Proposition 11 Let A be a set. The following are equivalent.

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1 Countable Sets Definition 6 Let A be a set It is countably infinite if N A and is countable if it is finite or countably infinite If it is not countable it is uncountable Note that if A B then A is countable iff B is countable Proposition 10 Every infinite subset of N is countably infinite Proof 1 Let A N, A not finite Define f : N A as follows: f(0) = mina, f(n + 1) = min(a\{f(0),f(1),,f(n)}) where for = X N, minx denotes the least element of X f is defined for all n N since A is not finite, it is clearly injective and also surjective since any a A must be f(n) for some n a Corollary 1 Every subset of N is countable Proposition 11 Let A be a set The following are equivalent (i) A is countable (ii) A N (iii) A = or N A Proof (i) = (iii) Assume A is countable and not equal to Then there is f : N n A for some n N, n > 0 or f : N A In the latter case clearly f : N A and in the former case we can obtain h : N A by choosing a point a A and defining { f(m) if m < n, h(m) = a if m n (iii) = (ii) If A is empty then trivially A N If f : N A then h : A N defined by h(a) = min{n f(n) = a} is an injection, so A N (ii) = (i) If A N then A B for some B N B is countable by Corollary 1 and hence A is countable Proposition 12 Let A B (i) If B is countable then so is A (ii) If A is uncountable then so is B 1 This proof (like some other ones in this section) gives the idea of why the result should hold A rigorous proof will follow from from a theorem on definition by recursion proved later on 11

2 Proof (i) Using Propositions 11 and 4(ii), if B is countable then B N and hence since A B, we also have A N, so A is countable (ii) follows from (i) Note It follows that any subset of a countable set is countable (by Propositions 4(i) and 12) Proposition 13 Let A B (i) If A is countable then so is B (ii) If B is uncountable then so is A Proof Exercise By a list we mean a sequence of objects x 0, x 1,x 2,, or x 0, x 1,x 2,,x n We write {x 0,x 1,x 2,} or {x 0,x 1, x 2, x n } for the sets containing (just) the objects in these lists An empty list does not feature any object Proposition 14 Let A be a set A is countable iff its elements can be arranged in a finite or infinite list, perhaps with repetitions If the set is nonempty we can assume that the list is infinite (even if the set is finite) Proof This follows from Proposition 11 For the empty set we have the empty list, and otherwise countability of A is equivalent to the existence of a surjective function f : N A, which in turn is equivalent to the existence of a list featuring each element of A at least once (via the correspondence of f and the list f(0),f(1),f(2), ) Proposition 15 If A and B are countable sets then A B and A B are countable Proof Assume A and B are countable If one of them is empty then A B is equal to the other one, and A B is empty, so the proposition holds If they are not empty then by proposition 14 A = {a 0, a 1,a 2,}, B = {b 0,b 1,b 2,} so the elements of A B can be arranged in a list: A B = {a 0, b 0 a 1,b 1,a 2,b 2,} and hence A B is countable by Proposition 14 again 12

3 The elements of A B can be arranged in an array: a 0,b 0 a 0, b 1 a 0, b 2 a 0,b 3 a 0,b 4 a 1,b 0 a 1, b 1 a 1, b 2 a 1,b 3 a 1,b 4 a 2,b 0 a 2, b 1 a 2, b 2 a 2,b 3 a 2,b 4 a 3,b 0 a 3, b 1 a 3, b 2 a 3,b 3 a 3,b 4 and subsequently, following the diagonals, listed: {}}{{}}{{}}{ A B = { a 0,b 0, a 0,b 1, a 1,b 0, a 0,b 2, a 1,b 1, a 2,b 0, a 0,b 3, a 1,b 2, } By Proposition 14, A B is countable Note It follows by induction that if A k for k {1, n} are countable then their union n k=1 A k is countable and their Cartesian product A 1 A 2 A n is countable In what follows we will often use the correspondence of countable sets and lists (Proposition 14) without further mention Proposition 16 Let A n for n N be countable sets Then countable set n=1 A n is a Proof 1 If some of the A n are empty then we can just leave them out If there are only finitely many non-empty sets left then the result follows by the above remark Otherwise assume A n are already the non-empty ones and let A 0 = {a 00,a 01,a 02,a 03,} A 1 = {a 10,a 11,a 12,a 13,} A 2 = {a 20,a 21,a 22,a 23,} Then following the diagonals again A n = {a 00, a 01, a 10, a 02,a 11,a 20,} n=1 Note that the above proof requires us to choose a list for each A n, simultaneously 1 This is the only proof in the naive part of the course that employs what is now known as the Axiom of Choice Cantor often used it 13

4 Theorem 1 The set N N and the sets Z of integers and Q of rational numbers are all countably infinite sets Proof For N N countability is an immediate consequence of Proposition 15 Z is equal to Z N, where Z are the negative integers We have Z N via the injection f(k) = k so Z is countable by Proposition 11 and consequently so is Z (by Proposition 15) Finally, the function f : Z N Q: { 0 if n = 0 f(m,n) = otherwise m n is a surjection so since Z is countable by the above and Z N is countable by Proposition 15, Q is countable by Proposition 13 We have N N N, N Z and N Q (via the injections f 1 (n) = n, 0, f 2 (n) = n, f 3 (n) = n respectively), so none of the sets is finite by Propositions 7 and 8 Hence they are all countably infinite Note Cantor knew the above facts some years before 1873 By that time, he has been trying for some years to prove also that the set of real numbers is countably infinite Some correspondence with Richard Dedekind helped him to switch focus and he then quickly proved that R is not countable Later, other sets were shown to be uncountable Theorem 2 (i) The set {0,1} N is uncountable (ii) The set R of real numbers is uncountable Proof (i) Clearly {0, 1} N is not empty since eg the function f : N {0, 1} defined by f(n) = 0 for all n N belongs to it, so assuming that {0, 1} N is countable means that we can arrange its elements in a list: {0, 1} N = {f 0,f 1,f 2,f 3,} (using Proposition 14) Define a function g : N {0, 1} as follows: g(n) = { 0 if fn (n) = 1, 1 if f n (n) = 0 Then g is a function from N to {0,1} but g differs from each f n, namely in the value it assigns to n: g(n) f n (n) This is a contradiction because f 0,f 1,f 2,f 3 is a list of all elements of {0, 1} N We must conclude that {0,1} N is not countable (ii) The above is an example of a diagonal argument due to Cantor (1891) To show that the set R is uncountable, we give Cantor s original proof of it (1874) However, see Example 25 for the better known diagonal argument proof 14

5 Lemma 1 Suppose that a set A contains at least 2 elements, it is linearly densely ordered 1 and has the following property: For any partition of A into two nonempty sets X and Y such that every element of X is less than every element of Y, there is c A so that every point less than c is in X and every point greater than c is in Y Then A is not countable The lemma is proved below, so since the set R satisfies the above conditions, it is uncountable Proof of Lemma 1 Assume to the contrary that A is countable, A = {a 0,a 1,a 2, a 3, } (by Proposition 14) Define x 0,x 1,x 2, and y 0,y 1,y 2, as follows: Choose x 0,y 0 A such that x 0 < y 0 Choose x n+1 A to be a m where m = min{k x n < a k < y n } Choose y n+1 A to be a m where m = min{k x n+1 < a k < y n } We have x 0 < x 1 < x 2 < < y 2 < y 1 < y 0 Let X = {a A a < x n for some n}, Y = A\X and let c A be the element guaranteed to exist by the property assumed in the lemma so x 0 < x 1 < x 2 < < c < < y 2 < y 1 < y 0 But c = a n for some n so by stage n, c must have been chosen as one of the x i or y i, contradiction Note The next goal that Cantor spent years trying to reach was showing that there is no bijection between, say, the line segment [0, 1) and the square [0, 1) [0, 1) From the geometric point of view this should be so, and many thought that it did not even require a proof, because of the sets having different dimensions A bijection between a line segment and a square seemed to contradict basic geometrical intuition Eventually, apparently incensed by the dismissive attitude of some of his fellow mathematicians towards the need for a proof, Cantor showed that such a bijection does exist! He was himself surprised by it and famously wrote to Dedekind I see it, but I do not believe it Cantor s original proof contained an error which he repaired at the cost of making it quite complicated However, his idea can be made to work easily with the help of the Schröder-Bernstein Theorem which we will prove in the next section 1 A is linearly ordered by if for all a, b, c A we have a b b a and (a b&b a) a = b and (a b&b c) a c; it is moreover densely ordered if between any two elements there is another (that is, for all a, b A such that a < b there is c A such that a < c < b) 15

6 Cardinals Recall that with each finite set A is associated a natural number A, called the size of the set A, such that for all finite sets A and B A = B A B, A B A B, The notion of the cardinal number of a set is intended to extend this idea of the size of a finite set to all sets, finite or infinite We will not say what cardinal numbers are until later and instead we will just use the term to describe that which is shared by equinumerous sets, and which generalizes the notion of size of finite sets Definition 7 Sets A and B have the same cardinal number iff A B We write A for the cardinal number of A If A is finite then the cardinal number of A is its size A as defined before, ie the number of its elements Consequently, all the natural numbers 0, 1,2, are cardinal numbers, the finite cardinal numbers We give two examples of infinite cardinal numbers Definition 8 ℵ 0 = N and c = R All countably infinite sets have the same cardinal number ℵ 0 as, by definition they are all equinumerous with N Also, as R is uncountable and so not equinumerous with N, we know that c ℵ 0 There is a natural ordering relation between cardinal numbers that extends the ordering on the natural numbers By Proposition 4 it can be easily seen that if A,B, C, D are sets such that A C and B D (ie A and C have the same cardinal number and B and D have the same cardinal number), then A B C D Hence we can define Definition 9 For cardinal numbers n,m let n m if A B for some/any sets A,B such that n = A and m = B Note that restricted to N, indeed agrees with the usual ordering on N The following reflexivity and transitivity properties of the relation on cardinal numbers follow from the corresponding properties for the relation on sets proved earlier Proposition 17 For cardinal numbers n,m, k 1 n n, 2 if n m and m k then n k 16

7 As is usual, we write n < m when n m & n m By the above we have ℵ 0 < c Diversion Consider sets of one-dimensional geometric shapes placed in a plane, circles or crosses (a cross consists of two lines of the same length crossing at right angle in the middle) There are either only circles or only crosses, of arbitrary sizes, and they do not intersect Show that for one of the shapes the set must be countable and for the other shape describe such a set with cardinality c The Schröder-Bernstein Theorem We state two further properties that we might expect of the relation on cardinal numbers If n m and m n then n = m, either n m or m n The first of these is a consequence of the Schröder-Bernstein theorem stated and proved below The second was often assumed by Cantor as obviously true; a discussion of it will have to wait until later, and we will see that it is in fact equivalent to the axiom of choice mentioned earlier in connection with Proposition 16 Theorem 3 (The Schröder-Bernstein Theorem) For sets A,B, if A B and B A then A B Proof Let f : A B and g : B A We need to define a bijection h : A B If X A then let fx = {f(x) x X} Similarily define gy for Y B Define the family of sets {A n } n N by the following iteration { A0 = A\gB A n+1 = g(fa n ) for n N Also let Ā = n N A n Note that A\Ā A\A 0 gb It follows that we may define h : A B as follows { f(x) if x Ā, h(x) = g 1 (x) if x A\Ā It only remains to show that h is a bijection; ie is both injective and surjective h is injective: Let x,x A such that h(x) = h(x ) We must show that x = x If x Ā and x Ā then f(x) = g 1 (x ) so that x = g(f(x)) As x Ā there is n N such that x A n But then x A n+1 Ā contradicting 17

8 that x Ā So we cannot have x Ā and x Ā Similarly we cannot have x Ā and x Ā So either x,x Ā or x,x Ā In the first case f(x) = f(x ) so that x = x as f is injective and in the second case g 1 (x) = g 1 (x ) so that x = x as g 1 is injective So x = x in either case h is surjective: Let y B We must find x A such that y = h(x) Let x = g(y) gb If x Ā: then h(x ) = g 1 (x ) = y so that we can let x = x If x Ā: then x A n for some n N As x gb, x A 0 So n > 0 and so g(y) = x = g(f(x)) for some x A n 1 Ā As g is injective y = f(x) = h(x) In either case we are done Note: Each x A determines a finite or infinite sequence x 0, x 1, of elements of A as follows { x0 = x, x n+1 = (g f) 1 (x n ) if this is defined The set Ā is the set of those x A such that the sequence above is finite and ends with x n A\gB Proposition 18 Let X R be such that X contains some open interval (a,b) (with a < b) Then R X Proof First note that the open interval ( 1, 1) and R are equinumerous, as witnessed for example by the bijection f : ( 1,1) R defined by f(x) = tan ( π 2 x), and that the intervals (a,b) and ( 1, 1) are equinumerous (since for example h : ( 1, 1) (a, b) defined by h(x) = a+b 2 +xb a 2 is a bijection) Hence the open interval (a,b) is equinumerous with R and so R X Furthermore, X R so also X R It follows by the Schröder Bernstein theorem that X R Corollary 2 Any non-trivial real interval is equinumerous with R Proposition 19 R P ow(n) Proof Using the above Corollary, it suffices to prove that the interval [0, 1) is equinumerous with P ow(n) We use the following property of real numbers: Any real number corresponds to a decimal expression, and conversely The correspondence is one to one if we add the condition that the decimal does not 18

9 end with 999 Aiming to use the Schröder Bernstein theorem, define f : Pow(N) [0, 1) as follows: Given X N, let { 1 if i X a i = 0 if i / X and let f(x) = 0a 0 a 1 a 2 a 3 Then f is an injection so Pow(N) [0,1) Now define h : [0, 1) Pow(N) as follows: for x [0, 1) we write x = 0b 0 b 1 b 2 b 3 as above and define h(x) = {10 n b n n N} It is easily checked that h is an injection, so [0, 1) Pow(N) and the result follows Proposition 20 The interval [0,1) R and the the square [0, 1) [0, 1) R R are equinumerous Proof We write [0, 1) 2 for [0, 1) [0, 1) Using the decimal representation of real numbers as in the proof of Proposition 19, define e : [0,1) 2 [0, 1) by e(0x 0 x 1 x 2 x 3, 0y 0 y 1 y 2 y 3 ) = 0x 0 y 0 x 1 y 1 x 2 y 2 x 3 y 3 Then e is an injection 1 since if 0x 0 x 1 x 2 x 3, 0y 0 y 1 y 2 y 3 are decimals not ending with 999 then so is 0x 0 y 0 x 1 y 1 x 2 y 2 x 3 y 3, and as this representation is unique, there cannot be another pair mapped onto the same number Hence [0, 1) 2 [0,1) and since clearly also [0,1) [0,1) 2, we have [0,1) [0,1) 2 by the Schröder Bernstein Theorem Diversion Show that if A and B are two bounded subsets of the plane (R 2 ) each with non-empty interior then there is a partition A 1,A 2 of A and a partition B 1,B 2 of B and affine 2 mappings f and g such that f : A 1 B 1 and g : A 2 B 2 Cantor s Theorem We now show that there are lots of infinite cardinal numbers Theorem 4 (Cantor s Theorem) For any set A, A < Pow(A) Proof Let A be a set We need to show that A Pow(A) but A Pow(A) (not A P ow(a)), so A P ow(a) but A = P ow(a) The function h : A Pow(A) defined by h(a) = {a} for a A 1 Note that it is not onto since eg is not in the range Claiming that e is a bijection was the error mentioned earlier, made by Cantor in his first attempt at proving the proposition 2 An affine mapping from R 2 to R 2 is a linear mapping followed by a translation 19

10 is an injection, so A Pow(A) does hold Assume A Pow(A) and let f : A Pow(A) Define X = {a A a / f(a)} Then X A, that is X Pow(A) so since f is a bijection, we must have X = f(b) for some b A If b X then since X = f(b) we have b f(b) and by definition of X, b / X, contradiction Hence we must have b / X, that is, b / f(b) but that means b X, contradiction again Hence there can be no f : A Pow(A), as required By Cantor s Theorem the strictly increasing sequence of cardinal numbers can be continued 0 < 1 < 2 < < ℵ 0 < c c = R < Pow(R) < Pow(Pow(R)) < The next result allows us to obtain even larger cardinal numbers Notation: For a set S we write S for its union, the set of all elements of its elements: S = {x x y for some y S} Theorem 5 Let S be a set of sets such that for every X S there is a set Y S such that X < Y Then for every set X S, X < S Proof Exercise So if we let S = {R,Pow(R),Pow(Pow(R)),} then S is a cardinal number strictly larger than the previous cardinal numbers Clearly we can continue to get larger cardinal numbers by repeatedly applying Cantor s Theorem to S etc 20