cardinality of disjoint union of finite sets

cardinality of disjoint union of finite sets

To begin we will need a lemma.

Lemma

.

Suppose

A

,

B

,

C

, and

D

are sets, with

A

∩

B

=

C

∩

D

=

∅

, and suppose there exist

bijective

maps

f

1

:

A

→

C

and

f

2

:

B

→

D

. Then
there exists a bijective map from

A

∪

B

to

C

∪

D

.

Proof.

Define the map

g

:

A

∪

B

→

C

∪

D

by

g

⁢

(

x

)

=

{

f

1

⁢

(

x

)

if

⁢

x

∈

A

f

2

⁢

(

x

)

if

⁢

x

∈

B

⁢

.

(1)

To see that

g

is

injective

, let

x

1

,

x

2

∈

A

∪

B

, where

x

1

≠

x

2

. If

x

1

,

x

2

∈

A

, then by the injectivity of

f

1

we have

g

⁢

(

x

1

)

=

f

1

⁢

(

x

1

)

≠

f

1

⁢

(

x

2

)

=

g

⁢

(

x

2

)

⁢

.

(2)

Similarly if

x

1

,

x

2

∈

B

,

g

⁢

(

x

1

)

≠

g

⁢

(

x

2

)

by the injectivity of

f

2

. If

x

1

∈

A

and

x

2

∈

B

, then

g

⁢

(

x

1

)

=

f

1

⁢

(

x

1

)

∈

C

, while

g

⁢

(

x

2

)

=

f

2

⁢

(

x

2

)

∈

D

, whence

g

⁢

(

x

1

)

≠

g

⁢

(

x

2

)

because

C

and

D

are

disjoint

. If

x

1

∈

B

and

x

2

∈

A

, then

g

⁢

(

x

1

)

≠

g

⁢

(

x

2

)

by the same reasoning. Thus

g

is injective. To see that

g

is

surjective

, let

y

∈

C

∪

D

. If

y

∈

C

, then by the surjectivity of

f

1

there exists some

x

∈

A

such that

f

1

⁢

(

x

)

=

y

, hence

g

⁢

(

x

)

=

y

. Similarly if

y

∈

D

, by the surjectivity of

f

2

there exists some

x

∈

B

such that

f

2

⁢

(

x

)

=

y

, hence

g

⁢

(

x

)

=

y

. Thus

g

is surjective.
∎

Theorem

.

If

A

and

B

are

finite sets

with

A

∩

B

=

∅

, then

∣

A

∪

B

∣

=

∣

A

∣

+

∣

B

∣

.

Proof.

Let

A

and

B

be finite,

disjoint sets

.
If either

A

or

B

is empty, the result holds trivially, so suppose

A

and

B

are nonempty with

∣

A

∣

=

n

∈

ℕ

and

∣

B

∣

=

m

∈

ℕ

. Then there exist bijections

f

:

ℕ

n

→

A

and

g

:

ℕ

m

→

B

. Define

h

:

ℕ

n

→

ℕ

n

+

m

∖

ℕ

m

by

h

⁢

(

i

)

=

m

+

i

for each

i

∈

ℕ

n

. To see that

h

is injective, let

i

1

,

i

2

∈

ℕ

, and suppose

h

⁢

(

i

1

)

=

h

⁢

(

i

2

)

. Then

m

+

i

1

=

m

+

i

2

, whence

i

1

=

i

2

. Thus

h

is injective. To see that

h

is surjective, let

k

∈

ℕ

n

+

m

∖

ℕ

m

. By construction,

m

+

1

≤

k

≤

m

+

n

, and consequently

1

≤

k

-

m

≤

n

, so

k

-

m

∈

ℕ

n

; therefore we may take

i

=

k

-

m

to have

h

⁢

(

i

)

=

k

, so

h

is surjective. Then, again by construction, the

composition

f

∘

h

-

1

is a bijection from

ℕ

n

+

m

∖

ℕ

m

to

A

, and as such, by the preceding lemma, the map

ϕ

:

ℕ

n

+

m

∖

ℕ

m

∪

ℕ

m

→

A

∪

B

defined by

ϕ

⁢

(

i

)

=

{

(

f

∘

h

-

1

)

⁢

(

i

)

if

⁢

i

∈

ℕ

n

+

m

∖

ℕ

m

g

⁢

(

i

)

if

⁢

i

∈

ℕ

m

⁢

,

(3)

is a bijection. Of course, the domain of

ϕ

is simply

ℕ

n

+

m

, so

∣

A

∪

B

∣

=

n

+

m

, as asserted.
∎

Corollary

.

Let

{

A

k

}

k

=

1

n

be a family of

mutually disjoint

, finite sets. Then

∣

⋃

k

=

1

n

A

k

∣

=

∑

k

=

1

n

∣

A

k

∣

.

Proof.

We proceed by

induction

on

n

. In the case

n

=

2

, the the preceding result applies, so let

n

≥

2

∈

ℕ

, and suppose

∣

⋃

k

=

1

n

A

k

∣

=

∑

k

=

1

n

∣

A

k

∣

. Then by our

inductive hypothesis

and the preceding result, we have

|

⋃

k

=

1

n

+

1

A

k

|

=

|

⋃

k

=

1

n

A

k

∪

A

n

+

1

|

=

|

⋃

k

=

1

n

A

k

|

+

∣

A

n

+

1

∣

=

∑

k

=

1

n

∣

A

k

∣

+

∣

A

n

+

1

∣

=

∑

k

=

1

n

+

1

∣

A

k

∣

⁢

.

(4)

Thus the result holds for

n

+

1

, and by the principle of induction, for all

n

.
∎

Title

cardinality of disjoint union of finite sets

Canonical

name

CardinalityOfDisjointUnionOfFiniteSets

Date of creation

2013-03-22 16:31:05

Last modified on

2013-03-22 16:31:05

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

17

Author

mathcam (2727)

Entry type

Theorem

Classification

msc 03E10

Related topic

Cardinality

Related topic

Bijection

Related topic

DisjointUnion

Related topic

Function

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