properties of set difference

properties of set difference

Let

A

,

B

,

C

,

D

,

X

be sets.

1.

A

∖

B

⊆

A

. This is

obvious

by

definition

.

2.

If

A

,

B

⊆

X

, then

A

∖

B

=

A

∩

B

∁

,

(

A

∖

B

)

∁

=

A

∁

∪

B

,

and

A

∁

∖

B

∁

=

B

∖

A

where

∁

denotes

complement

in

X

.

Proof.

For the first equation, see here (

http://planetmath.org/PropertiesOfComplement

). The second equation comes from the first:

(

A

∖

B

)

∁

=

(

A

∩

B

∁

)

∁

=

(

A

∁

)

∪

(

B

∁

)

∁

=

A

∁

∪

B

. The last equation also follows from the first:

A

∁

∖

B

∁

=

A

∁

∩

(

B

∁

)

∁

=

B

∩

A

∁

=

B

∖

A

.
∎

3.

A

⊆

B

iff

A

∖

B

=

∅

.

Proof.

Since

A

⊆

B

,

B

∁

⊆

A

∁

. Then

A

∖

B

=

A

∩

B

∁

⊆

A

∩

A

∁

=

∅

. On the other hand, suppose

A

∖

B

=

∅

. Then

A

∩

B

∁

=

∅

by

property

1, which means

A

⊆

(

B

∁

)

∁

=

B

.
∎

4.

A

∩

B

=

∅

iff

A

∖

B

=

A

.

Proof.

Suppose first that

A

∩

B

=

∅

. If

a

∈

A

, then

a

∉

B

, so

a

∈

A

∖

B

, and hence

A

⊆

A

∖

B

. The

equality

is shown by applying property 1. Next suppose

A

∖

B

=

A

. If

a

∈

A

, then

a

∈

A

∖

B

, so

a

∉

B

, which means

A

⊆

B

∁

, or

A

∩

B

=

∅

.
∎

5.

A

∖

∅

=

A

and

A

∖

A

=

∅

=

∅

∖

A

.

Proof.

The first equation follows from property 4 and the last two equations from property 3.
∎

6.

(de Morgan’s laws on

set difference

):

A

∖

(

B

∩

C

)

=

(

A

∖

B

)

∪

(

A

∖

C

)

and

A

∖

(

B

∪

C

)

=

(

A

∖

B

)

∩

(

A

∖

C

)

.

Proof.

These laws follow from property 2 and the de Morgan’s laws on set complement. For example,

A

∖

(

B

∩

C

)

=

(

A

∖

B

)

∪

(

A

∖

C

)

=

A

∩

(

B

∩

C

)

∁

=

A

∩

(

B

∁

∪

C

∁

)

=

(

A

∩

B

∁

)

∪

(

A

∩

C

∁

)

=

(

A

∖

B

)

∪

(

A

∖

C

)

. The other equation is proved similarly.
∎

7.

A

∖

(

A

∩

B

)

=

A

∖

B

=

(

A

∪

B

)

∖

B

.

Proof.

The first equation follows from property 6:

A

∖

(

A

∩

B

)

=

(

A

∖

A

)

∪

(

A

∖

B

)

=

A

∖

B

by property 5. Next,

(

A

∪

B

)

∖

B

=

(

A

∪

B

)

∩

B

∁

=

(

A

∩

B

∁

)

∪

(

B

∩

B

∁

)

=

A

∩

B

∁

=

A

∖

B

, proving the second equation.
∎

8.

(

A

∩

B

)

∖

C

=

(

A

∖

C

)

∩

(

B

∖

C

)

.

Proof.

Using property 2, we get

(

A

∩

B

)

∖

C

=

(

A

∩

B

)

∩

C

∁

=

(

A

∩

C

∁

)

∩

(

B

∩

C

∁

)

=

(

A

∖

C

)

∩

(

B

∖

C

)

.
∎

9.

A

∩

(

B

∖

C

)

=

(

A

∩

B

)

∖

(

A

∩

C

)

.

Proof.

(

A

∩

B

)

∖

(

A

∩

C

)

=

(

A

∩

B

)

∩

(

A

∩

C

)

∁

=

(

A

∩

B

)

∩

(

A

∁

∪

C

∁

)

=

(

(

A

∩

B

)

∩

A

∁

)

∪

(

(

A

∩

B

)

∩

C

∁

)

=

(

A

∩

B

)

∩

C

∁

=

A

∩

(

B

∩

C

∁

)

=

A

∩

(

B

∖

C

)

.
∎

10.

(

A

∖

B

)

∩

(

C

∖

D

)

=

(

C

∖

B

)

∩

(

A

∖

D

)

Proof.

Expanding the LHS, we get

A

∩

B

∁

∩

C

∩

D

∁

. Expanding the RHS, we get the same thing.
∎

11.

(

A

∖

B

)

∩

(

C

∖

D

)

=

(

A

∩

C

)

∖

(

B

∪

D

)

.

Proof.

Starting from the RHS:

(

A

∩

C

)

∖

(

B

∪

D

)

=

(

(

A

∩

C

)

∖

B

)

∩

(

(

A

∩

C

)

∖

D

)

=

(

A

∖

B

)

∩

(

C

∖

B

)

∩

(

A

∖

D

)

∩

(

C

∖

D

)

=

(

A

∖

B

)

∩

(

C

∖

D

)

, where the last equality comes from property 10.
∎

Remarks

.

1.

Many of the proofs above use the properties of the set complement. Please see this link (

http://planetmath.org/PropertiesOfComplement

) for more detail.

2.

All of the properties of

∖

on sets can be generalized to

Boolean

subtraction

(

http://planetmath.org/DerivedBooleanOperations

) on

Boolean algebras

.

Title

properties of set difference

Canonical

name

PropertiesOfSetDifference

Date of creation

2013-03-22 17:55:35

Last modified on

2013-03-22 17:55:35

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

7

Author

CWoo (3771)

Entry type

Derivation

Classification

msc 03E20

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