Chapter 7, Functions
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Function terminology

A relationship between elements of two sets such that no
element of the first set is related to more than one
element of the second set

Domain: the set which contains the values to which the
function is applied

Codomain: the set which contains the possible values
(results) of the function

Range (or image): the set of actual values produced
when applying the function to the values of the domain
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More function terminology

f: X  Y
–
–
–
–
–


f
is the function name
X
is the domain
Y
is the co-domain
xX yY
f sends x to y
f(x) = y f of x; the value of f at x ; the image of x under f
A total function is a relationship between elements of the
domain and elements of the co-domain where each and
every element of the domain relates to one and only one
value in the co-domain
A partial function does not need to map every element of
the domain
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Formal definitions
The range of f is {y  Y | (x  X)[f(x) = y]}

–where X is the domain and Y is the co-domain
The inverse image of y  Y is
{x  X | f(x) = y}

–the set of things in the domain X that map to y
Arrow diagrams

Determining if something is a function using an arrow diagram

Equality of functions
( functions f,g with the same domain X and codomain Y)
[f = g iff (x  X)[f(x) = g(x)] ]
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Discrete Structures
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Lecture 38
April 30, 2008
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Types of functions

F:X  Y is a one-to-one (or injective) function iff
(x1,x2  X)[F(x1) = F(x2)  x1 = x2], or alternatively
(x1,x2  X)[x1  x2  F(x1)  F(x2)]

F: X Y is not a one-to-one function iff
(x1,x2  X)[(F(x1) = F(x2)) ^ (x1  x2)]

F: X  Y is an onto (or surjective) function iff
(y  Y)(x  X)[F(x) = y]

F: X  Y is not an onto function iff
(y  Y)(x  X)[F(x)  y]
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Proving functions one-to-one and onto
f: R  R

f(x) = 3x  4
Prove or give a counterexample that f is one-to-one
– recall the definition (one of two definitions) of one-to-one is
(x1 , x2  R) [ f ( x1)  f ( x2 )  x1  x2 ]

Prove or give a counterexample that f is onto
– recall the definition of onto is
(y  R) (x  R) [ f ( x)  y ]
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One-to-one correspondence or bijection
F: X  Y is bijective iff F: X  Y is one-to-one and onto
If F: X  Y is bijective then it has an inverse function
1
(F ) [Y  X ]
1
(x  X ) [ F ( x)  y  F ( y )  x]
1
(y  Y ) [ F ( y )  x  F ( x)  y ]
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Proving something is a bijection

F: Q  Q
–
–
–
–
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F(x) = 5x + 1/2
prove it is one-to-one
prove it is onto
then it is a bijection
so it has an inverse function
• find F1
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The pigeonhole principle





Basic form:
A function from one finite set to a smaller finite set cannot be
one-to-one; there must be at least two elements in the domain
that have the same image in the codomain.
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Examples

Using this class as the domain:
– must two people share a birth month?
– must two people share a birthday?

Let A = {1,2,3,4,5,6,7,8}
– if I select 5 different integers at random from this set, must two
of the numbers sum exactly to 9?
– if I select 4 integers?

There exist two people in New York City who have the
same number of hairs on their heads.

There exist two subsets of {1,…,10} with three elements
which sum to the same value.
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Discrete Structures
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Lecture 39
May 2, 2008
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Another (more useful) form of the
pigeonhole principle

The generalized pigeonhole principle:
– For any function f from a finite set X to a finite set Y and for any
positive integer k, if n(X) > k * n(Y), then there is some y  Y such
that y is the image of at least k+1 distinct elements of X.

Contrapositive form:
– For any function f from a finite set X to a finite set Y and for any
positive integer k, if for each y  Y, f–1(y) has at most k elements,
then X has at most k  n(y) elements.
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Examples

Using the generalized form:
– assume 50 people in the room, how many must share the same birth
month?
– n(A)=5 n(B)=3 F: P (A)  P (B)
how many elements of P (A) must map to a single element of P (B)?
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Composition of functions

f: X  Y1 and g: Y  Z
– g ○ f: X  Z
where Y1  Y
where (x  X)[g(f(x)) = g ○ f(x)]
g(f(x))
x
z
y
f(x)
g(y)
Y1
X
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Z
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Composition on finite sets- example

Example
X = {1,2,3}, Y1 = {a,b,c,d}, Y = {a,b,c,d,e}, Z = {x,y,z}
f(1) = c
g(a) = y
g○f(1) = g(f(1)) = z
f(2) = b
g(b) = y
g○f(2) = g(f(2)) = y
f(3) = a
g(c) = z
g○f(3) = g(f(3)) = y
g(d) = x
g(e) = x
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Composition for infinite sets- example
f: Z  Z f(n) = n + 1
g: Z  Z g(n) = n2
g ○ f(n) = g(f(n)) = g(n+1) = (n+1)2
f ○ g(n) = f(g(n)) = f(n2) = n2 + 1
Note: g ○ f  f ○ g
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Identity function

iX
the identity function for the domain X
iX : X  X

iY
the identity function for the domain Y
iY : Y  Y

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(xX) [iX(x) = x]
(yY) [iY(y) = y]
composition with the identity functions
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Composition with inverse

Recall: if f is a bijection then f1 exists.

Let f: X  Y be a bijection.

What is f ○ f1?

What is f1 ○ f?
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One-to-one in composition

If f: X  Y and g: Y  Z are both one-to-one, then
g ○ f: X  Z is one-to-one.

If f: X  Y and g: Y  Z are both onto, then g ○ f: X  Z
is onto.
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Cardinality

Comparing the “sizes” of sets:
– finite sets ( or there is a positive integer n such that there is a
bijective function from the set to {1,2,…,n})
– infinite sets (there is no such n such that there is a bijective
function from the set to {1,2,…,n})

 sets A,B, A and B have the same cardinality iff there is
a one-to-one correspondence from A to B
In other words,
Cardinality(A) = Cardinality(B) 
( a function f ) [f: A  B  f is a bijection]
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Countable sets

A set S is called countably infinite iff Cardinalit(S) =
Cardinality(Z+).

A set is called countable iff it is finite or countably
infinite.

A set which is not countable is called uncountable.
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Discrete Structures
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Lecture 40
May 5, 2008
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Countability of sets of integers and the
rationals




N
is this a countably infinite set?
Z
is this countably infinite set?
Neven is this a countably infinite set?
Card(Q+) =?= Card(Z)
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Real numbers

We’ll take just a part of this infinite set

Reals between 0 and 1 (noninclusive)
X = {x  R | 0 < x < 1}

All elements of X can be written as
0.a1a2a3… an…
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Cantor’s proof
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


Assume the set X = {x  R | 0 < x < 1} is countable
Then the elements in the set can be listed
0.a11a12a13a14…a1n…
0.a21a22a23a24…a2n…
0.a31a32a33a34…a3n…
… ………
Select the digits on the diagonal
Build a number d, such that d differs in its nth position
from the nth number in the list
1 if ann  1
dn  
2 if ann  1
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All reals
Cardinality({x  R | 0 < x < 1}) = Cardinality(R)
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