properties of a function

properties of a function

Let

X

,

Y

be sets and

f

:

X

→

Y

be a

function

. For any

A

⊆

X

, define

f

⁢

(

A

)

:=

{

f

⁢

(

x

)

∈

Y

∣

x

∈

A

}

and any

B

⊆

Y

, define

f

-

1

⁢

(

B

)

:=

{

x

∈

X

∣

f

⁢

(

x

)

∈

B

}

.

So

f

⁢

(

A

)

is a subset of

Y

and

f

-

1

⁢

(

B

)

is a subset of

X

.

Let

A

,

A

1

,

A

2

,

A

i

be arbitrary subsets of

X

and

B

,

B

1

,

B

2

,

B

j

be arbitrary subsets of

Y

, where

i

belongs to the

index set

I

and

j

to the index set

J

.  We have the following

properties

:

1.

If

A

1

⊂

A

2

, then

f

⁢

(

A

1

)

⊆

f

⁢

(

A

2

)

. In particular,

f

⁢

(

A

)

⊆

f

⁢

(

X

)

.

2.

f

⁢

(

A

1

∪

A

2

)

=

f

⁢

(

A

1

)

∪

f

⁢

(

A

2

)

. More generally,

f

⁢

(

⋃

i

A

i

)

=

⋃

i

f

⁢

(

A

i

)

.

3.

f

⁢

(

A

1

∩

A

2

)

⊆

f

⁢

(

A

1

)

∩

f

⁢

(

A

2

)

. The equality fails in the example where

f

is a real function defined by

f

⁢

(

x

)

=

x

2

and

A

1

=

{

1

}

,

A

2

=

{

-

1

}

. Equality occurs iff

f

is

one-to-one

:

Suppose

f

⁢

(

x

)

=

f

⁢

(

y

)

=

z

. Pick

A

1

=

{

x

}

and

A

2

=

{

y

}

. Then

f

⁢

(

A

1

∩

A

2

)

=

f

⁢

(

A

1

)

∩

f

⁢

(

A

2

)

=

{

z

}

≠

∅

. This means that

A

1

∩

A

2

≠

∅

. Since both

A

1

and

A

2

are

singletons

,

A

1

=

A

2

, or

x

=

y

.

Conversely

, let’s show that

f

is one-to-one then

f

⁢

(

A

1

∩

A

2

)

=

f

⁢

(

A

1

)

∩

f

⁢

(

A

2

)

. To do this, we only need to show the

right hand side

is included in the left, and this follows since if

x

∈

f

⁢

(

A

1

)

∩

f

⁢

(

A

2

)

then for some

a

1

∈

A

1

and

a

2

∈

A

2

we have

x

=

f

⁢

(

a

1

)

=

f

⁢

(

a

2

)

. As

f

is one-to-one,

a

1

=

a

2

and so

a

1

lies in

A

1

∩

A

2

and

x

is in

f

⁢

(

A

1

∩

A

2

)

.

More generally,

f

⁢

(

⋂

i

A

i

)

⊆

⋂

i

f

⁢

(

A

i

)

.

4.

f

⁢

(

A

1

)

-

f

⁢

(

A

2

)

⊆

f

⁢

(

A

1

-

A

2

)

: If

y

∈

f

⁢

(

A

1

)

-

f

⁢

(

A

2

)

, then

y

=

f

⁢

(

x

)

for some

x

∈

A

1

. If

x

∈

A

2

, then

y

=

f

⁢

(

x

)

∈

f

⁢

(

A

2

)

as well, a

contradiction

. So

x

∈

A

1

-

A

2

, and

y

=

f

⁢

(

x

)

∈

f

⁢

(

A

1

-

A

2

)

. The inequality is strict in the case when

f

:

ℤ

→

ℤ

given by

f

⁢

(

x

)

=

1

, and

A

1

=

ℤ

and

A

2

=

{

2

}

.

5.

A

⊆

f

-

1

⁢

f

⁢

(

A

)

. Again, one finds that equality fails for the real function

f

⁢

(

x

)

=

x

2

by selecting

A

=

{

1

}

. Equality again holds iff

f

is injective:

Suppose

x

∈

f

-

1

⁢

f

⁢

(

A

)

. By

definition

this means that

f

⁢

(

x

)

=

f

⁢

(

a

)

for some

x

∈

A

, and since

f

is injective we have

x

=

a

∈

A

. It follows that

f

-

1

⁢

f

⁢

(

A

)

⊆

A

. Convserly, if

f

⁢

(

x

)

=

f

⁢

(

y

)

=

z

, then

{

x

,

y

}

=

f

-

1

⁢

f

⁢

(

{

x

,

y

}

)

=

f

-

1

⁢

(

{

z

}

)

. On the other hand

{

x

}

=

f

-

1

⁢

f

⁢

(

{

x

}

)

=

f

-

1

⁢

(

{

z

}

)

. So

{

x

,

y

}

=

{

x

}

,

x

=

y

.

6.

If

B

1

⊆

B

2

, then

f

-

1

⁢

(

B

1

)

⊆

f

-

1

⁢

(

B

2

)

. In particular,

f

-

1

⁢

(

B

)

⊆

f

-

1

⁢

(

Y

)

.

7.

f

-

1

⁢

(

B

1

∪

B

2

)

=

f

-

1

⁢

(

B

1

)

∪

f

-

1

⁢

(

B

2

)

. More generally,

f

-

1

⁢

(

⋃

j

B

j

)

=

⋃

j

f

-

1

⁢

(

B

j

)

.

8.

f

-

1

⁢

(

B

1

∩

B

2

)

=

f

-

1

⁢

(

B

1

)

∩

f

-

1

⁢

(

B

2

)

. More generally,

f

-

1

⁢

(

⋂

j

B

j

)

=

⋂

j

f

-

1

⁢

(

B

j

)

.

9.

f

-

1

⁢

(

Y

-

B

)

=

X

-

f

-

1

⁢

(

B

)

. As a result,

f

-

1

⁢

(

B

1

-

B

2

)

=

f

-

1

⁢

(

B

1

)

-

f

-

1

⁢

(

B

2

)

.

10.

f

⁢

f

-

1

⁢

(

B

)

⊆

B

. Yet again, one finds that equality fails for the real function

f

⁢

(

x

)

=

x

2

by selecting

B

=

[

-

1

,

1

]

. Equality holds iff

f

is

surjective

:

Suppose

f

is onto. Pick any

y

∈

B

⊂

Y

. Then

y

=

f

⁢

(

x

)

for some

x

∈

X

. In other words,

x

∈

f

-

1

⁢

(

B

)

and hence

y

=

f

⁢

(

x

)

∈

f

⁢

f

-

1

⁢

(

B

)

. Now suppose the convserse, then pick

B

=

Y

, and we have

Y

=

f

⁢

f

-

1

⁢

(

Y

)

=

f

⁢

(

X

)

.

11.

Combining

10

and

5

, we have that

f

⁢

f

-

1

⁢

f

⁢

(

A

)

=

f

⁢

(

A

)

and

f

-

1

⁢

f

⁢

f

-

1

⁢

(

B

)

=

f

-

1

⁢

(

B

)

. Let’s show the first equality:

From

5

,

A

⊆

f

-

1

⁢

f

⁢

(

A

)

, so that

f

⁢

(

A

)

⊆

f

⁢

f

-

1

⁢

f

⁢

(

A

)

(by 1). Set

B

=

f

⁢

(

A

)

. Then by

10

,

f

⁢

f

-

1

⁢

f

⁢

(

A

)

=

f

⁢

f

-

1

⁢

(

B

)

⊆

B

=

f

⁢

(

A

)

.

Remarks.

•

f

-

1

⁢

f

and

f

⁢

f

-

1

the

compositions

of the function and its

inverse

as defined at the beginning of the entry, so that

f

-

1

⁢

f

⁢

(

A

)

=

f

-

1

⁢

(

f

⁢

(

A

)

)

and

f

⁢

f

-

1

⁢

(

B

)

=

f

⁢

(

f

-

1

⁢

(

B

)

)

.

•

From the definition above, we see that a function

f

:

X

→

Y

induces two functions

[

f

]

and

[

f

-

1

]

defined by

[

f

]

:

2

X

→

2

Y

⁢

such that

⁢

[

f

]

⁢

(

A

)

:=

f

⁢

(

A

)

⁢

and

[

f

-

1

]

:

2

Y

→

2

X

⁢

such that

⁢

[

f

-

1

]

⁢

(

B

)

:=

f

-

1

⁢

(

B

)

.

The last property

11

says that

[

f

]

and

[

f

-

1

]

are

quasi-inverses

of each other.

•

f

is a

bijection

iff

[

f

]

and

[

f

-

1

]

are inverses of one another.

Title

properties of a function

Canonical

name

PropertiesOfAFunction

Date of creation

2013-03-22 16:21:38

Last modified on

2013-03-22 16:21:38

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

24

Author

CWoo (3771)

Entry type

Definition

Classification

msc 03E20

Related topic

PropertiesOfFunctions

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