the only compact metric spaces that admit a positively expansive homeomorphism are discrete spaces

the only compact metric spaces that admit a positively expansive homeomorphism are discrete spaces

Theorem

.

Let

(

X

,

d

)

be a

compact

metric space

. If there exists a

positively expansive

homeomorphism

f

:

X

→

X

, then

X

consists only of

isolated
points

, i.e.

X

is finite.

Lemma 1.

If

(

X

,

d

)

is a compact metric space and there is an expansive homeomorphism

f

:

X

→

X

such that every point is

Lyapunov stable

, then every point is asymptotically stable.

Proof.

Let

2

⁢

c

be the expansivity

constant

of

f

.
Suppose some point

x

is not asymptotically stable, and let

δ

be such
that

d

⁢

(

x

,

y

)

<

δ

implies

d

⁢

(

f

n

⁢

(

x

)

,

f

n

⁢

(

y

)

)

<

c

for all

n

∈

ℕ

. Then there exist

ϵ

>

0

, a point

y

with

d

⁢

(

x

,

y

)

<

δ

, and an

increasing sequence

{

n

k

}

such that

d

⁢

(

f

n

k

⁢

(

y

)

,

f

n

k

⁢

(

x

)

)

>

ϵ

for each

k

By

uniform expansivity

, there is

N

>

0

such that for every

u

and

v

such that

d

⁢

(

u

,

v

)

>

ϵ

there is

n

∈

ℤ

with

|

n

|

<

N

such that

d

⁢

(

f

n

⁢

(

x

)

,

f

n

⁢

(

y

)

)

>

c

.
Choose

k

so large that

n

k

>

N

. Then there is

n

with

|

n

|

<

N

such that

d

⁢

(

f

n

+

n

k

⁢

(

x

)

,

f

n

+

n

k

⁢

(

y

)

)

=

d

⁢

(

f

n

⁢

(

f

n

k

⁢

(

x

)

)

,

f

n

⁢

(

f

n

k

⁢

(

y

)

)

)

>

c

. But since

n

+

n

k

>

0

, this contradicts the choce of

δ

. Hence every point is asymptotically stable.

Lemma 2

If

(

X

,

d

)

is a compact metric space and

f

:

X

→

X

is a homeomorphism such that every point is asymptotically stable and Lyapunov stable, then

X

is finite.

Proof.

For each

x

∈

X

let

K

x

be a closed

neighborhood

of

x

such that for all

y

∈

K

x

we have

lim

n

→

∞

⁡

d

⁢

(

f

n

⁢

(

x

)

,

f

n

⁢

(

y

)

)

=

0

. We assert that

lim

n

→

∞

⁡

diam

⁡

(

f

n

⁢

(

K

x

)

)

=

0

. In fact,
if that is not the case, then there is an increasing sequence of

positive integers

{

n

k

}

, some

ϵ

>

0

and a

sequence

{

x

k

}

of points of

K

x

such that

d

⁢

(

f

n

k

⁢

(

x

)

,

f

n

k

⁢

(

x

k

)

)

>

ϵ

, and there is a

subsequence

{

x

k

i

}

converging to some point

y

∈

K

x

.

From the Lyapunov stability of

y

, we can find

δ

>

0

such that if

d

⁢

(

y

,

z

)

<

δ

, then

d

⁢

(

f

n

⁢

(

y

)

,

f

n

⁢

(

z

)

)

<

ϵ

/

2

for all

n

>

0

. In particular

d

⁢

(

f

n

k

i

⁢

(

x

k

i

)

,

f

n

k

i

⁢

(

y

)

)

<

ϵ

/

2

if

i

is large enough. But also

d

⁢

(

f

n

k

i

⁢

(

y

)

,

f

n

k

i

⁢

(

x

)

)

<

ϵ

/

2

if

i

is large enough, because

y

∈

K

x

. Thus, for large

i

, we have

d

⁢

(

f

n

k

i

⁢

(

x

k

i

)

,

f

n

k

i

⁢

(

x

)

)

<

ϵ

. That is a

contradiction

from our previous claim.

Now since

X

is compact, there are finitely many points

x

1

,

…

,

x

m

such that

X

=

⋃

i

=

1

m

K

x

i

,
so that

X

=

f

n

⁢

(

X

)

=

⋃

i

=

1

m

f

n

⁢

(

K

x

i

)

. To show that

X

=

{

x

1

,

…

,

x

m

}

, suppose there is

y

∈

X

such that

r

=

min

⁡

{

d

⁢

(

y

,

x

i

)

:

1

≤

i

≤

m

}

>

0

. Then there is

n

such that

diam

⁡

(

f

n

⁢

(

K

x

i

)

)

<

r

for

1

≤

i

≤

m

but since

y

∈

f

n

⁢

(

K

x

i

)

for some

i

, we have a contradiction.

Proof of the theorem.

Consider the sets

K

ϵ

=

{

(

x

,

y

)

∈

X

×

X

:

d

⁢

(

x

,

y

)

≥

ϵ

}

for

ϵ

>

0

and

U

=

{

(

x

,

y

)

∈

X

×

X

:

d

⁢

(

x

,

y

)

>

c

}

, where

2

⁢

c

is the expansivity constant of

f

, and let

F

:

X

×

X

→

X

×

X

be the

mapping

given by

F

⁢

(

x

,

y

)

=

(

f

⁢

(

x

)

,

f

⁢

(

y

)

)

. It is clear that

F

is a homeomorphism.
By uniform expansivity, we know that for each

ϵ

>

0

there is

N

ϵ

such that for all

(

x

,

y

)

∈

K

ϵ

, there is

n

∈

{

1

,

…

,

N

ϵ

}

such that

F

n

⁢

(

x

,

y

)

∈

U

.

We will prove that for each

ϵ

>

0

, there is

δ

>

0

such that

F

n

⁢

(

K

ϵ

)

⊂

K

δ

for all

n

∈

ℕ

. This is

equivalent

to say that every point of

X

is Lyapunov stable for

f

-

1

, and by the previous lemmas the proof will be completed.

Let

K

=

⋃

n

=

0

N

ϵ

F

n

⁢

(

K

ϵ

)

, and let

δ

0

=

min

⁡

{

d

⁢

(

x

,

y

)

:

(

x

,

y

)

∈

K

}

.
Since

K

is compact, the minimum distance

δ

0

is reached at some point of

K

; i.e. there exist

(

x

,

y

)

∈

K

ϵ

and

0

≤

n

≤

N

ϵ

such that

d

⁢

(

f

n

⁢

(

x

)

,

f

n

⁢

(

y

)

)

=

δ

0

.
Since

f

is

injective

, it follows that

δ

0

>

0

and letting

δ

=

δ

0

/

2

we have

K

⊂

K

δ

.

Given

α

∈

K

-

K

ϵ

, there is

β

∈

K

ϵ

and some

0

<

m

≤

N

ϵ

such that

α

=

F

m

⁢

(

β

)

, and

F

k

⁢

(

β

)

∉

K

ϵ

for

0

<

k

≤

m

.
Also, there is

n

with

0

<

m

<

n

≤

N

ϵ

such that

F

n

⁢

(

β

)

∈

U

⊂

K

ϵ

. Hence

m

<

N

ϵ

,
and

F

⁢

(

β

)

=

F

m

+

1

⁢

(

α

)

∈

F

m

+

1

⁢

(

K

ϵ

)

⊂

K

; On the other hand,

F

⁢

(

K

ϵ

)

⊂

K

. Therefore

F

⁢

(

K

)

⊂

K

, and inductively

F

n

⁢

(

K

)

⊂

K

for any

n

∈

ℕ

. It follows
that

F

n

⁢

(

K

ϵ

)

⊂

F

n

⁢

(

K

)

⊂

K

⊂

K

δ

for each

n

∈

ℕ

as required.

Title

the only compact metric spaces that admit a positively expansive homeomorphism are

discrete spaces

Canonical

name

TheOnlyCompactMetricSpacesThatAdmitAPositivelyExpansiveHomeomorphismAreDiscreteSpaces

Date of creation

2013-03-22 13:55:11

Last modified on

2013-03-22 13:55:11

Owner

Koro (127)

Last modified by

Koro (127)

Numerical id

10

Author

Koro (127)

Entry type

Theorem

Classification

msc 37B99

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