objective-C apps in the past but you are new to game development.

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The Sudoku puzzle is ideal for whenever you have a few spare minutes and want to indulge in a little bit of thinking power. Sudoku sometimes spelled "Su Doku", is a puzzle that originated in Japan.

The puzzle is known as a "placement" puzzle. It offers a systematic and precise exposition of classical logic with many examples and exercises, and only the necessary minimum of theory.. Thus for any positive n, there are 2n subsets of an n-element set. There is only one possibility if my highest number is 1, namely 1, 1. There are 2 possibilities for the case that my highest number is 2, namely 1, 2 and 2, 2. In general, for each positive integer n, there are only finitely many possibilities if the highest of my two numbers is n, namely the n pairs 1, n , 2, n ,.

Thus you first go through all the possibilities in which the highest number is 1, then all the possibilities in which the highest number is 2, and so forth. For each positive n, there are exactly 2n — 1 fractions in which n is the highest of the numerator and denominator, namely the fractions. Thus you first go through all the fractions in which the highest of the numerator and denominator is 1, then through all the fractions in which the highest of the numerator and denominator is 2, and so forth.

We can thus go through all the sets whose highest number is 1, then all the sets whose highest number is 2, and so forth. It is the set — call it S — of ordinary numbers that cannot be listed on any page. For each number n, let Sn be the set of natural numbers listed on page n. Reason: The number n is either ordinary or extraordinary.

If it is ordinary, then by definition it does not belong to Sn, but it must then belong to S, which is the set of all ordinary numbers. So in this case n belongs to S but not to Sn. Thus in this case, n belongs to Sn but not to S. This proves that S is different from every Sn, since n belongs to one of the two sets Sn or S, but not to both of them.

We already know that the set of all finite sets of natural numbers is denumerable, and that is a subset of. It was proved for finite sets in the solution to Problem 1 and remember that both the empty set and its power set are finite, the first having zero elements, and the second having a single element, the empty set. So we will prove this theorem here only for infinite sets.

Given an infinite set A, consider any correspondence which pairs each element x of A with a subset of A, and call the subset of A that x is paired with Sx. Again define x to be ordinary if x does not belong to Sx. Of course it is. Given an enumeration a1, a2,. Since the set of all finite subsets of N is denumerable, then if the set of all infinite subsets of N were denumerable, the union of these two sets would be denumerable as we saw in the last problem but this union is , which is non-denumerable.

Therefore the set of all infinite subsets of N cannot be denumerable, thus must be non-denumerable. Let D be the union of the denumerable sets D1, D2,. We know that the set of fractions is denumerable, and we can enumerate the elements of D in the same order as we did the fractions, i. We eliminate the duplicates that may arise at each step due to the fact that the sets Di are not required to be mutually disjoint although each one has denumerably many mutually distinct elements by itself.

Now consider a denumerable set D. We will show that the set of all finite sequences of elements of D is denumerable. Each set Sk n is finite as shown above. We can enumerate all elements of Sk by starting with the elements of Sk 1 of which there is only one followed by the finitely many elements of Sk 2 in any order, and deleting repetitions , followed by those of Sk 3, and so forth.

Thus each Sk is denumerable. Since each of the sets S1, S2,. As examples, the sequence 1, 0, 1, 0, 1, 0,. The sequence 1, 0, 1, 1, 0, 1, 0, 0. Obviously distinct sequences are then paired with distinct sets of natural numbers, and for each set of natural numbers A, there is one and only one sequence that is paired with A, namely the sequence in which for all n the nth term of the sequence is 1 if n is in A, and is 0 if n is not in A. Thus the pairing is a correspondence between the set of all the sequences to the set of all sets of positive integers, which of course is the same size as the set of all sets of natural number, , since the set of positive integers is the same size as the set of natural numbers.

Now, consider an infinite set A. A is obviously non-empty, hence we can remove an element a1. What remains is infinite, hence we can remove another element a2.

We thus generate a denumerable sequence a1, a2,. Hidden in the above proof is a principle known as the Axiom of Choice, which we will unfortunately not be able to consider in this work.

Consider an infinite set A. To augment this pairing of a subset of A with another subset of A to obtain a pairing from all of A to a subset of A, we let each element of A other than those in D correspond with itself.

Obviously every unloved man is in Group 1 and every unloved woman is in Group 2. Therefore every person in Group 3 is loved. Furthermore: a For every man in Group 1, the woman he loves is also in Group 1, and every woman in Group 1 is loved by some man since she is not in Group 2 , and this man is also in Group 1. Thus, if all the men in Group 1 marry the women they love, these women are all in Group 1 and include all the women in Group 1. Therefore we have the choice of either marrying all the men in Group 3 to the women they love, or marrying all the women to the men they love.

Since B is a subset of A, then it is obviously the same size as a subset of A namely B itself , and since A is the same size as a subset of B, it follows that A is the same size as B by the Bernstein Schroeder Theorem. One such paradox is this: Consider the set S of all sets.

Its power set is a subset of S, because every element of is a set, hence an element of the set S, the set of all sets. Also, S is the same size as a subset of by a proof similar to that given as the solution to Problem 7 of Chapter 2, taking S for N. Next there is the famous Russell Paradox discovered independently by Zermelo. Call a set ordinary if it is not a member of itself, and extraordinary if it is a member of itself. Whether or not extraordinary sets exist, ordinary sets obviously do.

Let M be the set of all ordinary sets. Is M ordinary or not? Either way we get a contradiction: Suppose M is ordinary. Then M is in M, which contains all ordinary sets, but being in M makes M extraordinary by definition. Thus it is paradoxical to assume that M is ordinary. On the other hand, suppose M is extraordinary.

This means that M is a member of itself, i. But the only sets in the set M are ordinary sets. This is again a contradiction. Russell subsequently made a popular version of this paradox. A male barber of a certain town shaved all men of the town who did not shave themselves, and only such men. Thus if a man of the town did not shave himself, the barber would shave him, but if a man of the town shaved himself, the barber would not shave him.

If he shaved himself, then he shaved someone who shaved himself, which he was not supposed to do. Thus either way we get a contradiction. The solution to the barber paradox is really very simple! What is it? I must tell you a very amusing incident about the barber paradox. I told it to a friend of mine who has an excellent sense of humor. For example, call an adjective autological if it has the property it describes, and call it heterological if it does not. Either way we get a contradiction.

Then there is the Berry paradox. The above description uses only ten words! What is the solution to the Berry paradox? I would now like to tell you of an amusing paradox I recently thought of.

Let us go back to the book of infinitely many pages, upon each of which is listed a description of a set of natural numbers. We recall that the set of ordinary numbers numbers n that do not belong to the set listed on Page n — this set cannot be any one of the sets listed on any page.

If it is ordinary, then it belongs to the set of all ordinary numbers, which is the set listed on Page 13, which makes 13 extraordinary! If 13 is extraordinary, then by definition it is a member of the set listed on page 13, which is the set of all ordinary numbers, and therefore 13 must be ordinary!

Either way we get a contradiction! What is the solution to that paradox? We consider only 2-person games. Call a game normal if it has to terminate in a finite number of moves. For example, tic-tac-toe is obviously normal. Chess is a normal game, if played by tournament rules.

Now, here is hypergame: The first move of hypergame is to choose what normal game should be played. Suppose, for example, that you and I are playing hypergame and that I have the first move. Then I must declare what normal game should be played. I can choose any normal game I like, but I am not allowed to choose a game that is not normal. Then the second player makes the first move in the game chosen by the first player, and the two players play that normal game, which must end sometime.

That is all there is to the rules of hypergame. The problem is this: Is hypergame normal or not? First I will prove that hypergame must be normal. For the first player must choose some normal game. Thus hypergame must be normal. And so the game never terminates, which proves that hypergame is not normal. Thus hypergame is both normal and not normal. This is the paradox! What is the solution?

Again we consider an infinite set A and a correspondence between A and its power set that assigns to each element x of A a subset of A that we will denote Sx. Here is how! By a path we mean any finite or infinite sequence constructed as follows: We start with any element x of A. We go to the set Sx. If Sx is empty that is the end of the path. If Sx is not empty we take some element y of Sx and then go to the set Sy. If Sy is empty that is the end of the path.

If Sy is not empty we take some element z of Sy and so forth. Either we eventually hit the empty set, or the path goes on forever. We now define an element x of A to be normal if all paths starting with x are finite, i. Well, as Zwicker proved and the proof is not hard , the set M of all normal elements of A cannot be Sx for any x.

Prove this. Before discussing the proposed remedies, I would like to present an amusing and perhaps annoying! Now, to make life more complicated, we consider another book with infinitely many pages, but this time, on each page is written either a genuine description or a pseudo-description. Let S be the set of all numbers described by this description.

For any n, if the description on Page n is not genuine, then n automatically belongs to the set S. If the description on Page n is genuine, then n belongs to S if and only if n does not belong to the set described on Page n.

Thus the above description specifies for each n whether n belongs to the set S or not, hence the description must be genuine, right? Now supposed this description in on Page What happens then? How do you get out of that one? Two Systems of Set Theory Coming back to the set paradoxes, their proposed resolutions led to two important systems of set theory, the Theory of Types, and the system of Zermelo set theory, later expanded by Fraenkel.

In this theory, one starts with a set of individual elements which are not necessarily sets. These elements are classified as of type 0 type zero. Any set of elements of type 0 is called a set of type 1. Any set of sets of type 1 is called a set of type 2, and so forth. This principle is known as the abstraction principle, and it is precisely the principle which led to a contradiction, namely the Russell paradox, since it can be applied to the property of sets that are not members of themselves, and thus we get the existence of the set of all ordinary sets, which we have seen leads to a contradiction.

The sad thing is that only after Frege completed his monumental work on set theory did Russell discover and communicate to Frege the inconsistency of his system! Frege was totally crestfallen by this discovery and regarded his whole system as a total failure! In actuality, his pessimism was quite unjustified, since all his other axioms were quite sound and of great importance to mathematical systems of today. It was merely his abstraction principle that needed modification, and this modification was carried out by Zermelo, who replaced the abstraction principle by the following axiom: Z0 [Limited abstraction principle].

For any property P and any set S, there exists the set of all elements of S having property P. This modification does not appear to lead to any inconsistencies. There is a set that contains no elements at all. Before stating other Zermelo axioms, let me tell you a funny incident concerning Z1. As a graduate student, I had a course in set theory. When the professor introduced the axioms, the first was the existence of the empty set.

Of course he was right. Why does it follow? For any pair of sets x and y, there is a set that contains both x and y. Prove that for any sets x and y, there is a set whose elements are just x and y.

For one thing, we can get a model of the natural numbers. For zero, Zermelo took the empty set. Thus n is denoted by 0 enclosed in n pairs of brackets. Von Neumann defined the natural numbers in such a way that each natural number was the set of all lessor natural numbers.

Solutions to the Problems of Chapter 3 1. Suppose I told you that there is a man who is more than six feet tall and also less than six feet tall. How would you explain that? The obvious answer is that I must be either mistaken or lying! Thus the answer to the paradox is that there was no such barber.

The answer is that the notion of describable is not well-defined. Thus it is not a genuine description, rather it is what is called a pseudo-description. The curious and interesting this is that if those same words are not written on any page of the book, it is a genuine description! But writing it on some page of the book makes it a pseudo- description. If Hypergame were well-defined, we would have a contradiction; hence it is not well-defined.

Yes, given a set of games not containing Hypergame, one can well-define a Hypergame for the set, but one cannot well define Hypergame for a set that already contains Hypergame. Suppose that in a correspondence between A and , there is an element a of A such that Sa is the set of all normal elements. If it were, then there would only be one element in the path starting from a, namely the path consisting of a itself, a path that terminates immediately. So a would have to be normal.

But a cannot be normal, because that would put it into the empty set Sa, which contains no elements. So assuming Sa empty led to a contradiction. On the other hand, if Sa were nonempty, a would again have to be normal, since in any path starting with a the next term of the path must be a member a1 of Sa, and a1 must be normal since we are assuming that all elements of Sa are. Thus all paths starting with a1 will terminate, so that a itself must be normal.

Hence we can construct the infinite path, a, a, a,. Hence, since the set of normal elements is a well- defined subset of A, there can be no correspondence between A and the full power set of A,. The answer is that the very notion of a genuine description is not well-defined!

If we assume the existence of a set S, then by the limited abstraction principle, there exists the set of all elements of S having this property P, and this set is the empty set. Alternatively, this follows from the last problem, by taking x and y to be the same element. In contrast to this, we need the notion of an ordered pair x, y [note the use of parentheses instead of curly brackets], which consists of two elements x and y, but x is designated as the first element of the ordered pair, and y is designated as the second element.

Now the order is crucial; in general, the ordered pair x, y is distinct from y, x [in fact, they are the same only if x and y are the same element]. The situation is similar with three or more elements. Consider now a binary relation R, i. The relation R determines a unique set of ordered pairs, namely the set of all ordered pairs x, y such that x stands in the relation R to y.

When we write R x, y , we will mean that x stands in the relation R to y. In many modern treatments of set theory, one identifies a binary relation R with the set of all ordered pairs x, y that stand in the relation R, and we shall sometimes do that.

For a trinary relation R also called a relation of three arguments, or a relation of degree 3 , we write R x, y, z to mean that the ordered triple x, y, z stands in the relation R. Again, one sometimes identifies a trinary relation R with the set of all ordered triples that stand in the relation. Similar remarks apply to n-ary relations, i. An ordered triple x, y, z is also sometimes called a three-tuple, and x1, x2,. Functions A function f of one argument is a correspondence through which to every element x of a set S1 there corresponds a unique element denoted f x in a set S2.

In many cases, the sets S1and S2 are the same. One can picture a numerical function f as a computing machine in which one feeds in a number x as input, and out comes a number denoted f x.

In modern treatments of set theory, one defines a function f of one argument to be a single-valued relation, i. More generally, for any positive n, a function f of n arguments is a correspondence which assigns to each n-tuple x1, x2,. For functions of two arguments, one sometimes writes xfy. Mathematical Induction Before explaining this important principle, here is a little story. A certain man was in quest of immortality. He read all the occult books on the subject, but none of them seemed to offer any practical solution.

Then he heard of a great sage in the East who knew the secret of immortality. Never make a false one. But your solution is not practical! No, your solution is not practical! I deal only in theory. Let us visit a land in which every inhabitant is one of two types, type T or type F.

Those of type T make only true statements; everything they say is true. Those of type F make only false statements; everything they say is false. Which type was the inhabitant who said that? Is he of type T or F? Both the above story and the above puzzle illustrate the principle of mathematical induction. As still another illustration, suppose I tell you that on a certain planet it is raining today, and that on any day on which it rains, it rains the next day as well — in other words, it never rains on any day without raining the next day as well.

Still another illustration, with an additional clever and amusing twist: Imagine that we are all immortal and that we live in the good old days in which the milk man would deliver milk to the house and the house wife would leave a note to the milk man telling him what to do.

Your order was to the effect that I should never leave milk on any day without leaving milk the next day as well. I never left milk at all, nor did you ever tell me I should! Her note was insufficient. That would indeed have guaranteed permanent delivery! When I told the above story to my friend the computer scientist Dr. Alan Tritter, he came up with the following delightful alternative, which illustrates what might be called the Turing Machine, or the recursive approach.

Complete Induction A variant of the principle of mathematical induction is the principle of complete mathematical induction, which is this: Suppose a property P of natural numbers is such that for every natural number n, if P holds for all natural numbers less than n, then P holds for n. Conclusion: P holds for all natural numbers. The least number principle is that every non-empty set of natural numbers contains a least number.

This principle is equivalent to the principle of mathematical induction in the sense that either one is a logical consequence of the other. The Principle of Finite Descent Suppose a property P is such that for any natural number n, if P holds for n, then P also holds for some natural number less than n. This is known as the principle of finite descent.

Show that the principle of finite decent is equivalent to the principle of mathematical induction. Limited Mathematical Induction Problem 5. Suppose that P is a property and n is a number such that the following two conditions hold: 1 P holds for 0. Using mathematical induction, prove that P holds for all numbers less than or equal to n.

A Cute Paradox Using mathematical induction, I will prove to you that given any finite non-empty set of billiard balls, they must all be of the same color! Here is the proof: For any positive n, let P n be the property that for every set of n billiard balls, they are all of the same color.

So suppose n is a number for which P holds. The balls 1 through n all have the same color by the assumption that P holds for n, i. What is the error in the above proof? Exercise 1. Prove that Q holds for all finite sets of natural numbers. Then use mathematical induction to show that P x holds for every natural number x. We earlier stated without proof that no finite set can be put into a correspondence with any of its proper subsets. Prove this by mathematical induction on the number of elements of the set.

The next two problems are not necessary for the rest of this book but are hopefully of independent interest. Prove that R x, y holds for every x and y. Prove by induction that every y is special. Prove that R x, y holds for all x and y. Then show by induction that every number is right normal hence also left normal. You are required to play the following game: There is an infinite supply of pool balls available, each bearing a positive integer.

For each positive n, there are infinitely many balls numbered n. In a certain box there are finitely many of these balls, but the box is expandable and has infinite capacity. If the box ever gets empty, you get executed! Is there a strategy by which you never get executed, or is execution inevitable sooner or later? A tree consists of an element a0 called the origin, which is connected to finitely many possibly zero or denumerably many elements called the successors of a0, each of which in turn has finitely or denumerably many successors, and so forth.

We say that an element x is a predecessor of y if y is a successor of x. The origin has no predecessor, and every other element of the tree has one and only one predecessor. The elements of the tree are called points. An element with no successor is called an end point, and a junction point otherwise. The origin of the tree is said to be of level 0, its successors are of level 1, the successors of these successors are of level 2, and so forth. Trees are displayed with the origin at the top and the tree grows downward.

By a path of the tree is meant a finite or denumerable sequence of points, beginning with the origin, and such that each of the other terms of the sequence is a successor of the preceding term. Since a point cannot have more than one predecessor, it follows that for any point x, there is one and only one path from x up to the origin a0 which implies that there is also only one path down from the origin a0 to the point x. By the length of a finite path is meant the level of its endpoint.

By the descendants of a point x are meant the successors of x, together with the successors of the successors, together with the successors of these, and so forth. Thus y is a descendant of x if and only if there is a path to y that goes through x. Suppose that we are given that for any positive integer n, there is at least one path of length n and thus every level is hit by at least one path. Does it necessarily follow that there must be at least one infinite path?

Finitely Generated Trees Let us note that if a point x of a tree has only finitely many successors x1,. A tree is said to be finitely generated if each point has only finitely many successors though it might have infinitely many descendants.

Suppose a tree is finitely generated. Now, in the solution to Problem 9, we had an example of a tree such that for each positive n, there is a path of length n, yet there was no infinite path.

Of course that tree was not finitely generated, since the origin had infinitely many successors. A finitely generated tree with infinitely many points must have an infinite path. The following result of the Dutch mathematician L. The Fan Theorem. If a tree is finitely generated and if all paths are finite, then the whole tree is finite.

Prove the Fan Theorem. There is a weaker system of logic known as intuitionist logic, of which Brouwer was a major representative, in which not all implications can be shown to be equivalent to their contrapositives.

Another proof of the Fan Theorem will be given shortly, one which is intuitionistically acceptable. Unfortunately, we will not be able to say more about intuitionist logic in this work. Here is a cute application of the Fan Theorem: Problem Can the reader find it?

Generalized Induction The principle of complete mathematical induction, which is a result about the natural numbers, has an important generalization to arbitrary sets, even non-denumerable ones.

In set theory the components of a set are its elements. In some applications of number theory, the components of a natural number n are the numbers less than n. In mathematical logic, the components of a formula are its so-called sub-formulas.

Thus for any finite sequence x1, x2. Generalized Induction Principle Given a component relation C x, y on a set A, a property P of elements of A is said to be inductive if for every element x of A, if P holds for all components of x, then P holds for x.

It is to be understood that if x has no components at all then P automatically holds for x, since P does hold for all components of x, of which there are none. Thus if C x, y obeys the generalized induction principle, then for any property P, to show that P holds for all elements of A, it suffices to show that for all elements x of A, if P holds for all components of x, then P holds for x as well. The following result is basic: Generalized Induction Theorem.

A sufficient condition for a component relation C x, y on a set A to obey the generalized induction principle is that all descending chains be finite. Thus if all descending chains are finite, then to show that a given property P holds for all elements of A, it suffices to show that for every element x of A, if P holds for all components of x, then P holds for x as well. Prove the above theorem. If all branches of the tree are finite, then to show that a given property P holds for all points of the tree, it suffices to show that for any point x, P holds for x provided P holds for all successors of x.

This is a special case of the Generalized Induction Theorem by taking the components of x to be the successors of x. The converse of the Generalized Induction Theorem also holds, that is, if the component relation obeys the generalized induction principle, then all descending chains must be finite.

Each of these two is in turn equivalent to the interesting third condition that we will now consider. Again we consider a relation C x, y on a set A. For any subset S of A, an element x of S is called an initial element of S with respect to the relation C x, y understood if x has no components in S. The relation C x, y is called well-founded if every non-empty subset of A has an initial element.

And now for a lovely result: Theorem. For any component relation C x, y on a set A, the following three conditions are all equivalent: 1 The generalized induction principle holds. Prove the above. Compactness We now turn to another principle that is quite useful in mathematical logic and set theory. Consider a universe V in which there are denumerably many people. The people 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, if we add one or more people to the set C the resulting set is not a club. Would that change the answer to a? Again we consider the case that V contains denumerably many inhabitants, and we assume that we are given the additional information that a set S of people is a club if and only if every finite subset of S is a club.

Then it does follow that if there is at least one club, there is a maximal club; better yet, every club is a subset of a maximal club. The problem is to prove this. Here are the key steps: 1 Show that under the given conditions, every subset of a club is a club.

Now, given any club C that exists, define the following infinite sequence C0, C1, C2,. Then consider x1. It is obvious that each Cn is a club. Consider now an arbitrary denumerable set A and a property P of subsets of A. The property P is called compact if it is the case that for every subset S of A, P holds for S if and only if P holds for all finite subsets of S.

In the last problem, we were given a denumerable universe V and a collection of subsets called clubs, and we were given that a subset S of V is a club if and only if all finite subsets of S are clubs, in other words, we were given that the property of being a club is compact, from which we could deduce that any club is a subset of a maximal club.

Now, there is nothing special about clubs that makes our argument go through. The same reasoning yields the following: Denumerable Compactness Theorem. For any compact property P of subsets of a denumerable set A, any subset S of A having property P is a subset of a maximal set that has property P.

Note: The above result actually holds even if A is a non-denumerable set whether finite or infinite , but the more advanced result for infinite non-denumerable sets is not needed for anything in this book. The reader should also be able to see that the proof just given can be modified slightly to work when V is finite. Discussion In mathematical logic, we deal with a denumerable set of symbolic sentences, some subsets of which are deemed inconsistent.

Now, since the mathematical logic we will be studying in this book considers only proofs employing a finite number of sentences, any proof that a given set of sentences is inconsistent uses only finitely many members of the set of those sentences. Thus a denumerable set S is defined to be consistent if and only if all finite subsets of S are consistent. In other words, consistency is a compact property, a fact that will prove very important later on!

Solutions to the Problems of Chapter 4 1. If he were of type T, then he really would have said that sometime before, as he said, and when he said that before, he was still of type T, hence he had said it sometime before that, hence sometime before even that, and so forth.

Thus unless he lived infinitely far back into the past, he cannot be of type T. Thus he is of type F. We consider a property P such that for every number n, if P holds for all numbers less than n, then P also holds for n. We are to show that P holds for all numbers.

Now, there are no natural numbers less than 0, so that the set of natural numbers less than zero is empty. Thus, by our assumption of the induction premise of complete mathematical induction i. We consider a property Q stronger than P in the sense that any number having property Q also has property P , and we use mathematical induction on the property Q, i.

In fact we define Q n to mean that P holds for n and all numbers less than n. Of course Q n implies P n. Now suppose n is a number for which Q holds. We are then to infer the principle of mathematical induction.

We are to show that P holds for every n. To do this, it suffices to show that for any number n, if P holds for all numbers less than n. Well, suppose P holds for all numbers less than n.