gluing together continuous functions

gluing together continuous functions

\PMlinkescapephrase

closed sets

\PMlinkescapephrase

open set

\PMlinkescapephrase

open sets

\PMlinkescapephrase

open subset

Introduction

Suppose we have a

collection

𝒮

of subsets of a

topological space

X

,
and for each

A

∈

𝒮

we have a

continuous function

f

A

:

A

→

Y

,
where

Y

is another topological space.
If the

functions

f

A

agree wherever their domains

intersect

,
then we can glue them together in the obvious way to form a new function

f

:

∪

𝒮

→

Y

.
The

theorems

in this entry
give some

sufficient

conditions for

f

to be

continuous

.

Theorems

Theorem 1

.

Let

X

and

Y

be topological spaces,
let

S

be a

locally finite collection

of closed subsets (

http://planetmath.org/ClosedSet

) of

X

,
and let

f

:

∪

S

→

Y

be a function
such that the

restriction

(

http://planetmath.org/RestrictionOfAFunction

)

f

|

A

is continuous for all

A

∈

S

.
Then

f

is continuous.

Theorem 2

.

Let

X

and

Y

be topological spaces,
let

S

be a collection of open subsets (

http://planetmath.org/OpenSet

) of

X

,
and let

f

:

∪

S

→

Y

be a function
such that the restriction

f

|

A

is continuous for all

A

∈

S

.
Then

f

is continuous.

Notes

Note that the theorem for closed subsets
requires the collection to be

locally finite

.
To see that this is condition cannot be omitted,
notice that any function

f

:

ℝ

→

ℝ

restricts to a continuous function on each singleton,
yet need not be continuous itself.

Proofs

The two theorems are proved in essentially in the same way,
but for the first theorem we need to make use of the fact that
the union of a locally finite collection of closed sets is closed.

Proof of Theorem 1.

Let

C

be a closed subset of

Y

.
Then

f

-

1

⁢

(

C

)

=

⋃

A

∈

𝒮

(

A

∩

f

-

1

⁢

(

C

)

)

=

⋃

A

∈

𝒮

(

f

|

A

)

-

1

⁢

(

C

)

.
By

continuity

, each

(

f

|

A

)

-

1

⁢

(

C

)

is closed in

A

.
But by

assumption

each

A

is closed in

X

,
so it follows that each

(

f

|

A

)

-

1

⁢

(

C

)

is closed in

X

.
Thus

f

-

1

⁢

(

C

)

is the union of a locally finite collection of closed sets,
and is therefore closed in

X

, and so closed in

∪

𝒮

.
So

f

is continuous.
∎

Proof of Theorem 2.

Let

U

be an open subset of

Y

.
Then

f

-

1

⁢

(

U

)

=

⋃

A

∈

𝒮

(

A

∩

f

-

1

⁢

(

U

)

)

=

⋃

A

∈

𝒮

(

f

|

A

)

-

1

⁢

(

U

)

.
By continuity, each

(

f

|

A

)

-

1

⁢

(

U

)

is open in

A

.
But by assumption each

A

is open in

X

,
so it follows that each

(

f

|

A

)

-

1

⁢

(

U

)

is open in

X

.
Thus

f

-

1

⁢

(

U

)

is the union of a collection of open sets,
and is therefore open in

X

, and so open in

∪

𝒮

.
So

f

is continuous.
∎

Title

gluing together continuous functions

Canonical name

GluingTogetherContinuousFunctions

Date of creation

2013-03-22 15:17:20

Last modified on

2013-03-22 15:17:20

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

16

Author

yark (2760)

Entry type

Theorem

Classification

msc 54C05

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