principle of inclusion-exclusion, proof of

principle of inclusion-exclusion, proof of

The proof is by

induction

. Consider a single set

A

1

. Then the

principle of inclusion-exclusion

states that

|

A

1

|

=

|

A

1

|

, which is trivially true.

Now consider a

collection

of exactly two sets

A

1

and

A

2

. We know that

A

∪

B

=

(

A

∖

B

)

∪

(

B

∖

A

)

∪

(

A

∩

B

)

Furthermore, the three sets on the right-hand side of that equation must be

disjoint

. Therefore, by the addition principle, we have

|

A

∪

B

|

=

|

A

∖

B

|

+

|

B

∖

A

|

+

|

A

∩

B

|

=

|

A

∖

B

|

+

|

A

∩

B

|

+

|

B

∖

A

|

+

|

A

∩

B

|

-

|

A

∩

B

|

=

|

A

|

+

|

B

|

-

|

A

∩

B

|

So the principle of inclusion-exclusion holds for any two sets.

Now consider a collection of

N

>

2

finite sets

A

1

,

A

2

,

…

⁢

A

N

. We assume that the principle of inclusion-exclusion holds for any collection of

M

sets where

1

≤

M

<

N

. Because the union of sets is associative, we may break up the union of all sets in the collection into a union of two sets:

⋃

i

=

1

N

A

i

=

(

⋃

i

=

1

N

-

1

A

i

)

∪

A

N

By the principle of inclusion-exclusion for two sets, we have

|

⋃

i

=

1

N

A

i

|

=

|

⋃

i

=

1

N

-

1

A

i

|

+

|

A

N

|

-

|

(

⋃

i

=

1

N

-

1

A

i

)

∩

A

N

|

Now, let

I

k

be the collection of all

k

-fold

intersections

of

A

1

,

A

2

,

…

⁢

A

N

-

1

, and let

I

k

′

be the collection of all

k

-fold intersections of

A

1

,

A

2

,

…

⁢

A

N

that include

A

N

. Note that

A

N

is included in every member of

I

k

′

and in no member of

I

k

, so the two sets do not duplicate one another.

We then have

|

⋃

i

=

1

N

A

i

|

=

∑

j

=

1

N

(

(

-

1

)

(

j

+

1

)

⁢

∑

S

∈

I

j

|

S

|

)

+

|

A

N

|

-

|

(

⋃

i

=

1

N

-

1

A

i

)

∩

A

N

|

by the principle of inclusion-exclusion for a collection of

N

-

1

sets. Then, we may distribute set intersection over set union to find that

|

⋃

i

=

1

N

A

i

|

=

∑

j

=

1

N

(

(

-

1

)

(

j

+

1

)

⁢

∑

S

∈

I

j

|

S

|

)

+

|

A

N

|

-

|

⋃

i

=

1

N

-

1

(

A

i

∩

A

N

)

|

Note, however, that

(

A

x

∩

A

N

)

∪

(

A

y

∩

A

N

)

=

(

A

x

∩

A

y

∩

A

N

)

Hence we may again apply the principle of inclusion-exclusion for

N

-

1

sets, revealing that

|

⋃

i

=

1

N

A

i

|

=

∑

j

=

1

N

-

1

(

(

-

1

)

(

j

+

1

)

⁢

∑

S

∈

I

j

|

S

|

)

+

|

A

N

|

-

∑

j

=

1

N

-

1

(

(

-

1

)

(

j

+

1

)

⁢

∑

S

∈

I

j

|

S

∩

A

N

|

)

=

∑

j

=

1

N

-

1

(

(

-

1

)

(

j

+

1

)

⁢

∑

S

∈

I

j

|

S

|

)

+

|

A

N

|

-

∑

j

=

1

N

-

1

(

(

-

1

)

(

j

+

1

)

⁢

∑

S

∈

I

j

+

1

′

|

S

|

)

=

∑

j

=

1

N

-

1

(

(

-

1

)

(

j

+

1

)

⁢

∑

S

∈

I

j

|

S

|

)

+

|

A

N

|

-

∑

j

=

2

N

(

(

-

1

)

(

j

)

⁢

∑

S

∈

I

j

′

|

S

|

)

=

∑

j

=

1

N

-

1

(

(

-

1

)

(

j

+

1

)

⁢

∑

S

∈

I

j

|

S

|

)

+

|

A

N

|

+

∑

j

=

2

N

(

(

-

1

)

(

j

+

1

)

⁢

∑

S

∈

I

j

′

|

S

|

)

The second sum does not include

I

1

′

. Note, however, that

I

1

′

=

{

A

N

}

, so we have

|

⋃

i

=

1

N

A

i

|

=

∑

j

=

1

N

-

1

(

(

-

1

)

(

j

+

1

)

⁢

∑

S

∈

I

j

|

S

|

)

+

∑

j

=

1

N

(

(

-

1

)

(

j

+

1

)

⁢

∑

S

∈

I

j

′

|

S

|

)

=

∑

j

=

1

N

-

1

[

(

-

1

)

(

j

+

1

)

⁢

(

∑

S

∈

I

j

|

S

|

+

∑

S

∈

I

j

′

|

S

|

)

]

+

(

-

1

)

N

+

1

⁢

|

⋂

i

=

1

N

A

i

|

Combining the two sums yields the principle of inclusion-exclusion for

N

sets.

Title

principle of inclusion-exclusion, proof of

Canonical name

PrincipleOfInclusionexclusionProofOf

Date of creation

2013-03-22 12:33:27

Last modified on

2013-03-22 12:33:27

Owner

mps (409)

Last modified by

mps (409)

Numerical id

6

Author

mps (409)

Entry type

Proof

Classification

msc 05A99

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