compactness is preserved under a continuous map

compactness is preserved under a continuous map

Theorem

[

1

,

2

]

Suppose

f

:

X

→

Y

is a

continuous map

between

topological spaces

X

and

Y

.
If

X

is

compact

and

f

is

surjective

, then

Y

is compact.

The

inclusion map

[

0

,

1

]

↪

[

0

,

2

)

shows that the requirement for

f

to be surjective cannot be omitted.
If

X

is compact and

f

is

continuous

we can always conclude, however, that

f

⁢

(

X

)

is compact, since

f

:

X

→

f

⁢

(

X

)

is continuous (

http://planetmath.org/IfFcolonXtoYIsContinuousThenFcolonXtoFXIsContinuous

).

Proof of theorem.

(Following

[

1

]

.)
Suppose

{

V

α

∣

α

∈

I

}

is an arbitrary

open cover

for

f

⁢

(

X

)

. Since

f

is continuous, it follows
that

{

f

-

1

⁢

(

V

α

)

∣

α

∈

I

}

is a

collection

of

open sets

in

X

.
Since

A

⊆

f

-

1

⁢

f

⁢

(

A

)

for any

A

⊆

X

,
and since the

inverse

commutes with unions
(see this page (

http://planetmath.org/InverseImage

)),
we have

X

⊆

f

-

1

⁢

f

⁢

(

X

)

=

f

-

1

⁢

(

⋃

α

∈

I

(

V

α

)

)

=

⋃

α

∈

I

f

-

1

⁢

(

V

α

)

.

Thus

{

f

-

1

⁢

(

V

α

)

∣

α

∈

I

}

is an open cover for

X

.
Since

X

is compact, there exists a finite subset

J

⊆

I

such that

{

f

-

1

⁢

(

V

α

)

∣

α

∈

J

}

is a
finite open cover for

X

.
Since

f

is a surjection, we have

f

⁢

f

-

1

⁢

(

A

)

=

A

for any

A

⊆

Y

(see this page (

http://planetmath.org/InverseImage

)). Thus

f

⁢

(

X

)

=

f

⁢

(

⋃

i

∈

J

f

-

1

⁢

(

V

α

)

)

=

f

⁢

f

-

1

⁢

⋃

i

∈

J

f

-

1

⁢

(

V

α

)

=

⋃

i

∈

J

V

α

.

Thus

{

V

α

∣

α

∈

J

}

is an open cover for

f

⁢

(

X

)

,
and

f

⁢

(

X

)

is compact.

□

A shorter proof can be given using the

characterization

of

compactness

by the

finite intersection
property

(

http://planetmath.org/ASpaceIsCompactIfAndOnlyIfTheSpaceHasTheFiniteIntersectionProperty

):

Shorter proof.

Suppose

{

A

i

∣

i

∈

I

}

is a collection of

closed
subsets

of

Y

with the finite intersection property.
Then

{

f

-

1

⁢

(

A

i

)

∣

i

∈

I

}

is a collection of closed subsets of

X

with the finite intersection property,
because if

F

⊆

I

is finite then

⋂

i

∈

F

f

-

1

⁢

(

A

i

)

=

f

-

1

⁢

(

⋂

i

∈

F

A

i

)

,

which is nonempty as

f

is a surjection.
As

X

is compact, we have

f

-

1

⁢

(

⋂

i

∈

I

A

i

)

=

⋂

i

∈

I

f

-

1

⁢

(

A

i

)

≠

∅

and so

⋂

i

∈

I

A

i

≠

∅

.
Therefore

Y

is compact.

□

References

1

I.M. Singer, J.A.Thorpe,

Lecture Notes on

Elementary

Topology and Geometry

,
Springer-Verlag, 1967.

2

J.L. Kelley,

General Topology

, D. van Nostrand Company, Inc., 1955.

3

G.J. Jameson,

Topology

and

Normed Spaces

,
Chapman and Hall, 1974.

Title

compactness is preserved under a continuous map

Canonical name

CompactnessIsPreservedUnderAContinuousMap

Date of creation

2013-03-22 13:55:50

Last modified on

2013-03-22 13:55:50

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

13

Author

yark (2760)

Entry type

Theorem

Classification

msc 54D30

Related topic

ContinuousImageOfACompactSpaceIsCompact

Related topic

ContinuousImageOfACompactSetIsCompact

Related topic

ConnectednessIsPreservedUnderAContinuousMap

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