testing for continuity via closure operation

testing for continuity via closure operation

Proposition 1

.

Let

X

,

Y

be

topological spaces

, and

f

:

X

→

Y

a

function

. Then the following are

equivalent

:

1.

f

is

continuous

,

2.

for any

closed set

D

⊆

Y

, the set

f

-

1

⁢

(

D

)

is closed in

X

,

3.

f

⁢

(

A

¯

)

⊆

f

⁢

(

A

)

¯

, where

A

¯

is the

closure

of

A

,

4.

f

-

1

⁢

(

B

)

¯

⊆

f

-

1

⁢

(

B

¯

)

,

5.

f

-

1

⁢

(

C

∘

)

⊆

f

-

1

⁢

(

C

)

∘

, where

C

∘

is the

interior

of

C

.

Proof.

•

(

1

)

⇔

(

2

)

. Use the

identity

f

-

1

⁢

(

A

-

B

)

=

f

-

1

⁢

(

A

)

-

f

-

1

⁢

(

B

)

for any function

f

. Then

f

-

1

⁢

(

Y

-

D

)

=

X

-

f

-

1

⁢

(

D

)

. So if

D

is closed (or open),

f

-

1

⁢

(

Y

-

D

)

is open (or closed), whence

f

-

1

⁢

(

D

)

is closed (or open).

•

(

2

)

⇔

(

3

)

. Suppose first that

f

:

X

→

Y

is continuous. Since

f

⁢

(

A

)

¯

=

⋂

{

C

∣

C

⁢

closed in

⁢

Y

,

and

⁢

f

⁢

(

A

)

⊆

C

}

,

A

⊆

f

-

1

⁢

f

⁢

(

A

)

⊆

f

-

1

⁢

(

C

)

, which is closed in

X

. So

A

¯

⊆

f

-

1

⁢

(

C

)

, and therefore

f

⁢

(

A

¯

)

⊆

f

⁢

f

-

1

⁢

(

C

)

⊆

C

. As a result,

f

⁢

(

A

¯

)

⊆

⋂

{

C

∣

C

⁢

closed in

⁢

Y

,

and

⁢

f

⁢

(

A

)

⊆

C

}

=

f

⁢

(

A

)

¯

.

Conversely

, let

V

be closed in

Y

. Then

V

¯

=

V

. Let

U

=

f

-

1

⁢

(

V

)

. So

f

⁢

(

U

)

=

V

. Let

W

=

U

¯

. Then

f

⁢

(

W

)

=

f

⁢

(

U

¯

)

⊆

f

⁢

(

U

)

¯

=

V

¯

=

V

. So

W

⊆

f

-

1

⁢

f

⁢

(

W

)

⊆

f

-

1

⁢

(

V

)

=

U

⊆

U

¯

=

W

. As a result,

U

=

W

is closed.

•

(

3

)

⇔

(

4

)

. First, assume

(

2

)

. Let

B

⊆

Y

and

A

=

f

-

1

⁢

(

B

)

. So

f

⁢

(

A

)

⊆

B

. Then

f

⁢

(

A

¯

)

⊆

f

⁢

(

A

)

¯

⊆

B

¯

. As a result,

f

-

1

⁢

(

B

)

¯

=

A

¯

⊆

f

-

1

⁢

f

⁢

(

A

¯

)

⊆

f

-

1

⁢

(

B

¯

)

.

Conversely, assume

(

3

)

. Let

A

⊆

X

and

B

=

f

⁢

(

A

)

. So

A

⊆

f

-

1

⁢

(

B

)

. Then

f

⁢

(

A

¯

)

⊆

f

⁢

(

f

-

1

⁢

(

B

)

¯

)

⊆

f

⁢

f

-

1

⁢

(

B

¯

)

⊆

B

¯

=

f

⁢

(

A

)

¯

.

•

(

4

)

⇔

(

5

)

. First, assume

(

3

)

. We use the identity:

C

∘

=

Y

-

Y

-

C

¯

. Then

f

-

1

⁢

(

C

∘

)

=

f

-

1

⁢

(

Y

-

Y

-

C

¯

)

=

f

-

1

⁢

(

Y

)

-

f

-

1

⁢

(

Y

-

C

¯

)

⊆

X

-

f

-

1

⁢

(

Y

-

C

)

¯

=

X

-

f

-

1

⁢

(

Y

)

-

f

-

1

⁢

(

C

)

¯

=

X

-

X

-

f

-

1

⁢

(

C

)

¯

=

f

-

1

⁢

(

C

)

∘

.

Conversely, assume

(

4

)

. We use the identity

B

¯

=

Y

-

(

Y

-

B

)

∘

. Then

f

-

1

⁢

(

B

)

¯

=

X

-

(

X

-

f

-

1

⁢

(

B

)

)

∘

=

X

-

f

-

1

⁢

(

Y

-

B

)

∘

⊆

X

-

f

-

1

⁢

(

(

Y

-

B

)

∘

)

=

f

-

1

⁢

(

Y

-

(

Y

-

B

)

∘

)

=

f

-

1

⁢

(

B

¯

)

.

∎

Title

testing for continuity via closure operation

Canonical name

TestingForContinuityViaClosureOperation

Date of creation

2013-03-22 19:09:11

Last modified on

2013-03-22 19:09:11

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

8

Author

CWoo (3771)

Entry type

Result

Classification

msc 26A15

Classification

msc 54C05

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