Cardinality
The “cardinality” of a set provides a precise meaning to “the number of elements” in the set,
even when the set contains infinitely many elements. We start with the following very reasonable
definitions of “the set
S
contains the same number of elements as the set
T
”, “the set
S
contains
fewer elements than the set
T
” and “the set
S
contains more elements then the set
T
”.
Definition 1
Let
S
and
T
be nonempty sets. Then
card (
S
) = card (
T
)
⇐⇒
there exists an
f
:
S
→
T
which is one–to–one and onto
card (
S
)
≤
card (
T
)
⇐⇒
there exists an
f
:
S
→
T
which is one–to–one
card (
S
)
≥
card (
T
)
⇐⇒
there exists an
f
:
S
→
T
which is onto
card (
S
)
<
card (
T
)
⇐⇒
card (
S
)
≤
card (
T
) but card (
S
)
6
= card (
T
)
card (
S
)
>
card (
T
)
⇐⇒
card (
S
)
≥
card (
T
) but card (
S
)
6
= card (
T
)
Definition 2
Let
S
be a nonempty set. Then
(a) The set
S
has card (
S
) =
n
∈
IN if card (
S
) = card
(
{
1
,
2
,
3
,
· · ·
, n
}
)
.
(b) The set
S
is countable if card (
S
)
≤
card (IN). The set
S
is countably infinite if card (
S
) =
card (IN).
(c) The set
S
is uncountable if it is not countable.
(d) The set
S
has the cardinality of the continuum if card (
S
) = card (IR).
We now verify that
≤
and
≥
, in the sense of cardinality, work as expected.
Proposition 3
Let
S
,
T
and
U
be nonempty sets. Then
(a)
card (
S
)
≤
card (
T
)
if and only if
card (
T
)
≥
card (
S
)
.
(b) Either
card (
S
)
≤
card (
T
)
or
card (
S
)
≥
card (
T
)
.
(c) If
card (
S
)
≤
card (
T
)
and
card (
S
)
≥
card (
T
)
, then
card (
S
) = card(
T
)
.
(d) If
card (
S
)
≤
card (
T
)
and
card (
T
)
≤
card (
U
)
, then
card (
S
)
≤
card (
U
)
.
Proof:
(a) If card (
S
)
≤
card (
T
), then there is a one–to–one function
f
:
S
→
T
. Fix any
s
0
∈
S
.
Then
g
(
t
) =
f

1
(
t
)
if
t
is in the range of
f
s
0
otherwise
maps
T
onto
S
, so that card (
T
)
≥
card (
S
).
If card (
T
)
≥
card (
S
), then there is a function
g
that maps
T
onto
S
. So the sets
g

1
(
s
) =
t
∈
T
g
(
t
) =
s
,
s
∈
S
are disjoint and nonempty.
For each
s
∈
S
select one
t
s
∈
g

1
(
s
).
Then
f
(
s
) =
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 Spring '08
 AMJAD
 Empty set, Natural number, iI Si

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