Things you should know about sets and mappings:
Notation:
N = the natural numbers (i.e., the positive integers)
Z = the integers (we may also use Z)
Q = the rational numbers
R = the real numbers (we may also use R)
C = the complex numbers (we may also use C)
R+ = the positive real numbers (similarly, Z+, Q+ etc.)
R = the nonnegative real numbers, etc.
Set builder notation
{ x | p(x) } i.e., the set of all x such that p(x) is true
{ x Z | p(x) } i.e., the set of all integers x such that p(x) is true
etc.
Definition of a mapping (function)
f:AB
( A is the domain; B is the codomain)
To each element a in A, there corresponds a unique element b in B; f(a) = b
Definition of one-to-one (injective)
If a  c, then f(a)  f(c) ; equivalently, if f(a) = f(c), then a = c
Equivalently, for all b in B, there is at most one a in A such that f(a) = b
Definition of onto (surjective)
For all b in B, there exists at least one a in A such that f(a) = b
That is, the codomain of f and the image (range) of f coincide
Definition of a one to one correspondence (bijection)
f : A  B is a one to one correspondence (also called a bijection) if f is both
one to one and onto, i.e., if f is both an injection and a surjection
How to show that a mapping is injective
Suppose that for a and c in A, f(a) = f(c), and show that this forces a = c
How to show that a mapping is surjective
Select a general element b in B, and find (or show that there exists) an a in A
such that f(a) = b. [Note the chosen element of B must be generic; you cannot
choose a specific element of B.]
Composition of mappings
(f  g)(x) = f(g(x)) [Note: The mapping on the right, g, is performed first]
The identity map
i(x) = x
What it means for a map to be invertible
A mapping f : A  B is invertible if there exists a mapping g : B  A such
that f  g = i = g  f ,
i.e., for all b in B, f(g(b)) = b, and, for all a in A, g(f(a)) = a
When a map is invertible
f : A  B is invertible if and only if f is a bijection
The composition of bijections is a bijection
If f : A  B and g : B  C are both injective, then so is g  f : A  C
If f : A  B and g : B  C are both surjective, then so is g  f : A  C
If f : A  B and g : B  C are both bijective, then so is g  f : A  C
The converse of these three results are not true:
I.e., if g  f is injective, it does not mean that f and g are both injective. (It
does force f to be injective.)
Similarly, if g  f is surjective, it does not mean that f and g are both
injective. (It does force g to be surjective.)
And if g  f is bijective, it does not mean that f and g are both bijective. (It
does force f to be injective and g to be surjective.)
The inverse of a composed map:
(g  f)-1 = f -1  g-1
What is meant by the cardinality of a set
Two sets A and B are said to have the same cardinality, denoted |A| = |B|, if
there exists a bijection f : A  B