intervals are connected

intervals are connected

We wish to show that

intervals

(with

standard topology

) are

connected

. In order to this, we will prove that the space of

real numbers

ℝ

is connected. First we need a lemma.

Let

(

X

,

d

)

be a

metric space

. Recall that for

x

∈

X

and

r

∈

ℝ

+

we have

B

⁢

(

x

,

r

)

=

{

y

∈

X

|

d

⁢

(

x

,

y

)

<

r

}

.

Lemma

. Let

(

X

,

d

)

be a metric space and

R

⊂

ℝ

+

such that

R

is nonempty and

bounded

. Then for any

x

∈

X

we have

⋃

r

∈

R

B

⁢

(

x

,

r

)

=

B

⁢

(

x

,

sup

⁢

(

R

)

)

.

Proof

. Assume that

y

∈

⋃

r

∈

R

B

⁢

(

x

,

r

)

. Then there is

r

0

∈

R

such that

d

⁢

(

x

,

y

)

<

r

0

and thus

d

⁢

(

x

,

y

)

<

sup

⁢

(

R

)

, so

y

∈

B

⁢

(

x

,

sup

⁢

(

R

)

)

.

Now assume that

y

∈

B

⁢

(

x

,

sup

⁢

(

R

)

)

. Then

d

⁢

(

x

,

y

)

<

sup

⁢

(

R

)

and it follows (from the

definition

of

supremum

) that there is

r

0

∈

R

such that

d

⁢

(

x

,

y

)

<

r

0

and therefore

y

∈

B

⁢

(

x

,

r

0

)

⊂

⋃

r

∈

R

B

⁢

(

x

,

r

)

, which

completes

the proof.

□

Proposition

. The space of real numbers is connected.

Proof

. Assume that

U

,

V

⊆

ℝ

are

open subsets

of

ℝ

such that

U

∩

V

=

∅

and

U

∪

V

=

ℝ

. Furthermore assume that

U

≠

∅

and take any

x

0

∈

U

. Then (since

U

is open) there is

r

0

∈

ℝ

such that the open ball

B

⁢

(

x

0

,

r

0

)

=

{

x

∈

ℝ

|

|

x

-

x

0

|

<

r

0

}

is

contained

in

U

. Consider the set

R

=

{

r

∈

ℝ

+

|

B

⁢

(

x

0

,

r

)

⊆

U

}

.

Thus

R

is nonempty.

Assume that

R

is bounded. Denote by

s

=

sup

⁢

(

R

)

<

∞

. We can apply the lemma:

⋃

r

∈

R

B

⁢

(

x

0

,

r

)

=

B

⁢

(

x

0

,

s

)

.

Thus (due to the definition of

R

)

B

⁢

(

x

0

,

s

)

is a maximal open ball (with the center in

x

0

) which is contained in

U

. Now

B

⁢

(

x

0

,

s

)

=

(

a

,

b

)

for some

a

,

b

∈

ℝ

. Since

(

a

,

b

)

is maximal then

a

∉

U

or

b

∉

U

. Indeed, if both

a

∈

U

and

b

∈

U

, then (since

U

is open) small

neighbourhoods

of

a

and

b

are also contained in

U

, so

(

a

-

ϵ

,

b

+

ϵ

)

is contained in

U

(for some

ϵ

>

0

), but

(

a

,

b

)

was maximal.

Contradiction

.

Without loss of generality we can assume that

b

∉

U

. Then

b

∈

V

, because

U

∪

V

=

ℝ

. But then (since

V

is open) there is

c

∈

ℝ

such that

a

<

c

<

b

and

c

∈

V

. Thus

U

∩

V

≠

∅

. Contradiction. Therefore

R

is

unbounded

.

Take any unbounded

sequence

(

a

n

)

n

=

1

∞

from

R

. Then we have

ℝ

=

⋃

n

=

1

∞

B

⁢

(

x

0

,

a

n

)

⊆

U

and thus

U

=

ℝ

, so

V

=

∅

. This completes the proof.

□

Corollary

. For any

a

,

b

∈

ℝ

such that

a

<

b

intervals

(

a

,

b

)

,

[

a

,

b

)

,

(

a

,

b

]

and

[

a

,

b

]

are connected.

Proof

. One can easily show that intervals are continous image of

ℝ

and therefore

intervals are connected

.

Title

intervals are connected

Canonical

name

IntervalsAreConnected

Date of creation

2013-03-22 18:32:49

Last modified on

2013-03-22 18:32:49

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

5

Author

joking (16130)

Entry type

Example

Classification

msc 54D05

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