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Homeomorphic spaces are uniformly isomorphic

1

$begingroup$

A continuous function $f$ is a homeomorphism if it is bijective, and open.

A uniformly continuous function $f$ is a uniform isomorphism if it is bijective and $f^{-1}$ is uniformly continuous.

Is it true that a homeomorphism is a uniform isomorphism? I know the converse is true since each uniformly continuous function is continuous.

I think it doesn't always hold since not every continuous function is uniformly continuous, which doesn't guarantee that a continuous function $f^{-1}$ is uniformly continuous. But I can't think of a counterexample.

Please help.

general-topology examples-counterexamples uniform-continuity uniform-spaces

share|cite|improve this question

edited Mar 24 at 11:19

José Carlos Santos

175k24134243

asked Mar 24 at 11:16

PercyPercy

808

$endgroup$

add a comment | 

1

$begingroup$

A continuous function $f$ is a homeomorphism if it is bijective, and open.

A uniformly continuous function $f$ is a uniform isomorphism if it is bijective and $f^{-1}$ is uniformly continuous.

Is it true that a homeomorphism is a uniform isomorphism? I know the converse is true since each uniformly continuous function is continuous.

I think it doesn't always hold since not every continuous function is uniformly continuous, which doesn't guarantee that a continuous function $f^{-1}$ is uniformly continuous. But I can't think of a counterexample.

Please help.

general-topology examples-counterexamples uniform-continuity uniform-spaces

share|cite|improve this question

edited Mar 24 at 11:19

José Carlos Santos

175k24134243

asked Mar 24 at 11:16

PercyPercy

808

$endgroup$

add a comment | 

$begingroup$

A continuous function $f$ is a homeomorphism if it is bijective, and open.

A uniformly continuous function $f$ is a uniform isomorphism if it is bijective and $f^{-1}$ is uniformly continuous.

Is it true that a homeomorphism is a uniform isomorphism? I know the converse is true since each uniformly continuous function is continuous.

I think it doesn't always hold since not every continuous function is uniformly continuous, which doesn't guarantee that a continuous function $f^{-1}$ is uniformly continuous. But I can't think of a counterexample.

Please help.

general-topology examples-counterexamples uniform-continuity uniform-spaces

share|cite|improve this question

edited Mar 24 at 11:19

José Carlos Santos

175k24134243

asked Mar 24 at 11:16

PercyPercy

808

$endgroup$

share|cite|improve this question

edited Mar 24 at 11:19

José Carlos Santos

175k24134243

asked Mar 24 at 11:16

PercyPercy

808

share|cite|improve this question

edited Mar 24 at 11:19

José Carlos Santos

175k24134243

asked Mar 24 at 11:16

PercyPercy

808

edited Mar 24 at 11:19

José Carlos Santos

175k24134243

edited Mar 24 at 11:19

José Carlos Santos

175k24134243

asked Mar 24 at 11:16

PercyPercy

808

asked Mar 24 at 11:16

PercyPercy

808

3 Answers
3

active

oldest

votes

0

$begingroup$

The map$$begin{array}{ccc}mathbb R&longrightarrow&left(-fracpi2,fracpi2right)\x&mapsto&arctan xend{array}$$is such a counterexample. It's a homeomorphism, but the inverse is not uniformly continuous.

share|cite|improve this answer

answered Mar 24 at 11:19

José Carlos SantosJosé Carlos Santos

175k24134243

$endgroup$

add a comment | 

1

$begingroup$

In topology we have a notion of open map, which allows us to reformulate the condition of "$f^{-1}$ is continuous" (for a bijection) as "$f$ is open". The formal definition is the category-theory one: $f$ must be an isomorphism in the category $mathrm{Top}$, which means that there is an inverse that is also continuous and the uniform notion is defined in the same way. There is no notion (that I have seen) of "uniformly open map" that allows us to reformulate a uniform isomorphism as a bijective uniformly continuous map that is uniformly open, e.g.

share|cite|improve this answer

edited Mar 24 at 15:55

answered Mar 24 at 12:37

Henno BrandsmaHenno Brandsma

117k349127

$endgroup$

add a comment | 

0

$begingroup$

$x to x^{2}$ is a homeomorphsm of $(0,infty)$ but it is not uniformly continuous.

share|cite|improve this answer

answered Mar 24 at 11:33

Kavi Rama MurthyKavi Rama Murthy

75.3k53270

$endgroup$

add a comment | 

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0

$begingroup$

The map$$begin{array}{ccc}mathbb R&longrightarrow&left(-fracpi2,fracpi2right)\x&mapsto&arctan xend{array}$$is such a counterexample. It's a homeomorphism, but the inverse is not uniformly continuous.

share|cite|improve this answer

answered Mar 24 at 11:19

José Carlos SantosJosé Carlos Santos

175k24134243

$endgroup$

add a comment | 

0

$begingroup$

The map$$begin{array}{ccc}mathbb R&longrightarrow&left(-fracpi2,fracpi2right)\x&mapsto&arctan xend{array}$$is such a counterexample. It's a homeomorphism, but the inverse is not uniformly continuous.

share|cite|improve this answer

answered Mar 24 at 11:19

José Carlos SantosJosé Carlos Santos

175k24134243

$endgroup$

add a comment | 

$begingroup$

The map$$begin{array}{ccc}mathbb R&longrightarrow&left(-fracpi2,fracpi2right)\x&mapsto&arctan xend{array}$$is such a counterexample. It's a homeomorphism, but the inverse is not uniformly continuous.

share|cite|improve this answer

answered Mar 24 at 11:19

José Carlos SantosJosé Carlos Santos

175k24134243

$endgroup$

share|cite|improve this answer

answered Mar 24 at 11:19

José Carlos SantosJosé Carlos Santos

175k24134243

answered Mar 24 at 11:19

José Carlos SantosJosé Carlos Santos

175k24134243

answered Mar 24 at 11:19

José Carlos SantosJosé Carlos Santos

175k24134243

1

$begingroup$

In topology we have a notion of open map, which allows us to reformulate the condition of "$f^{-1}$ is continuous" (for a bijection) as "$f$ is open". The formal definition is the category-theory one: $f$ must be an isomorphism in the category $mathrm{Top}$, which means that there is an inverse that is also continuous and the uniform notion is defined in the same way. There is no notion (that I have seen) of "uniformly open map" that allows us to reformulate a uniform isomorphism as a bijective uniformly continuous map that is uniformly open, e.g.

share|cite|improve this answer

edited Mar 24 at 15:55

answered Mar 24 at 12:37

Henno BrandsmaHenno Brandsma

117k349127

$endgroup$

add a comment | 

1

$begingroup$

In topology we have a notion of open map, which allows us to reformulate the condition of "$f^{-1}$ is continuous" (for a bijection) as "$f$ is open". The formal definition is the category-theory one: $f$ must be an isomorphism in the category $mathrm{Top}$, which means that there is an inverse that is also continuous and the uniform notion is defined in the same way. There is no notion (that I have seen) of "uniformly open map" that allows us to reformulate a uniform isomorphism as a bijective uniformly continuous map that is uniformly open, e.g.

share|cite|improve this answer

edited Mar 24 at 15:55

answered Mar 24 at 12:37

Henno BrandsmaHenno Brandsma

117k349127

$endgroup$

add a comment | 

$begingroup$

In topology we have a notion of open map, which allows us to reformulate the condition of "$f^{-1}$ is continuous" (for a bijection) as "$f$ is open". The formal definition is the category-theory one: $f$ must be an isomorphism in the category $mathrm{Top}$, which means that there is an inverse that is also continuous and the uniform notion is defined in the same way. There is no notion (that I have seen) of "uniformly open map" that allows us to reformulate a uniform isomorphism as a bijective uniformly continuous map that is uniformly open, e.g.

share|cite|improve this answer

edited Mar 24 at 15:55

answered Mar 24 at 12:37

Henno BrandsmaHenno Brandsma

117k349127

$endgroup$

share|cite|improve this answer

edited Mar 24 at 15:55

answered Mar 24 at 12:37

Henno BrandsmaHenno Brandsma

117k349127

answered Mar 24 at 12:37

Henno BrandsmaHenno Brandsma

117k349127

answered Mar 24 at 12:37

Henno BrandsmaHenno Brandsma

117k349127

0

$begingroup$

$x to x^{2}$ is a homeomorphsm of $(0,infty)$ but it is not uniformly continuous.

share|cite|improve this answer

answered Mar 24 at 11:33

Kavi Rama MurthyKavi Rama Murthy

75.3k53270

$endgroup$

add a comment | 

0

$begingroup$

$x to x^{2}$ is a homeomorphsm of $(0,infty)$ but it is not uniformly continuous.

share|cite|improve this answer

answered Mar 24 at 11:33

Kavi Rama MurthyKavi Rama Murthy

75.3k53270

$endgroup$

add a comment | 

$begingroup$

$x to x^{2}$ is a homeomorphsm of $(0,infty)$ but it is not uniformly continuous.

share|cite|improve this answer

answered Mar 24 at 11:33

Kavi Rama MurthyKavi Rama Murthy

75.3k53270

$endgroup$

share|cite|improve this answer

answered Mar 24 at 11:33

Kavi Rama MurthyKavi Rama Murthy

75.3k53270

answered Mar 24 at 11:33

Kavi Rama MurthyKavi Rama Murthy

75.3k53270

answered Mar 24 at 11:33

Kavi Rama MurthyKavi Rama Murthy

75.3k53270