Section 1.2:
Equivalent and Countable Sets
In essence, this section deals with the “size” of a given set. For example, it we consider the sets
{
}
, ,
A
a b c
=
and
{
}
1,2,3
B
=
, we would not say that the two sets are the same, but we could say that the sets are of the same
“size.” On the other hand, let
{
}
, , , ,
C
α β δ ε λ
=
. In this case, we can say that
A
is “smaller” than
C
. We will
make this notion of “size” more precise.
Definition:
Let
A
and
B
be sets. Then we say that
A
and
B
are
equivalent
, denoted
A
B
≈
, if there exists a one
toone function
f
from
A
onto
B
. That is, if we can exhibit a function
1 1
:
onto
f
A
B

→
. Unsurprisingly, if
A
B
≈
it is
also true that
B
A
≈
.
Example:
Let
A
and
B
be defined as above. Then define the function
f
from
A
to
B
by
( )
1
f a
=
,
( )
2
f b
=
, and
( )
3
f c
=
. Then
f
is a bijection and so
A
B
≈
. On the other hand, while it is certainly possible to exhibit a
function that maps onetoone from
A
to
C
, it is impossible for this function to be onto. Hence,
A
C
≈
/
.
So far, this all seems very basic. The problem occurs when the sets we are considering are infinite. One might
suspect that if
A
is a proper subset of
B
, then there is no way that
A
B
≈
; and in fact, this is indeed the case if
A
is a
finite
set. However, as the next example will show, this is not the case with infinite sets.
Example:
The sets
N
and
Z
are equivalent. To see this, define the function
:
f
→
N
Z
by
1
2
1
2
(
1)
if
is odd
( )
if
is even
n
n
f n
n
n

=

We must show that
f
is indeed a bijection. Suppose
( )
( )
f n
f m
=
. Then it must be that
n
and
m
are both even or
both odd. For if
n
is odd and
m
is even, then
( )
0
f n
≥
and
( )
0
f m
<
; it follows that they could not possibly be
equal in this case. Therefore, assume that
n
and
m
are both odd. It follows that
1
1
2
2
(
1)
(
1)
n
m

=

. Some simple
arithmetic shows that
n
m
=
; a similar argument can be used in the case that both
m
and
n
are even. Hence,
f
maps onetoone. Now, consider some integer
z
. Then either
0,
0,
z
z
=
or
0
z
<
. If
0
z
=
, let
n
= 1 and note
that
( )
0
f n
z
=
=
. If
0
z
, let
2
1
n
z
= +
. Again,
( )
f n
z
=
. Finally, if
0
z
<
, let
2
n
z
= 
, and once more, we
find that
( )
f n
z
=
. Thus,
f
maps onto
Z
and so
f
is a bijection from
N
to
Z
. We conclude that
≈
N
Z
.
The question of the size of infinity and if there are different sized infinities is quite perplexing. To explore this
question, we begin with the counting numbers,
N
. Sets that are equivalent to the set of positive integers deserve
a special name.
Notation:
We will let
m
N
denote the set
{
}
1,2,.
..,
m
m
=
N