testing for continuity via nets

testing for continuity via nets

Proposition 1

.

Let

X

,

Y

be

topological spaces

and

f

:

X

→

Y

. Then the following are

equivalent

:

1.

f

is

continuous

;

2.

If

(

x

i

)

is a net in

X

converging to

x

, then

(

f

⁢

(

x

i

)

)

is a net in

Y

converging to

f

⁢

(

x

)

.

3.

Whenever two nets

(

x

i

)

and

(

y

j

)

in

X

converge

to the same point, then

(

f

⁢

(

x

i

)

)

and

(

f

⁢

(

y

j

)

)

converge to the same point in

Y

.

Proof.

(

1

)

⇔

(

2

)

. Let

A

be the (directed)

index set

for

i

. Suppose

f

⁢

(

x

)

∈

U

is open in

Y

. Then

x

∈

f

-

1

⁢

(

U

)

is open in

X

since

f

is continuous. By

assumption

,

(

x

i

)

is a net, so there is

b

∈

A

such that

x

j

∈

f

-

1

⁢

(

U

)

for all

j

≥

b

. This means that

f

⁢

(

x

j

)

∈

U

for all

i

≥

b

, so

(

f

⁢

(

x

i

)

)

is a net too.

Conversely

, suppose

f

is not continuous, say, at a point

x

∈

X

. Then there is an

open set

V

containing

f

⁢

(

x

)

such that

f

-

1

⁢

(

V

)

does not contain any open set containing

x

. Let

A

be the set of all open sets containing

x

. Then under reverse

inclusion

,

A

is a

directed set

(if

U

1

,

U

2

∈

A

, then

U

1

∩

U

2

∈

A

). Define a

relation

R

⊆

A

×

X

as follows:

(

U

,

x

)

∈

R

iff

x

∈

U

-

f

-

1

⁢

(

V

)

.

Then for each

U

∈

A

, there is an

x

∈

X

such that

(

U

,

x

)

∈

R

, since

U

⊈

f

-

1

⁢

(

V

)

. By the

axiom of choice

, we get a

function

d

⊆

R

from

A

to

X

. Write

d

⁢

(

U

)

:=

x

U

. Since

A

is directed,

(

x

U

)

is a net. In

addition

,

(

x

U

)

converges to

x

(just pick any

U

∈

A

, then for any

W

≥

U

, we have

x

∈

W

by the definition of

A

). However,

(

f

⁢

(

x

U

)

)

does not converge to

f

⁢

(

x

)

, since

x

U

∉

f

-

1

⁢

(

V

)

for any

U

∈

A

.

(

2

)

⇔

(

3

)

. Suppose nets

(

x

i

)

and

(

y

j

)

both converge to

z

∈

X

. Then, by assumption,

(

f

⁢

(

x

i

)

)

and

(

f

⁢

(

y

j

)

)

are nets converging to

f

⁢

(

z

)

∈

Y

.

Conversely, suppose

(

x

i

)

converges to

x

, and

i

is indexed by a directed set

A

. Define a net

(

y

i

)

such that

y

i

=

x

for all

i

∈

A

. Then

(

y

i

)

=

(

x

)

clearly converges to

x

. Hence both

(

f

⁢

(

x

i

)

)

and

(

f

⁢

(

y

i

)

)

converge to the same point in

Y

. But

(

f

⁢

(

y

i

)

)

=

(

f

⁢

(

x

)

)

converges to

f

⁢

(

x

)

, we see that

(

f

⁢

(

x

i

)

)

converges to

f

⁢

(

x

)

as well.
∎

Remark

. In particular, if

X

,

Y

are

first countable

, we may replace nets by

sequences

in the

proposition

. In other words,

f

is continuous iff it preserves converging sequences.

Title

testing for continuity via nets

Canonical name

TestingForContinuityViaNets

Date of creation

2013-03-22 19:08:58

Last modified on

2013-03-22 19:08:58

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

7

Author

CWoo (3771)

Entry type

Result

Classification

msc 54C05

Classification

msc 26A15

Related topic

Net

Related topic

FirstCountable

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