Download 04-function

COMP2121
Discrete Mathematics
Function
Hubert Chan (Chapter 2.1, 2.2)
[O1 Abstract Concepts]
[O2 Proof Techniques]
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What you will learn…
• Z+ = {1,2,3…..}
• E+ = {2,4,6,8,…}
• Real numbers
• Z+ and E+ are countable sets
• Rational numbers are countable
• Real numbers are uncountable
• To prove that they have different kinds of infinity, we
need the concept of function
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What is a Function?
Let us consider some mathematical functions
 f(x) = 3x-2
 square-area(x) = x2
 Function represented by table
Price (HK$)
5
5.5
6
6.5
7
7.5
Demand (no.)
5000
4700
4500
4350
4250
4200

Function represented by graph
5200
5000
4800
4600
4400
4200
4000
3800
5.0
5.5
6.0
6.5
7.0
7.5
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Functions and Sets [O1]
Let A, B be sets. A function f from A to B is an assignment
of exactly one element of B to each element of A.
 We write f : A  B
 We can also write f(a) = b where aA and bB
 A function is also called a mapping.
Example: （Labeled balls into labeled bins）
A function f represented by labeled balls into labeled bins
from A (the set of balls) to B (the set of bins) is defined
as follows:
f(i)=j means put ball i into bin j.
Hence the function is putting balls into bins and each ball is
put into exactly one bin.
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Basic Terminology
Given a function f : A  B ( f maps A to B ),
 A is the domain of f (‘balls’)
B
 B is the codomain of f (‘bins’)
A
 If f(a) = b,
b is the image of a.
 The range of f is the set of all images of elements of A.
{ b | b  B and (  a (f (a) = b) ) } (‘non-empty bins’)
Example:
Let f : Z  Z and f(x) = x2, where Z = set of integers
Domain = Z, codomain = Z, range = Z ?
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Injective Functions (One-to-one)
A function f is injective,
iff for every distinct x, y in the domain,
f(x)  f(y).
A
B
“Different inputs imply different outputs”
Examples of injective function:
f: Z  Z: f(x)=2x, f(x)=x3,
balls into bins function such that each bin has at most
one ball.
Examples of not injective function:
f: Z  Z: f(x)=x4
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Surjective Functions (Onto)
A function f from A to B is called surjective,
iff for all b  B, there exists
an element a  A such that f(a) = b.
A
B
“Every element in the codomain is the output of some input”
Examples of surjective function:
f: Z  Z: f(x)=x+2
balls into bins function such that each bin has at least
one ball.
Examples of not surjective function
f: Z  Z: f(x)=2x
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Bijective Functions
A function f that is both injective and
surjective is called a bijection.
A
B
Examples:
f: Z  Z: f(x)=x, f(x)=x+2
balls into bins function such that each bin has exactly
one ball.
which means the number of balls = the number of bins if there
are finite number of balls and bins.
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Examples
Determine whether the following functions are injection,
surjection, or bijection?
f1 : Z  Z, f1(x) = x2
 Not injection, since f(-1) = f(1) = 1.
 Not surjection, since there is no integer x such that
x2 = -1.
f2 : Z  Z, f2(x) = x + 1
 Injection, since x+1  y+1 implies x  y.
 Surjection, since f(x) = y  x + 1 = y  x = y – 1.
 Bijection, since it is both one-to-one and onto.
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Different Kinds of Infinity
• Two sets A and B have the same cardinality
iff there exists a bijection f:A→B
• Example:
 Z+ = {1,2,3,4,…..}
 E+ = {2,4,6,8,…}
 Does E+
Z+ imply |E+| < |Z+|? NO!
 Bijection f : Z+→E+ such that f(x)=2x
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Different Kinds of Infinity
• f : Z+→E+ such that f(x)=2x is
 Injective since if x1, x2  Z+ are different, f(x1) ≠ f(x2)
 Surjective since for any y  E+, there exists x=y/2
such that x  Z+ and f(x)=y
 f(x) is a bijection
•
Z+ and E+ have the same cardinality:
 1 2 3 4 5 ….
 ↓↓↓↓↓…
 2 4 6 8 10 ….
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Countable Sets
Definition. A set S is countable if either:
(i) the set S is finite, or
(ii) there is a bijection from the set S to Z+.
Examples of Countable Sets
(i) Z+ (positive integers) is countable.
(ii) Set of all integers
(iii) Even numbers
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Some Results
• Countable and uncountable:
1. All subsets of a countable set are countable. Proof?
2. If there is a injection from a set A to another set B
 If
A is uncountable, then B is uncountable.
 If B is countable, then A is countable, too.
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Q (rational numbers) is countable [O2]
• Simple case: Q+ is countable.
• Each element of Q+ can be represented by a/b, where a
and b are relatively-prime integers and a,b>0.
b
1
2
3
4
…
1
1/1
1/2
1/3
1/4
…
2
2/1
2/2
2/3
…
…
3
3/1
3/2
…
…
…
4
4/1
…
…
…
…
…
…
…
…
…
…
a
Q+
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Q (rational numbers) is countable
• Simple case: Q+ is countable.
• Each element of Q+ can be represented by a/b, where a
and b are relatively-prime integers and a,b>0.
b
1
2
3
4
…
1
1
1/1
1/2
2
1/3
4
1/4
6
…
2
2/1
3
2/3
7
…
…
3
3/1
5
3/2
8
…
…
…
4
4/1
9
…
…
…
…
…
…
…
…
…
…
a
Q
Z++
Removing
duplicates
Mapping to
integers
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Question:
All infinite sets are countable?
• Examples:
 Z+ = {1,2,3,4,…..}
 E+ = {2,4,6,8,…}
 Q = rational numbers
 R = real numbers
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Question:
Do Z+ and P(Z+) have the same cardinality?
• They are all infinite sets.
• But they have different kinds of infinities.
 P(Z+) contains “more” elements and is uncountable.
Theorem:
There is NO bijection mapping from
any non-empty set S to P(S)
P(S) has a larger cardinality than S
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Theorem:
There is NO bijection mapping any non-empty set S to P(S)
Proof: (proof by cases and contradiction)
Case 1: If S is a finite set. Then |P(S)| = 2|S| > |S|.
Case 2: If S is a infinite set.
Assume there is a bijection f: S → P(S). Then for each xS, f(x)P(S).
Define
A  {x  S : x  f ( x)} , then A  P(S).
condition*
Since f is also a surjection, for each element X in P(S), there exists x in S such
that f(x)=X. Hence there exists an element a in S such that f(a)=A.
If a  f (a ) , then condition* of A is not satisfied, a  A  f (a ) . Contradiction.
If a  f (a ) , then condition* of A is satisfied, a  A  f (a ) . Contradiction.
Hence, such a bijection does not exist.
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R (real numbers) is uncountable
• Intuition: P(Z+) is uncountable and is a “subset” of R.
• There is a injection from P(Z+) to R:
 Input: a subset of Z+, S={s1, s2, s3,…}
 Output: a real number X in [0,1] such that
 X=0.x1x2x3x4…
where xi=1 if iS and 0 if iS
• Examples:
 f({1,3,4,7,10}) = 0.1011001001
 f({2,5,6,8,9}) = 0.0100110110
 f(Ø)=0, f(Z+)=0.111111…
• R has at least the cardinality as P(Z+)
• R is uncountable!
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Axiom of Choice
(Something sounds trivial but has profound implication later)
f : C   Si
Si C
f ( Si )  Si
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