EE1J2 – Discrete Maths Lecture 9 Isomorphic sets Cardinality Countability and 0 Countability of ℤ and ℚ

Isomorphic Sets If there exists a bijection between two sets A and B, f: A  B, then: The elements of A and B are in ‘one-to-one’ correspondence A and B are basically the same set A and B are isomorphic f 1-1 and onto - bijection

Example 1 Let A = {0,1,2,3} and B = {a,b,c,d} The function f :A  B defined by {(0,a),(1,b),(2,c),(3,d)} is a bijection The sets A and B are isomorphic. B is just a ‘re-labelled’ version of A

Example 2 Recall that ℕ is the set of positive whole numbers and ℤ is the set of all whole numbers. Then the function: f : ℕ  ℤ, f(n) = n/2 if n is even, f(n)=-(n+1)/2 if n is odd is a bijection Hence ℤ and ℕ are isomorphic

Example 2 continued f : ℕ  ℤ, f(n) = n/2 if n is even, f(n)=-(n+1)/2 if n is odd 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 ….                  …. 0 -1 1 -2 2 -3 3 -4 4 -5 5 -6 6 -7 7 -8 8 ….

Cardinality Revisited Recall that for a finite set A={a1,…,an}, the cardinality of A is simply the number of members which A has. In this case |A|=n For infinite sets the notion of cardinality is more complex. But, if two infinite sets A and B are isomorphic, then surely |A|=|B|

Cardinality Revisited In other words, if we can find a bijection f:AB, then |A|=|B| Because, in this case B is just ‘A with different labels’

Countable Sets The simplest infinite set is the set of natural numbers ℕ = {0,1,2,3,…} E.g: if n ℕ then we can talk about the next biggest member of ℕ, i.e. n+1 The same is not true of the real numbers ℝ or even the rational numbers ℚ Given any n  ℕ , we also know that by counting through the numbers, starting at 0, we will eventually reach n. We can count ℕ

Countability A set A is called countable if it is isomorphic with ℕ i.e A is countable if there exists a bijection f: ℕA The cardinality of ℕ, |ℕ|, is denoted by 0 – pronounced aleph zero

Countability of ℤ It is easy to show that the set ℤ of integers is countable. Remember, ℤ = {…,-3,-2,-1,0,1,2,3,…} Define f: ℕ ℤ (as in previous example) by:

Countability of ℤ So, ℤ is countable, and |ℤ|= 0 Claim: f is a bijection f is 1-1 Suppose f(m)=f(n). If f(m)=f(n) is positive, then m=2f(m)=2f(n)=n. If f(m)=f(n) is negative, then m=2f(m)-1=2f(n)-1=n f is onto, by definition Hence f is a bijection So, ℤ is countable, and |ℤ|= 0

Countability and ‘listing’ Sometimes it is difficult to write down the bijection f : ℕA which shows that A is countable The first step in constructing a bijection with ℕ is to show that A can be written as a list Take the integers in the previous example: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 ….                  …. 0 -1 1 -2 2 -3 3 -4 4 -5 5 -6 6 -7 7 -8 8 ….

Countability of ℕ  ℕ Recall that ℕ  ℕ is the set {(n,m) : n,m  ℕ} Claim: ℕ  ℕ is countable How can we write ℕ  ℕ as a list? (0,1) (0,2) (0,3) … (0,n) … ? This is no good – we’ll never get to (1,1) for example

Countability of ℕ  ℕ 0 1 2 3 4 5 6 …. 0 (0,0) (0,1) (0,2) (0,3) (0,4) (0,5) (0,6) … 1 (1,0) (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) … 2 (2,0) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) … 3 (3,0) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) … 4 (4,0) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) … :

Countability of ℕ  ℕ So the countability of ℕ  ℕ is demonstrated by the list: (0,0) (1,0) (1,1) (2,0) (2,1) (2,2) (3,0) (3,1) (3,2) … All pairs containing just 0 All pairs containing just 0 and 1 and not in ‘red’ section All pairs containing just 0, 1 or 2 and not in ‘red’ or ‘blue’ sections

Countability of ℚ It is not surprising that ℤ is countable … or that ℕ  ℕ is countable A much more surprising result, demonstrated by the mathematician George Cantor, is that ℚ is countable This is counter-intuitive, but it is true. In other words, in some sense there are ‘no more’ rational numbers than there are natural numbers

Cantor’s Proof of the Countability of ℚ Cantor’s proof is based on a particular ordering of ℚ For each n let Sn be the set of rational numbers x=a/b such that n=max{|a|,b}and such that xSm for m<n (assume a and b have no common factors), so that: S0={0}, S1={-1,1}, S2={-2,-1/2,1/2,2}, S3={-3,-3/2,-2/3,-1/3,1/3,2/3,3/2,3} etc

Countability of ℚ Now list the elements of Sn in order of n: S0 = 0 ... = …

Countability of ℚ Every member of ℚ eventually appears on list. The list is ‘counted’ as follows: S0 = 0 S1 = -1,1 S2 = -2, -1/2, 1/2, 2 S3 = -3, -3/2, -2/3, -1/3, 1/3, 2/3, 3/2, 3 S4 = -4, -4/3, -3/4,-1/4,1/4, 3/4, 4/3, 4 ... = …

Generalisation of Cantor’s Proof Let A={a1,…,aN} be a finite set The set FS(A) of finite sequences of elements of A is countable To construct a proof, let Sn be the set of sequences of elements in A of length n Order Sn ‘alphabetically’ according to the alphabet a1,…,aN

Generalisation of Cantor’s Proof (continued) Now list the sets S0, S1, S2,… Clearly every finite sequence of elements from A eventually appears in the list, and The list can be counted using Cantor’s method

Summary of Lecture 9 Cardinality Countability Cardinality of ℕ, definition of 0 Countability of ℚ