topological condition for a set to be uncountable

topological condition for a set to be uncountable

Theorem

.

A nonempty

compact

Hausdorff space

with no

isolated points

is

uncountable

.

Proof.

Let

X

be a nonempty compact Hausdorff space with no isolated points.
To each finite

0

,

1

-

sequence

α

associate a point

x

α

and an open

neighbourhood

U

α

as follows. First, since

X

is nonempty, let

x

0

be a point of

X

. Second, since

x

0

is not isolated, let

x

1

be another point of

X

. The fact that

X

is

Hausdorff

implies that

x

0

and

x

1

can be

separated

by

open sets

. So let

U

0

and

U

1

be

disjoint

open neighborhoods of

x

0

and

x

1

respectively.

Now suppose for

induction

that

x

α

and a neighbourhood

U

α

of

x

α

have been constructed for all

α

of length less than

n

. A

0

,

1

-sequence of length

n

has the form

(

α

,

0

)

or

(

α

,

1

)

for some

α

of length

n

-

1

. Define

x

(

α

,

0

)

=

x

α

. Since

x

(

α

,

0

)

is not isolated, there is a point in

U

α

besides

x

(

α

,

0

)

; call that point

x

(

α

,

1

)

. Now apply the Hausdorff property to find disjoint open neighbourhoods

U

(

α

,

0

)

and

U

(

α

,

1

)

of

x

(

α

,

0

)

and

x

(

α

,

1

)

respectively. The neighbourhoods

U

(

α

,

0

)

and

U

(

α

,

1

)

can be chosen to be

proper subsets

of

U

α

. Proceed by induction to find an

x

α

∈

U

α

for each finite

0

,

1

-sequence

α

.

Figure 1:

The induction step:

separating

points in

U

α

.

Now define a

function

f

:

2

ω

→

X

as follows. If

α

is
eventually zero, put

f

⁢

(

α

)

=

x

α

. Otherwise, consider the sequence

(

x

(

α

0

)

,

x

(

α

0

,

α

1

)

,

x

(

α

0

,

α

1

,

α

2

)

,

…

)

of points in

X

. Since

X

is compact and Hausdorff, it is closed and

limit point compact

, so the sequence has a

limit point

in

X

. Let

f

⁢

(

α

)

be such a limit point. Observe that for each finite prefix

(

α

0

,

…

,

α

n

)

of

α

, the point

f

⁢

(

α

)

is in

U

(

α

0

,

…

,

α

n

)

.

Figure 2:

Defining

f

⁢

(

α

)

as a limit point of the

x

(

α

0

,

…

,

α

n

)

.

Suppose

α

and

β

are distinct sequences in

2

ω

. Let

n

be the first position where

α

n

≠

β

n

. Then

f

⁢

(

α

)

∈

U

(

α

0

,

…

,

α

n

)

and

f

⁢

(

β

)

∈

U

(

β

0

,

…

,

β

n

)

, and by construction

U

(

α

0

,

…

,

α

n

)

and

U

(

β

0

,

…

,

β

n

)

are disjoint. Hence

f

⁢

(

α

)

≠

f

⁢

(

β

)

, implying that

f

is an

injective function

. Since the set

2

ω

is uncountable and

f

is an injective function from

2

ω

into

X

,

X

is also uncountable.
∎

Corollary

.

The set

[

0

,

1

]

is uncountable.

Proof.

Being closed and

bounded

,

[

0

,

1

]

is compact by the

Heine-Borel Theorem

; because

[

0

,

1

]

is a

subspace

of the Hausdorff space

ℝ

, it too is Hausdorff; finally, since

[

0

,

1

]

has no isolated points, the preceding

theorem

implies that it is uncountable.
∎

Title

topological condition for a set to be uncountable

Canonical name

TopologicalConditionForASetToBeUncountable

Date of creation

2013-03-22 16:15:15

Last modified on

2013-03-22 16:15:15

Owner

mps (409)

Last modified by

mps (409)

Numerical id

15

Author

mps (409)

Entry type

Theorem

Classification

msc 54D10

Classification

msc 54A25

Classification

msc 54D30

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