a space is connected under the ordered topology if and only if it is a linear continuum.

a space is connected under the ordered topology if and only if it is a linear continuum.

Let

X

be

totally ordered

by the strict

total order

<

, and let it have the

order topology

.

Suppose

X

is not a

linear continuum

. Then either

X

is not bounded-complete, or the order on

X

is not a

dense total order

.

Suppose

X

is not bounded-complete. Let

A

be a nonempty subset of

X

that is bounded above by

b

, but has no least

upper bound

. Let

U

be the set of upper bounds of

A

. If

x

∈

U

, then

x

is not a least upper bound of

A

, so there is a

z

∈

X

such that

z

<

x

and

z

is an upper bound of

A

. Then the set

{

y

∈

X

⁢

∣

y

>

⁢

z

}

is open and contains

x

. Furthermore, all of its

elements

exceed

z

, so it is a subset of

U

. Thus,

U

is open.

U

contains

b

, so it is not empty. Let

x

∈

X

∖

U

. Then

x

is not an upper bound of

A

, so there is an

a

∈

A

such that

x

<

a

. The set

{

y

∈

X

∣

y

<

a

}

is open, and contains no upper bounds of

A

, so it is a subset of

X

∖

U

. Thus

X

∖

U

contains a

neighborhood

of each of its points, and is therefore open. Since

U

and

X

∖

U

are open, X is not

connected

.

Suppose the

ordering

of

X

is not dense, so there are

a

and

b

in

X

with

a

<

b

so that there is no

c

in

X

with

a

<

c

<

b

(there is a gap between

a

and

b

). Let

U

=

{

x

∈

X

∣

x

<

b

}

and let

V

=

{

x

∈

X

∣

x

>

a

}

. Because there are no elements between

a

and

b

,

U

∩

V

=

∅

. By

transitivity

and

trichotomy

,

U

∪

V

=

X

.

U

and

V

are both open.

a

∈

U

and

b

∈

V

, so neither

U

nor

V

is empty. Thus,

U

and

V

separate

X

, so

X

is not connected.

Therefore, if

X

is connected, then

X

is a linear continuum.

Now suppose that

X

is disconnected and bounded-complete, and that

U

and

V

are (nonempty, open and closed) sets

separating

X

. Suppose that

a

∈

U

, and suppose also that there is an element

b

∈

V

such that

a

<

b

(if there is none, swap the names of

U

and

V

, or reverse the ordering).

Z

=

{

x

∈

V

∣

x

>

a

}

is open (it is the

intersection

of two

open sets

), and contains

b

(so it is not empty).

Z

is bounded below by

a

, so it has a

greatest lower bound

z

.

If

z

∈

U

, then, since

U

is open, there is an

interval

in

U

containing

z

, which must, to exclude

b

, be of the form

{

x

∣

x

<

k

}

or of the form

{

x

∣

j

<

x

<

k

}

, for some

k

and perhaps

j

. But then

k

would be a

lower bound

of

Z

, contradicting the fact that

z

is the

infimum

of

Z

.

If

z

∈

V

, then, since

V

is open, there is an interval in

V

containing

z

, which must, to exclude

a

, be of the form

{

x

∣

j

<

x

}

or of the form

{

x

∣

j

<

x

<

k

}

for some

j

and perhaps

k

. In either case,

a

<

j

<

z

and the set

W

=

{

x

∣

j

<

x

<

z

}

is a subset of

V

. If

x

∈

W

, then

x

∈

V

and

x

>

a

, so

x

∈

Z

, contradicting the fact that

z

is the infimum of

Z

. Thus, there are no

elements of

X

between

j

and

z

, so the order on

X

is not dense. This proves that if

X

is a linear continuum, then

X

is connected.

Title

a space is connected under the ordered

topology

if and only if it is a linear continuum.

Canonical name

ASpaceIsConnectedUnderTheOrderedTopologyIfAndOnlyIfItIsALinearContinuum

Date of creation

2013-03-22 17:38:38

Last modified on

2013-03-22 17:38:38

Owner

dfeuer (18434)

Last modified by

dfeuer (18434)

Numerical id

7

Author

dfeuer (18434)

Entry type

Result

Classification

msc 54B99

Classification

msc 06F30

Related topic

LinearContinuum

Related topic

OrderTopology

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