aarjaneiro/planetmath huggingface dataset:
equivalent formulations for continuity

equivalent formulations for continuity

Suppose

f

:

X

→

Y

is a

function

between

topological spaces

X

,

Y

. Then the following are

equivalent

:

1.

f

is

continuous

.

2.

If

B

is open in

Y

, then

f

-

1

⁢

(

B

)

is open in

X

.

3.

If

B

is closed in

Y

, then

f

-

1

⁢

(

B

)

is closed in

X

.

4.

f

⁢

(

A

¯

)

⊆

f

⁢

(

A

)

¯

for all

A

⊆

X

.

5.

If

(

x

i

)

is a net in

X

converging to

x

, then

(

f

⁢

(

x

i

)

)

is a net in

Y

converging to

f

⁢

(

x

)

. The

concept

of net
can be replaced by the more familiar one of

sequence

if the spaces

X

and

Y

are

first countable

.

6.

Whenever two nets

S

and

T

in

X

converge

to the same point, then

f

∘

S

and

f

∘

T

converge to the same point in

Y

.

7.

If

𝔽

is a filter on

X

that converges to

x

, then

f

⁢

(

𝔽

)

is a filter on

Y

that converges to

f

⁢

(

x

)

. Here,

f

⁢

(

𝔽

)

is the

filter generated by

the filter base

{

f

⁢

(

F

)

∣

F

∈

𝔽

}

.

8.

If

B

is any element of a

subbase

(

http://planetmath.org/Subbasis

)

𝒮

for the

topology

of

Y

,
then

f

-

1

⁢

(

B

)

is open in

X

.

9.

If

B

is any element of a basis

ℬ

for the topology of

Y

, then

f

-

1

⁢

(

B

)

is open in

X

.

10.

If

x

∈

X

, and

N

is any

neighborhood

of

f

⁢

(

x

)

, then

f

-

1

⁢

(

N

)

is a neighborhood of

x

.

11.

f

is continuous at every point in

X

.

Title

equivalent formulations for continuity

Canonical name

EquivalentFormulationsForContinuity

Date of creation

2013-03-22 15:18:23

Last modified on

2013-03-22 15:18:23

Owner

matte (1858)

Last modified by

matte (1858)

Numerical id

14

Author

matte (1858)

Entry type

Theorem

Classification

msc 26A15

Classification

msc 54C05

Related topic

Characterization

Generated on Fri Feb 9 19:48:24 2018 by

LaTeXML
example of jump discontinuity

example of jump discontinuity

The

elementary

(

http://planetmath.org/ElementaryFunction

) real function

f

:

x

↦

1

1

+

e

1

x

has a

jump discontinuity

at the origin, since

lim

x

→

0

-

⁡

f

⁢

(

x

)

=

1

and

lim

x

→

0

+

⁡

f

⁢

(

x

)

=

0

.

Indeed,

•

if

x

→

0

-

,  then

1

x

→

-

∞

,

e

1

x

→

0

,

1

1

+

e

1

x

→

1

;

•

if

x

→

0

+

,  then

1

x

→

∞

,

e

1

x

→

∞

,

1

1

+

e

1

x

→

0

.

These results can be seen also from the series of the

function

gotten by performing the divisions:  for

x

<

0

we obtain the converging (

http://planetmath.org/

Converge

)

alternating series

(

http://planetmath.org/LeibnizEstimateForAlternatingSeries

)

1

:

(

1

+

e

1

x

)

=

∑

k

=

0

∞

(

-

1

)

k

e

k

x

=

1

-

e

1

x

+

e

2

x

-

e

3

x

+

-

…

and for

x

>

0

the series

1

:

(

e

1

x

+

1

)

=

∑

k

=

1

∞

(

-

1

)

k

+

1

e

-

k

x

=

e

-

1

x

-

e

-

2

x

+

e

-

3

x

-

+

…

Note.

The

derivative

of the function may be written as

f

′

⁢

(

x

)

=

1

x

2

⁢

(

e

-

1

x

+

1

)

⁢

(

1

+

e

1

x

)

,

and thus we have the

one-sided limits

lim

x

→

0

±

⁡

f

′

⁢

(

x

)

=

0

(see

growth of exponential function

).

Figure 1:

Graph of the function

f

with jump discontinuity

Title

example of jump discontinuity

Canonical name

ExampleOfJumpDiscontinuity

Date of creation

2013-03-22 16:25:02

Last modified on

2013-03-22 16:25:02

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

16

Author

pahio (2872)

Entry type

Example

Classification

msc 26A15

Classification

msc 54C05

Related topic

ExponentialFunction

Related topic

ImproperLimits

Generated on Fri Feb 9 19:48:27 2018 by

LaTeXML
finite limit implying uniform continuity

finite limit implying uniform continuity

Theorem.

If the

real function

f

is

continuous

on the

interval

[

0

,

∞

)

and the limit

lim

x

→

∞

⁡

f

⁢

(

x

)

exists as a finite number

a

, then

f

is

uniformly continuous

on that interval.

Proof.

Let

ε

>

0

.  According to the limit condition, there is a positive number

M

such that

|

f

⁢

(

x

)

-

a

|

<

ε

2

∀

x

>

M

.

(1)

The

function

is continuous on the finite interval

[

0

,

M

+

1

]

;  hence

f

is also uniformly continuous on this

compact

interval.  Consequently, there is a positive number

δ

<

1

such that

|

f

⁢

(

x

1

)

-

f

⁢

(

x

2

)

|

<

ε

∀

x

1

,

x

2

∈

[

0

,

M

+

1

]

⁢

with

⁢

|

x

1

-

x

2

|

<

δ

.

(2)

Let

x

1

,

x

2

be nonnegative numbers and

|

x

1

-

x

2

|

<

δ

.  Then

|

x

1

-

x

2

|

<

1

and thus both numbers either belong to

[

0

,

M

+

1

]

or are

greater

than

M

.  In the latter case, by (1) we have

|

f

⁢

(

x

1

)

-

f

⁢

(

x

2

)

|

=

|

f

⁢

(

x

1

)

-

a

+

a

-

f

⁢

(

x

2

)

|

≦

|

f

⁢

(

x

1

)

-

a

|

+

|

f

⁢

(

x

2

)

-

a

|

<

ε

2

+

ε

2

=

ε

.

(3)

So, one of the conditions (2) and (3) is always in , whence the assertion is true.

Title

finite limit implying uniform continuity

Canonical name

FiniteLimitImplyingUniformContinuity

Date of creation

2013-03-22 19:00:20

Last modified on

2013-03-22 19:00:20

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

5

Author

pahio (2872)

Entry type

Theorem

Classification

msc 26A15

Generated on Fri Feb 9 19:48:30 2018 by

LaTeXML
function continuous at only one point

function continuous at only one point

Let us show that the

function

f

:

ℝ

→

ℝ

,

f

⁢

(

x

)

=

{

x

,

when

x

is rational

,

-

x

,

when

x

is irrational

,

is

continuous at

x

=

0

, but

discontinuous

for all

x

∈

ℝ

∖

{

0

}

[

1

]

.

We shall use the following

characterization

of

continuity

for

f

:

f

is continuous at

a

∈

ℝ

if and only if

lim

k

→

∞

⁡

f

⁢

(

x

k

)

=

f

⁢

(

a

)

for all

sequences

(

x

k

)

⊂

ℝ

such that

lim

k

→

∞

⁡

x

k

=

a

.

It is not difficult to see that

f

is continuous at

x

=

0

. Indeed, if

x

k

is a
sequence converging to

0

. Then

lim

k

→

∞

⁡

|

f

⁢

(

x

k

)

|

=

lim

k

→

∞

⁡

|

f

⁢

(

x

k

)

|

=

lim

k

→

∞

⁡

|

x

k

|

=

0

.

Suppose

a

≠

0

. Then there exists a sequence of irrational numbers

x

1

,

x

2

,

…

converging to

a

. For

instance

, if

a

is irrational, we can take

x

k

=

a

+

1

/

k

, and
if

a

is

rational

, we can take

x

k

=

a

+

2

/

k

. For this sequence we have

lim

k

→

∞

⁡

f

⁢

(

x

k

)

=

-

lim

k

→

∞

⁡

x

k

=

-

a

.

On the other hand, we can also construct a sequence of

rational numbers

y

1

,

y

2

,

…

converging to

a

. For example, if

a

is irrational, this follows as the rational numbers
are dense in

ℝ

, and if

a

is rational, we can set

y

k

=

x

k

+

1

/

k

.
For this sequence we have

lim

k

→

∞

⁡

f

⁢

(

y

k

)

=

lim

k

→

∞

⁡

y

k

=

a

.

In

conclusion

f

is not continuous at

a

.

References

1

Homepage of Thomas Vogel,

http://www.math.tamu.edu/ tom.vogel/gallery/node3.html

A function which is continuous at only one point.

Title

function continuous at only one point

Canonical name

FunctionContinuousAtOnlyOnePoint

Date of creation

2013-03-22 14:56:19

Last modified on

2013-03-22 14:56:19

Owner

Andrea Ambrosio (7332)

Last modified by

Andrea Ambrosio (7332)

Numerical id

7

Author

Andrea Ambrosio (7332)

Entry type

Example

Classification

msc 26A15

Classification

msc 54C05

Related topic

DirichletsFunction

Related topic

FunctionDifferentiableAtOnlyOnePoint

Generated on Fri Feb 9 19:48:33 2018 by

LaTeXML
If f⁢m⁢MERROR(\nonscript):X→Y is continuous then f⁢m⁢MERROR(\nonscript):X→f⁢(X) is continuous

If

f

⁢

m

⁢

M

\nonscript

:

X

→

Y

is continuous then

f

⁢

m

⁢

M

\nonscript

:

X

→

f

⁢

(

X

)

is continuous

Theorem 1

.

Suppose

X

,

Y

are

topological spaces

and

f

:

X

→

Y

is a

continuous function

. Then

f

:

X

→

f

⁢

(

X

)

is

continuous

when

f

⁢

(

X

)

is equipped with
the

subspace topology

.

Proof.

Let us first note that using a
property on this page (

http://planetmath.org/InverseImage

), we have

X

=

f

-

1

⁢

f

⁢

(

X

)

.

For the proof, suppose that

A

is open
in

f

⁢

(

X

)

, that is,

A

=

U

∩

f

⁢

(

X

)

for some

open set

U

⊂

Y

. From the properties of
the

inverse image

, we have

f

-

1

⁢

(

A

)

=

f

-

1

⁢

(

U

)

∩

f

-

1

⁢

(

f

⁢

(

X

)

)

=

f

-

1

⁢

(

U

)

so

f

-

1

⁢

(

A

)

is open in

X

.
∎

Title

If

f

⁢

m

⁢

M

\nonscript

:

X

→

Y

is continuous then

f

⁢

m

⁢

M

\nonscript

:

X

→

f

⁢

(

X

)

is continuous

Canonical name

IfFcolonXtoYIsContinuousThenFcolonXtoFXIsContinuous

Date of creation

2013-03-22 15:16:28

Last modified on

2013-03-22 15:16:28

Owner

matte (1858)

Last modified by

matte (1858)

Numerical id

6

Author

matte (1858)

Entry type

Theorem

Classification

msc 26A15

Classification

msc 54C05

Related topic

ContinuityIsPreservedWhenCodomainIsExtended

Generated on Fri Feb 9 19:48:35 2018 by

LaTeXML
integral of limit function

integral of limit function

Theorem.

If a

sequence

f

1

,

f

2

,

…

of

real functions

,

continuous

on the

interval

[

a

,

b

]

,

converges uniformly

on this interval to the limit

function

f

, then

∫

a

b

f

⁢

(

x

)

⁢

𝑑

x

=

lim

n

→

∞

⁡

∫

a

b

f

n

⁢

(

x

)

⁢

𝑑

x

.

(1)

Proof.

Let

ε

>

0

.  The uniform continuity implies the existence of a positive integer

n

ε

such that

|

f

n

⁢

(

x

)

-

f

⁢

(

x

)

|

<

ε

b

-

a

∀

x

∈

[

a

,

b

]

when

⁢

n

>

n

ε

.

The function

f

is continuous (see

http://planetmath.org/node/7191

this) and thus

Riemann integrable

(

http://planetmath.org/RiemannIntegral

) (see

http://planetmath.org/node/4461

this) on the interval.  Utilising the

estimation theorem of integral

, we obtain

|

∫

a

b

f

n

⁢

(

x

)

⁢

𝑑

x

-

∫

a

b

f

⁢

(

x

)

⁢

𝑑

x

|

=

|

∫

a

b

(

f

n

⁢

(

x

)

-

f

⁢

(

x

)

)

⁢

𝑑

x

|

≦

∫

a

b

|

f

n

⁢

(

x

)

-

f

⁢

(

x

)

|

⁢

𝑑

x

<

ε

b

-

a

⁢

(

b

-

a

)

=

ε

as soon as

n

>

n

ε

.  Consequently, (1) is true.

Remark 1.

The equation (1) may be written in the form

∫

a

b

lim

n

→

∞

⁡

f

n

⁢

(

x

)

⁢

d

⁢

x

=

lim

n

→

∞

⁡

∫

a

b

f

n

⁢

(

x

)

⁢

𝑑

x

,

(2)

i.e. under the assumptions of the theorem, the integration and the limit process can be interchanged.

Remark 2.

Considering the

partial sums

of a series

∑

n

=

1

∞

f

n

⁢

(

x

)

with continuous terms and converging uniformly on

[

a

,

b

]

,  one gets from the theorem the result analogous to (2):

∫

a

b

∑

n

=

1

∞

f

n

⁢

(

x

)

⁢

d

⁢

x

=

∑

n

=

1

∞

∫

a

b

f

n

⁢

(

x

)

⁢

𝑑

x

.

(3)

Title

integral of limit function

Canonical name

IntegralOfLimitFunction

Date of creation

2013-03-22 19:01:41

Last modified on

2013-03-22 19:01:41

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

10

Author

pahio (2872)

Entry type

Theorem

Classification

msc 26A15

Classification

msc 40A30

Related topic

TermwiseDifferentiation

Generated on Fri Feb 9 19:48:37 2018 by

LaTeXML
integral over a period interval

integral over a period interval

Theorem.

If the real function

f

is

periodic

and

integrable

(

http://planetmath.org/RiemannIntegrable

) over a period (

http://planetmath.org/Periodic

)

interval

, the value of integral over a period interval is always the same, i.e.

∫

a

a

+

p

f

⁢

(

x

)

⁢

𝑑

x

=

∫

0

p

f

⁢

(

x

)

⁢

𝑑

x

∀

a

∈

ℝ

(1)

where

p

is the period of

f

.

Proof.

The

right hand side

of the equation (1) is manipulated, with one substitution (

http://planetmath.org/ChangeOfVariableInDefiniteIntegral

)

x

=

t

+

p

:

∫

0

p

f

⁢

(

x

)

⁢

𝑑

x

=

∫

0

a

f

⁢

(

x

)

⁢

𝑑

x

+

∫

a

p

f

⁢

(

x

)

⁢

𝑑

x

=

∫

0

a

f

⁢

(

x

)

⁢

𝑑

x

+

∫

a

a

+

p

f

⁢

(

x

)

⁢

𝑑

x

-

∫

p

a

+

p

f

⁢

(

x

)

⁢

𝑑

x

=

∫

0

a

f

⁢

(

x

)

⁢

𝑑

x

+

∫

a

a

+

p

f

⁢

(

x

)

⁢

𝑑

x

-

∫

0

a

f

⁢

(

t

+

p

)

⁢

𝑑

t

=

∫

0

a

f

⁢

(

x

)

⁢

𝑑

x

+

∫

a

a

+

p

f

⁢

(

x

)

⁢

𝑑

x

-

∫

0

a

f

⁢

(

t

)

⁢

𝑑

t

=

∫

a

a

+

p

f

⁢

(

x

)

⁢

𝑑

x

References

1

Ernst Lindelöf

:

Johdatus korkeampaan analyysiin

. Fourth edition. Werner Söderström Osakeyhtiö, Porvoo ja Helsinki (1956).

2

Fråga Lund om matematik

,

http://www.maths.lth.se/query/

here.

Title

integral over a period interval

Canonical name

IntegralOverAPeriodInterval

Date of creation

2013-03-22 18:43:57

Last modified on

2013-03-22 18:43:57

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

7

Author

pahio (2872)

Entry type

Theorem

Classification

msc 26A15

Classification

msc 26A42

Synonym

integral over a period

Synonym

integral of periodic function

Related topic

DefiniteIntegral

Related topic

IntegralsOfEvenAndOddFunctions

Generated on Fri Feb 9 19:48:40 2018 by

LaTeXML
least and greatest zero

least and greatest zero

Theorem

.

If a real function

f

is

continuous

on the

interval

[

a

,

b

]

and has zeroes on this interval, then

f

has a least zero and a greatest zero.

Proof.

If

f

⁢

(

a

)

=

0

then the assertion concerning the least zero is true.  Let’s assume therefore, that

f

⁢

(

a

)

≠

0

.

The set

A

=

{

x

∈

[

a

,

b

]

⋮

f

(

x

)

=

0

}

is

bounded from below

since all numbers of

A

are

greater

than

a

.  Let the

infimum

(

http://planetmath.org/InfimumAndSupremumForRealNumbers

) of

A

be

ξ

.  Let us make the antithesis, that

f

⁢

(

ξ

)

≠

0

.  Then, by the

continuity

of

f

, there is a positive number

δ

such that

f

⁢

(

x

)

≠

0

always

⁢

when

⁢

|

x

-

ξ

|

<

δ

.

Chose a number

x

1

between

ξ

and

ξ

+

δ

; then

f

⁢

(

x

1

)

≠

0

,  but this number

x

1

is not a

lower bound

of

A

.  Therefore there exists a member

a

1

of

A

which is less than

x

1

(

ξ

<

a

1

<

x

1

).  Now

|

a

1

-

ξ

|

<

|

x

1

-

ξ

|

<

δ

,  whence this member of

A

ought to

satisfy

that

f

⁢

(

a

1

)

=

0

.  This a

contradiction

.  Thus the antithesis is wrong, and

f

⁢

(

ξ

)

=

0

.

This that

ξ

∈

A

and

ξ

is the least number of

A

.

Analogically one shows that the

supremum

of

A

is the greatest zero of

f

on the interval.

Title

least and greatest zero

Canonical name

LeastAndGreatestZero

Date of creation

2013-03-22 16:33:22

Last modified on

2013-03-22 16:33:22

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

6

Author

pahio (2872)

Entry type

Theorem

Classification

msc 26A15

Synonym

zeroes of continuous function

Related topic

ZeroesOfAnalyticFunctionsAreIsolated

Generated on Fri Feb 9 19:48:42 2018 by

LaTeXML
limit function of sequence

limit function of sequence

Theorem 1

.

Let

f

1

,

f

2

,

…

be a

sequence

of

real functions

all defined in the

interval

[

a

,

b

]

.  This

function

sequence

converges uniformly

to the

limit function

f

on the interval

[

a

,

b

]

if and only if

lim

n

→

∞

sup

{

|

f

n

(

x

)

-

f

(

x

)

|

⋮

a

≦

x

≦

b

}

=

0

.

If all functions

f

n

are

continuous

in the interval

[

a

,

b

]

and

lim

n

→

∞

⁡

f

n

⁢

(

x

)

=

f

⁢

(

x

)

in all points

x

of the interval, the limit function needs not to be continuous in this interval; example

f

n

⁢

(

x

)

=

sin

n

⁡

x

in

[

0

,

π

]

:

Theorem 2

.

If all the functions

f

n

are continuous and the sequence

f

1

,

f

2

,

…

converges uniformly to a function

f

in the interval

[

a

,

b

]

,  then the limit function

f

is continuous in this interval.

Note.

The notion of can be extended to the sequences of complex functions (the interval is replaced with some subset

G

of

ℂ

).  The limit function of a

uniformly convergent

sequence of

continuous functions

is continuous in

G

.

Title

limit function of sequence

Canonical name

LimitFunctionOfSequence

Date of creation

2013-03-22 14:37:45

Last modified on

2013-03-22 14:37:45

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

22

Author

pahio (2872)

Entry type

Theorem

Classification

msc 26A15

Classification

msc 40A30

Related topic

LimitOfAUniformlyConvergentSequenceOfContinuousFunctionsIsContinuous

Related topic

PointPreventingUniformConvergence

Defines

function sequence

Defines

limit function

Generated on Fri Feb 9 19:48:45 2018 by

LaTeXML
non-uniformly continuous function

non-uniformly continuous function

We assert that the

real function

x

↦

sin

⁡

1

x

is not

uniformly continuous

on the

open interval

(

0

,

1

)

.

For proving this, we make the antithesis that there exists a positive number

δ

such that

|

f

⁢

(

x

1

)

-

f

⁢

(

x

2

)

|

<

1

always when

x

1

,

x

2

∈

(

0

,

1

)

⁢

and

⁢

|

x

1

-

x

2

|

<

δ

.

Choose

x

1

=

1

π

2

+

2

⁢

n

⁢

π

,

x

2

=

1

3

⁢

π

2

+

2

⁢

n

⁢

π

where the integer

n

is so great that

x

1

<

δ

2

,

x

2

<

δ

2

.  Then we have

|

x

1

-

x

2

|

≦

|

x

1

|

+

|

x

2

|

<

δ

.

However,

f

⁢

(

x

1

)

-

f

⁢

(

x

2

)

=

1

-

(

-

1

)

=

2

.

This contradictory result shows that the antithesis is wrong.

Title

non-uniformly continuous function

Canonical name

NonuniformlyContinuousFunction

Date of creation

2013-03-22 19:00:07

Last modified on

2013-03-22 19:00:07

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

9

Author

pahio (2872)

Entry type

Example

Classification

msc 26A15

Related topic

PointPreventingUniformConvergence

Related topic

ReductioAdAbsurdum

Generated on Fri Feb 9 19:48:56 2018 by

LaTeXML
polynomial equation of odd degree

polynomial equation of odd degree

Theorem

.

The equation

a

0

⁢

x

n

+

a

1

⁢

x

n

-

1

+

⋯

+

a

n

-

1

⁢

x

+

a

n

=

0

(1)

with odd degree

n

and real

coefficients

a

i

(

a

0

≠

0

) has at least one real root

x

.

Proof.

Denote by

f

⁢

(

x

)

the left hand side of (1).  We can write

f

⁢

(

x

)

=

a

0

⁢

x

n

⁢

[

1

+

g

⁢

(

x

)

]

where

g

⁢

(

x

)

:=

a

1

x

+

⋯

+

a

n

-

1

x

n

-

1

+

a

n

x

n

.  But we have

lim

|

x

|

→

∞

⁡

g

⁢

(

x

)

=

0

because

lim

|

x

|

→

∞

⁡

a

i

x

i

=

0

for all

i

=

1

,

…

,

n

.  Thus there exists an

M

>

0

such that

|

g

⁢

(

x

)

|

<

1

⁢

for

⁢

|

x

|

≧

M

.

Accordingly

1

+

g

⁢

(

±

M

)

>

0

and

sign

⁢

f

⁢

(

±

M

)

=

(

sign

⁢

a

0

)

⁢

(

sign

⁢

(

±

M

)

)

n

⋅

1

=

(

sign

⁢

a

0

)

⁢

(

±

1

)

since

n

is odd.  Therefore the real polynomial function

f

has opposite signs in the

end points

of the

interval

[

-

M

,

M

]

.  Thus the

continuity

of

f

guarantees, according to Bolzano’s theorem, at least one zero

x

of

f

in that interval.  So (1) has at least one real root

x

.

Title

polynomial equation of odd degree

Canonical name

PolynomialEquationOfOddDegree

Date of creation

2013-03-22 15:39:19

Last modified on

2013-03-22 15:39:19

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

7

Author

pahio (2872)

Entry type

Theorem

Classification

msc 26A15

Classification

msc 26A09

Classification

msc 12D10

Classification

msc 26C05

Related topic

AlgebraicEquation

Related topic

ExampleOfSolvingACubicEquation

Generated on Fri Feb 9 19:49:01 2018 by

LaTeXML
semi-continuous

semi-continuous

A real function

f

:

A

→

ℝ

, where

A

⊆

ℝ

is said to be

lower semi-continuous

in

x

0

if

∀

ε

>

0

⁢

∃

δ

>

0

⁢

∀

x

∈

A

⁢

|

x

-

x

0

|

<

δ

⇒

f

⁢

(

x

)

>

f

⁢

(

x

0

)

-

ε

,

and

f

is said to be

upper semi-continuous

if

∀

ε

>

0

⁢

∃

δ

>

0

⁢

∀

x

∈

A

⁢

|

x

-

x

0

|

<

δ

⇒

f

⁢

(

x

)

<

f

⁢

(

x

0

)

+

ε

.

Remark

A real function is

continuous

in

x

0

if and only if it is both upper and lower semicontinuous in

x

0

.

We can generalize the definition to arbitrary

topological spaces

as follows.

Let

A

be a topological space.

f

:

A

→

ℝ

is lower semicontinuous at

x

0

if, for each

ε

>

0

there is a

neighborhood

U

of

x

0

such that

x

∈

U

implies

f

⁢

(

x

)

>

f

⁢

(

x

0

)

-

ε

.

Theorem

Let

f

:

[

a

,

b

]

→

ℝ

be a lower (upper) semi-continuous function. Then

f

has a

minimum

(

maximum

) in

[

a

,

b

]

.

Title

semi-continuous

Canonical name

Semicontinuous

Date of creation

2013-03-22 12:45:41

Last modified on

2013-03-22 12:45:41

Owner

drini (3)

Last modified by

drini (3)

Numerical id

6

Author

drini (3)

Entry type

Definition

Classification

msc 26A15

Classification

msc 54-XX

Synonym

semicontinuous

Generated on Fri Feb 9 19:49:03 2018 by

LaTeXML
semicontinuous

semicontinuous

Suppose

X

is a

topological space

, and

f

is a

function

from

X

into the

extended real numbers

ℝ

*

;

f

:

X

→

ℝ

*

.
Then:

1.

If

f

-

1

⁢

(

(

α

,

∞

]

)

=

{

x

∈

X

∣

f

⁢

(

x

)

>

α

}

is an

open set

in

X

for all

α

∈

ℝ

,
then

f

is said to be

lower semicontinuous

.

2.

If

f

-

1

⁢

(

[

-

∞

,

α

)

)

=

{

x

∈

X

∣

f

⁢

(

x

)

<

α

}

is an open set in

X

for all

α

∈

ℝ

,
then

f

is said to be

upper semicontinuous

.

In other words,

f

is lower semicontinuous, if

f

is

continuous

with
respect to the

topology

for

ℝ

*

containing

∅

and
open sets

U

⁢

(

α

)

=

(

α

,

∞

]

,

α

∈

ℝ

∪

{

-

∞

}

.

It is not difficult to see that this is a topology. For example,
for a union of sets

U

⁢

(

α

i

)

we have

∪

i

U

⁢

(

α

i

)

=

U

⁢

(

inf

⁡

α

i

)

.
Obviously, this topology is much

coarser

than
the

usual topology

for the extended numbers.
However,
the sets

U

⁢

(

α

)

can be seen as

neighborhoods

of

infinity

, so
in some sense, semicontinuous functions are ”continuous at infinity”
(see example 3 below).

0.0.1

Examples

1.

A function

f

:

X

→

ℝ

*

is continuous if and only if
it is lower and upper semicontinuous.

2.

Let

f

be the

characteristic function

of a set

Ω

⊆

X

.
Then

f

is lower (upper)

semicontinuous

if and only if

Ω

is open (closed).
This also holds for the function that
equals

∞

in the set and

0

outside.

It follows that the characteristic function of

ℚ

is not
semicontinuous.

3.

On

ℝ

, the function

f

⁢

(

x

)

=

1

/

x

for

x

≠

0

and

f

⁢

(

0

)

=

0

, is not
semicontinuous. This example illustrate how semicontinuous ”at infinity”.

0.0.2

Properties

Let

f

:

X

→

ℝ

*

be a function.

1.

Restricting

f

to a

subspace

preserves

semicontinuity.

2.

Suppose

f

is upper (lower) semicontinuous,

A

is a topological space, and

Ψ

:

A

→

X

is a

homeomorphism

. Then

f

∘

Ψ

is upper (lower) semicontinuous.

3.

Suppose

f

is upper (lower) semicontinuous, and

S

:

ℝ

*

→

ℝ

*

is a sense preserving homeomorphism.
Then

S

∘

f

is upper (lower) semicontinuous.

4.

f

is lower semicontinuous if and only if

-

f

is upper semicontinuous.

References

1

W. Rudin,

Real and complex analysis

, 3rd ed., McGraw-Hill Inc., 1987.

2

D.L. Cohn,

Measure Theory

, Birkhäuser, 1980.

Title

semicontinuous

Canonical

name

Semicontinuous1

Date of creation

2013-03-22 14:00:16

Last modified on

2013-03-22 14:00:16

Owner

bwebste (988)

Last modified by

bwebste (988)

Numerical id

13

Author

bwebste (988)

Entry type

Definition

Classification

msc 26A15

Defines

lower semicontinuous

Defines

upper semicontinuous

Defines

lower semi-continuous

Defines

upper semi-continuous

Generated on Fri Feb 9 19:49:06 2018 by

LaTeXML