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Countable Sets

Definition 6 Let A be a . 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 of N is countably infinite.

Proof 1 Let A ⊆ N, A not finite. Define f : N → A as follows:

f(0) = min A, f (n + 1) = min( A\{ f(0) , f (1) ,...,f (n)}) where for ∅ 6= X ⊆ N, min X denotes the least 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 ¹ N.

(iii) A = ∅ or N º∗ A.

Proof (i) = ⇒ (iii) Assume A is countable and not equal to ∅. Then there is f : Nn ∼ A for some n ∈ N, n > 0 or f : N ∼ A. In the latter case clearly f : N º∗ A and in the former case we can obtain h : N º∗ A by choosing a point a ∈ A and defining f(m) if m < n, h(m) = ½ a if m ≥ n.

(iii) = ⇒ (ii) If A is empty then trivially A ¹ N. If f : N º∗ A then h : A → N defined by h(a) = min {n | f(n) = a} is an injection, so A ¹ N. (ii) = ⇒ (i) If A ¹ N then A ∼ B for some B ⊆ N. B is countable by Corollary 1 and hence A is countable.

Proposition 12 Let A ¹ B.

(i) If B is countable then so is A.

(ii) If A is uncountable then so is B.

1This 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 on definition by proved later on.

11 Proof (i) Using 11 and 4(ii), if B is countable then B ¹ N and hence since A ¹ B, we also have A ¹ N, so A is countable. (ii) follows from (i).

Note It follows that any subset of a is countable (by Propositions 4(i) and 12).

Proposition 13 Let A º∗ 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 of objects x0, x 1, x 2,... , or x0, x 1, x 2,...,x n. We write {x0, x 1, x 2,... } or {x0, 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 we have the empty list, and otherwise countability of A is equivalent to the existence of a surjective f : N º∗ 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 = {a0, a 1, a 2,... }, B = {b0, b 1, b 2,... } so the elements of A ∪ B can be arranged in a list:

A ∪ B = {a0, b 0.a 1, b 1, a 2, b 2,... } and hence A ∪ B is countable by Proposition 14 again.

12 The elements of A × B can be arranged in an array:

ha0, b 0i h a0, b 1i h a0, b 2i h a0, b 3i h a0, b 4i . . .

ha1, b 0i h a1, b 1i h a1, b 2i h a1, b 3i h a1, b 4i . . .

ha2, b 0i h a2, b 1i h a2, b 2i h a2, b 3i h a2, b 4i . . .

ha3, b 0i h a3, b 1i h a3, b 2i h a3, b 3i h a3, b 4i . . .

...... and subsequently, following the diagonals, listed:

A × B = {ha0, b 0i, ha0, b 1i, ha1, b 0i, ha0, b 2i, ha1, b 1i, ha2, b 0i, ha0, b 3i, ha1, b 2i,... } z }| { z }| { z }| { By Proposition 14, A × B is countable. ¥

Note It follows by induction that if Ak for k ∈ { 1, . . . n } are countable then n their k=1 Ak is countable and their A1 ×A2 ×. . . ×An is countable.S In what follows we will often use the correspondence of countable sets and lists (Proposition 14) without further mention.

∞ Proposition 16 Let An for n ∈ N be countable sets. Then n=1 An is a countable set. S

1 Proof If some of the An 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 An are already the non-empty ones and let

A0 = {a00 , a 01 , a 02 , a 03 ,... }

A1 = {a10 , a 11 , a 12 , a 13 ,... }

A2 = {a20 , a 21 , a 22 , a 23 ,... } . .

Then following the diagonals again ∞

An = {a00 , 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 An, simultaneously.

1This is the only proof in the naive part of the course that employs what is now known as the of Choice. Cantor often used it.

13 Theorem 1 The set N×N and the sets Z of and Q of rational 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− ¹ 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 ) = m ½ n otherwise. 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 ¹ N × N, N ¹ Z and N ¹ Q (via the injections f1(n) = hn, 0i, f2(n) = n, f3(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 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 e.g 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 count- N able means that we can arrange its elements in a list: {0, 1} = {f0, f 1, f 2, f 3,... } (using Proposition 14). Define a function g : N → { 0, 1} as follows:

0 if fn(n) = 1 , g(n) = ½ 1 if fn(n) = 0 .

Then g is a function from N to {0, 1} but g differs from each fn, namely in the value it assigns to n: g(n) 6= fn(n). This is a contradiction because N N f0, f 1, f 2, f 3 . . . is a list of all elements of {0, 1} . We must conclude that {0, 1} is not countable. (ii) The above is an example of a diagonal due to Cantor (1891). To show that the set R is uncountable, we give Cantor’s original proof of it (1874). However, see Example 2.5 for the better known diagonal argument proof.

14 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 = {a0, a 1, a 2, a 3,... } (by Proposition 14). Define x0, x 1, x 2,... and y0, y 1, y 2,... as follows:

Choose x0, y 0 ∈ A such that x0 < y 0. Choose xn+1 ∈ A to be am where m = min {k | xn < a k < y n}. Choose yn+1 ∈ A to be am where m = min {k | xn+1 < a k < y n}.

We have x0 < x 1 < x 2 <......

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 x0 < x 1 < x 2 <...

Note. The next goal that Cantor spent years trying to reach was showing that there is no 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 differ- ent 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 Schr¨oder-BernsteinTheorem which we will prove in the next section.

1A 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

15 Cardinals

Recall that with each finite set A is associated a natural |A|, called the size of the set A, such that for all finite sets A and B

|A| = |B| ⇐⇒ A ∼ B, |A| ≤ | B| ⇐⇒ A ¹ B,

The notion of the 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, i.e. 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 ℵ0 = |N| and c = |R|.

All countably infinite sets have the same cardinal number ℵ0 as, by definition they are all equinumerous with N. Also, as R is uncountable and so not equinu- merous with N, we know that c 6= ℵ0. 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 (i.e. A and C have the same cardinal number and B and D have the same cardinal number), then A ¹ B ⇐⇒ C ¹ D. Hence we can define

Definition 9 For cardinal numbers n, m let n ≤ m if A ¹ 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 ¹ 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 As is usual, we write n < m when n ≤ m & n 6= m. By the above we have ℵ0 < 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 c.

The Schr¨oder-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 Schr¨oder-Bernsteintheorem 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 mentioned earlier in connection with Proposition 16. Theorem 3 (The Schr¨oder-Bernstein Theorem) For sets A, B , if A ¹ B and B ¹ A then A ∼ B. Proof Let f : A ¹ B and g : B ¹ 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 {An}n∈N by the following iteration.

A0 = A\gB ½ An+1 = g(fA n) for n ∈ N ¯ ¯ Also let A = n∈N An. Note that A\A ⊆ A\A0 ⊆ gB . It follows that we may define h : A→SB as follows. f(x) if x ∈ A,¯ h(x) = − ½ g 1(x) if x ∈ A\A.¯

It only remains to show that h is a bijection; i.e. 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 ∈ A¯ and x′ 6∈ A¯ then f(x) = g−1(x′) so that x′ = g(f(x)). As x ∈ A¯ ′ there is n ∈ N such that x ∈ An. But then x ∈ An+1 ⊆ A¯ contradicting

17 that x′ 6∈ A¯. So we cannot have x ∈ A¯ and x′ 6∈ A¯. Similarly we cannot have x′ ∈ A¯ and x 6∈ A¯. So either x, x ′ ∈ A¯ or x, x ′ 6∈ A¯. 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′ 6∈ A¯: then h(x′) = g−1(x′) = y so that we can let x = x′. ′ ′ ′ ′ If x ∈ A¯: then x ∈ An for some n ∈ N. As x ∈ gB , x 6∈ A0. So n > 0 ′ and so g(y) = x = g(f(x)) for some x ∈ An−1 ⊆ A¯. 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 x0, x 1,... of elements of A as follows.

x0 = x, −1 ½ xn+1 = ( g ◦ f) (xn) if this is defined.

The set A¯ is the set of those x ∈ A such that the sequence above is finite and ends with xn ∈ 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) = π tan 2 x , and that the intervals ( a, b ) and ( −1, 1) are equinumerous (since for a+b b−a example¡ ¢h : ( −1, 1) → (a, b ) defined by h(x) = 2 + x 2 is a bijection). Hence the open interval ( a, b ) is equinumerous with R and so R ¹ X. Furthermore, X ⊆ R so also X ¹ R. It follows by the Schr¨oderBernstein 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 corresponds to a decimal expression, and conversely. The correspondence is one to one if we add the condition that the decimal does not

18 end with 999 . . . . Aiming to use the Schr¨oderBernstein theorem, define f : P ow (N) → [0 , 1) as follows: Given X ⊆ N, let

1 if i ∈ X ai = ½ 0 if i∈ / X and let f(X) = 0 .a 0a1a2a3 . . . . Then f is an injection so P ow (N) ¹ [0 , 1). Now define h : [0 , 1) → P ow (N) as follows: for x ∈ [0 , 1) we write x = 0 .b 0b1b2b3 . . . n as above and define h(x) = {10 bn | n ∈ N}. It is easily checked that h is an injection, so [0 , 1) ¹ P ow (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(0 .x 0x1x2x3 ..., 0.y 0y1y2y3 . . . ) = 0 .x 0y0x1y1x2y2x3y3 . . .

1 Then e is an injection since if 0 .x 0x1x2x3 ..., 0.y 0y1y2y3 . . . are decimals not ending with 999 . . . then so is 0 .x 0y0x1y1x2y2x3y3 . . . , and as this representation is unique, there cannot be another pair mapped onto the same number. Hence [0 , 1) 2 ¹ [0 , 1) and since clearly also [0 , 1) ¹ [0 , 1) 2, we have [0 , 1) ∼ [0 , 1) 2 by the Schr¨oderBernstein Theorem.

Diversion Show that if A and B are two bounded of the plane ( R2) each with non-empty interior then there is a partition A1, A 2 of A and a partition 2 B1, B 2 of B and affine mappings f and g such that f : A1 ∼ B1 and g : A2 ∼ B2.

Cantor’s Theorem

We now show that there are lots of infinite cardinal numbers.

Theorem 4 (Cantor’s Theorem) For any set A, |A| < |P ow (A)|.

Proof Let A be a set. We need to show that A ¹ P ow (A) but A ≁ P ow (A) (not A ∼ P ow (A)), so |A| ≤ | P ow (A)| but |A| 6= |P ow (A)|. The function h : A → P ow (A) defined by

h(a) = {a} for a ∈ A

1Note that it is not onto since e.g 0 .909090909 . . . 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. 2An affine mapping from R2 to R2 is a linear mapping followed by a translation.

19 is an injection, so A ¹ P ow (A) does hold. Assume A ∼ P ow (A) and let f : A ∼ P ow (A). Define

X = {a ∈ A | a∈ / f(a)}.

Then X ⊆ A, that is X ∈ P ow (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 ∼ P ow (A), as required.

By Cantor’s Theorem the strictly increasing sequence of cardinal numbers

0 < 1 < 2 < · · · < ℵ0 < c can be continued

c = |R| < |P ow (R)| < |P ow (P ow (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 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|. S Proof Exercise.

So if we let S = {R, P ow (R), P ow (P ow (R)) ,... } then | S| is a cardinal number strictly larger than the previous cardinal numbers. ClearlyS we can continue to get larger cardinal numbers by repeatedly applying Cantor’s Theorem to S etc.

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