uniqueness of cardinality

uniqueness of cardinality

Theorem

.

The

cardinality

of a set is unique.

Proof.

We will verify the result for

finite sets

only.
Suppose

A

is a finite set with cardinality

∣

A

∣

=

n

and

∣

A

∣

=

m

. Then there exist

bijections

f

:

ℕ

n

→

A

and

g

:

ℕ

m

→

A

. Since

g

is a bijection, it is

invertible

, and

g

-

1

:

A

→

ℕ

m

is a bijection. Then the

composition

g

-

1

∘

f

is a bijection from

ℕ

n

to

ℕ

m

. We will show by

induction

on

n

that

m

=

n

. In the case

n

=

1

we have

ℕ

n

=

ℕ

1

=

{

1

}

, and it must be that

ℕ

m

=

{

1

}

as well, whence

m

=

1

=

n

. Now let

n

≥

1

∈

ℕ

, and suppose that, for all

m

∈

ℕ

, the

existence

of a bijection

f

:

ℕ

n

→

ℕ

m

implies

n

=

m

. Let

m

∈

ℕ

and suppose

h

:

ℕ

n

+

1

→

ℕ

m

is a bijection. Let

k

=

h

⁢

(

n

+

1

)

, and notice that

1

≤

k

≤

m

. Since

h

is onto there exists some

j

∈

ℕ

n

+

1

such that

f

⁢

(

j

)

=

m

. There are two cases to consider. If

j

=

n

+

1

, then we may define

ϕ

:

ℕ

n

→

ℕ

m

-

1

by

ϕ

⁢

(

i

)

=

h

⁢

(

i

)

for all

i

∈

ℕ

n

, which is clearly a bijection, so by the

inductive hypothesis

n

=

m

-

1

, and thus

n

+

1

=

m

. Now suppose

j

≠

n

+

1

, and define

ϕ

:

ℕ

n

→

ℕ

m

-

1

by

ϕ

⁢

(

i

)

=

{

h

⁢

(

i

)

if

⁢

i

≠

j

k

if

⁢

i

=

j

⁢

.

We first show that

ϕ

is

one-to-one

. Let

i

1

,

i

2

∈

ℕ

n

, where

i

1

≠

i

2

. First consider the case where neither

i

1

nor

i

2

is equal to

j

. Then, since

h

is one-to-one, we have

ϕ

⁢

(

i

1

)

=

h

⁢

(

i

1

)

≠

h

⁢

(

i

2

)

=

ϕ

⁢

(

i

2

)

⁢

.

In the case

i

1

=

j

, again because

h

is one-to-one, we have

ϕ

⁢

(

i

1

)

=

k

=

h

⁢

(

n

+

1

)

≠

h

⁢

(

i

2

)

=

ϕ

⁢

(

i

2

)

⁢

.

Similarly it can be shown that, if

i

2

=

j

,

ϕ

⁢

(

i

1

)

≠

ϕ

⁢

(

i

2

)

, so

ϕ

is one-to-one. We now show that

ϕ

is onto. Let

l

∈

ℕ

m

-

1

. If

l

=

k

, then we may take

i

=

j

to have

ϕ

⁢

(

i

)

=

ϕ

⁢

(

j

)

=

k

=

l

. If

l

≠

k

, then, because

h

is onto, there exists some

i

∈

ℕ

n

such that

h

⁢

(

i

)

=

l

, so

g

⁢

(

i

)

=

h

⁢

(

i

)

=

l

. Thus

g

is onto, and our inductive hypothesis again gives

n

=

m

-

1

, hence

n

+

1

=

m

. So by the principle of induction, the result holds for all

n

. Returning now to our set

A

and our bijection

g

-

1

∘

f

:

ℕ

n

→

ℕ

m

, we may conclude that

m

=

n

, and consequently that the cardinality of

A

is unique, as desired.
∎

Corollary

.

There does not exist a bijection from a finite set to a

proper subset

of itself.

Proof.

Without a loss of generality we may assume the proper subset is nonempty, for if the set is empty then the

corollary

holds

vacuously

. So let

A

be as above with cardinality

n

∈

ℕ

,

B

a proper subset of

A

with cardinality

m

<

n

∈

ℕ

, and suppose

f

:

A

→

B

is a bijection. By

hypothesis

there exist bijections

g

:

ℕ

n

→

A

and

h

:

ℕ

m

→

B

; but then

h

-

1

∘

(

f

∘

g

)

is a bijection from

ℕ

n

to

ℕ

m

, whence by the preceding result

n

=

m

, contrary to

assumption

.
∎

The corollary is also known more popularly as the

pigeonhole principle

.

Title

uniqueness of cardinality

Canonical

name

UniquenessOfCardinality

Date of creation

2013-03-22 16:26:52

Last modified on

2013-03-22 16:26:52

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

27

Author

mathcam (2727)

Entry type

Theorem

Classification

msc 03E10

Related topic

cardinality

Related topic

bijection

Related topic

function

Related topic

set

Related topic

subset

Related topic

Subset

Related topic

Bijection

Related topic

Set

Related topic

PigeonholePrinciple

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