EE1J2 - Slide 1
EE1J2 – Discrete Maths Lecture 9
Isomorphic sets Cardinality Countability and 0
Countability of ℤ and ℚ
EE1J2 - Slide 2
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
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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
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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 and are isomorphicℤ ℕ
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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 ….
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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|
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Cardinality Revisited
In other words, if we can find a bijection f:AB, then |A|=|B|
Because, in this case B is just ‘A with different labels’
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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
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Countability
A set A is called countable if it is isomorphic with ℕ
i.e A is countable if there exists a bijection f: ℕA
The cardinality of ℕ, |ℕ|, is denoted by 0 –
pronounced aleph zero
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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:
odd is if 21n-
even is if 2n
0 if 0
n
n
n
nf
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Countability of ℤ 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
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Countability and ‘listing’ Sometimes it is difficult to write down the bijection f : ℕA 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 1 -2 2 -3 3 -4 4 -5 5 -6 6 -7 7 -8 8 ….
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 …. ….
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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
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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) …
:
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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
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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
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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 xSm 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
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Countability of ℚ Now list the elements of Sn in order of n:
S0 = 0
S1 = -1,1
S2 = -2, -1/2, 1/2, 2
S3 = -3, -3/2, -1/3, 1/3, 2/3, 3
S4 = -4, -4/3, -3/4,-1/4,1/4, 3/4, 4/3, 4
... = …
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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
... = …
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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
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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
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Summary of Lecture 9
Cardinality Countability Cardinality of ℕ, definition of 0
Countability of ℚ
