Mixing
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induced homeomorphism between $mathbb R^n$ and $mathbb R^m$ from the hypothetical existence of a homeomorphism between respective open subsets
Clash Royale CLAN TAG#URR8PPP
up vote
1
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I know that $mathbb R^n$ and $mathbb R^m$ are not homeomorphic, and from this fact I want to show there is no homeomorphism between any pair of respective open subsets reasoning by contraction. Does it have a simple argument or should I use a specific theorem?
real-analysis general-topology
asked Jul 31 at 17:29
Selflearner
157110
add a comment |Â
up vote
1
down vote
favorite
I know that $mathbb R^n$ and $mathbb R^m$ are not homeomorphic, and from this fact I want to show there is no homeomorphism between any pair of respective open subsets reasoning by contraction. Does it have a simple argument or should I use a specific theorem?
real-analysis general-topology
asked Jul 31 at 17:29
Selflearner
157110
3
Hint: Any open set in $mathbbR^n$ contains an open ball, which is itself homeomorphic to $mathbbR^n$.
– Daniel Mroz
Jul 31 at 17:32
@DanielMroz but this only shows that $mathbb R^n$ is homeomorphic to an open subset of $mathbb R^m$ which in turn may not be homeomorphic to $mathbb R^m$, do I miss a point?
– Selflearner
Jul 31 at 18:11
@selflearner: the open subset of $Bbb R^m$ contains a ball homeomorphic to $Bbb R^m$. You can look at restrictions to concoct a homeomorphism between $Bbb R^n$ and $Bbb R^m$.
– Clayton
Jul 31 at 18:25
@Clayton restrictions to open balls in the domain lead to images that may not be open balls themselves
– Selflearner
Jul 31 at 18:38
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I know that $mathbb R^n$ and $mathbb R^m$ are not homeomorphic, and from this fact I want to show there is no homeomorphism between any pair of respective open subsets reasoning by contraction. Does it have a simple argument or should I use a specific theorem?
real-analysis general-topology
asked Jul 31 at 17:29
Selflearner
157110
I know that $mathbb R^n$ and $mathbb R^m$ are not homeomorphic, and from this fact I want to show there is no homeomorphism between any pair of respective open subsets reasoning by contraction. Does it have a simple argument or should I use a specific theorem?
real-analysis general-topology
asked Jul 31 at 17:29
Selflearner
157110
asked Jul 31 at 17:29
Selflearner
157110
asked Jul 31 at 17:29
Selflearner
157110
asked Jul 31 at 17:29
Selflearner
157110
157110
3
Hint: Any open set in $mathbbR^n$ contains an open ball, which is itself homeomorphic to $mathbbR^n$.
– Daniel Mroz
Jul 31 at 17:32
@DanielMroz but this only shows that $mathbb R^n$ is homeomorphic to an open subset of $mathbb R^m$ which in turn may not be homeomorphic to $mathbb R^m$, do I miss a point?
– Selflearner
Jul 31 at 18:11
@selflearner: the open subset of $Bbb R^m$ contains a ball homeomorphic to $Bbb R^m$. You can look at restrictions to concoct a homeomorphism between $Bbb R^n$ and $Bbb R^m$.
– Clayton
Jul 31 at 18:25
@Clayton restrictions to open balls in the domain lead to images that may not be open balls themselves
– Selflearner
Jul 31 at 18:38
add a comment |Â
3
Hint: Any open set in $mathbbR^n$ contains an open ball, which is itself homeomorphic to $mathbbR^n$.
– Daniel Mroz
Jul 31 at 17:32
@DanielMroz but this only shows that $mathbb R^n$ is homeomorphic to an open subset of $mathbb R^m$ which in turn may not be homeomorphic to $mathbb R^m$, do I miss a point?
– Selflearner
Jul 31 at 18:11
@selflearner: the open subset of $Bbb R^m$ contains a ball homeomorphic to $Bbb R^m$. You can look at restrictions to concoct a homeomorphism between $Bbb R^n$ and $Bbb R^m$.
– Clayton
Jul 31 at 18:25
@Clayton restrictions to open balls in the domain lead to images that may not be open balls themselves
– Selflearner
Jul 31 at 18:38
3
3
Hint: Any open set in $mathbbR^n$ contains an open ball, which is itself homeomorphic to $mathbbR^n$.
– Daniel Mroz
Jul 31 at 17:32
Hint: Any open set in $mathbbR^n$ contains an open ball, which is itself homeomorphic to $mathbbR^n$.
– Daniel Mroz
Jul 31 at 17:32
@DanielMroz but this only shows that $mathbb R^n$ is homeomorphic to an open subset of $mathbb R^m$ which in turn may not be homeomorphic to $mathbb R^m$, do I miss a point?
– Selflearner
Jul 31 at 18:11
@DanielMroz but this only shows that $mathbb R^n$ is homeomorphic to an open subset of $mathbb R^m$ which in turn may not be homeomorphic to $mathbb R^m$, do I miss a point?
– Selflearner
Jul 31 at 18:11
@selflearner: the open subset of $Bbb R^m$ contains a ball homeomorphic to $Bbb R^m$. You can look at restrictions to concoct a homeomorphism between $Bbb R^n$ and $Bbb R^m$.
– Clayton
Jul 31 at 18:25
@selflearner: the open subset of $Bbb R^m$ contains a ball homeomorphic to $Bbb R^m$. You can look at restrictions to concoct a homeomorphism between $Bbb R^n$ and $Bbb R^m$.
– Clayton
Jul 31 at 18:25
@Clayton restrictions to open balls in the domain lead to images that may not be open balls themselves
– Selflearner
Jul 31 at 18:38
@Clayton restrictions to open balls in the domain lead to images that may not be open balls themselves
– Selflearner
Jul 31 at 18:38
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
Of course
(1) If $n ne m$, then there is no homeomorphism between an open subset of $mathbbR^n$ and an open subset of $mathbbR^m$
implies
(2) If $n ne m$, then there is no homeomorphism between $mathbbR^n$ and $mathbbR^m$.
I doubt that you can obtain (1) as an immediate corollary of (2). Both results are usually proved using the machinery of algebraic topology, but proving two results by the same method does not mean that there is simple deduction of one from the other.
A nice overview is contained in
https://terrytao.wordpress.com/tag/invariance-of-domain/
answered yesterday
Paul Frost
3,553420
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Of course
(1) If $n ne m$, then there is no homeomorphism between an open subset of $mathbbR^n$ and an open subset of $mathbbR^m$
implies
(2) If $n ne m$, then there is no homeomorphism between $mathbbR^n$ and $mathbbR^m$.
I doubt that you can obtain (1) as an immediate corollary of (2). Both results are usually proved using the machinery of algebraic topology, but proving two results by the same method does not mean that there is simple deduction of one from the other.
A nice overview is contained in
https://terrytao.wordpress.com/tag/invariance-of-domain/
answered yesterday
Paul Frost
3,553420
add a comment |Â
up vote
0
down vote
accepted
Of course
(1) If $n ne m$, then there is no homeomorphism between an open subset of $mathbbR^n$ and an open subset of $mathbbR^m$
implies
(2) If $n ne m$, then there is no homeomorphism between $mathbbR^n$ and $mathbbR^m$.
I doubt that you can obtain (1) as an immediate corollary of (2). Both results are usually proved using the machinery of algebraic topology, but proving two results by the same method does not mean that there is simple deduction of one from the other.
A nice overview is contained in
https://terrytao.wordpress.com/tag/invariance-of-domain/
answered yesterday
Paul Frost
3,553420
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Of course
(1) If $n ne m$, then there is no homeomorphism between an open subset of $mathbbR^n$ and an open subset of $mathbbR^m$
implies
(2) If $n ne m$, then there is no homeomorphism between $mathbbR^n$ and $mathbbR^m$.
I doubt that you can obtain (1) as an immediate corollary of (2). Both results are usually proved using the machinery of algebraic topology, but proving two results by the same method does not mean that there is simple deduction of one from the other.
A nice overview is contained in
https://terrytao.wordpress.com/tag/invariance-of-domain/
answered yesterday
Paul Frost
3,553420
Of course
(1) If $n ne m$, then there is no homeomorphism between an open subset of $mathbbR^n$ and an open subset of $mathbbR^m$
implies
(2) If $n ne m$, then there is no homeomorphism between $mathbbR^n$ and $mathbbR^m$.
I doubt that you can obtain (1) as an immediate corollary of (2). Both results are usually proved using the machinery of algebraic topology, but proving two results by the same method does not mean that there is simple deduction of one from the other.
A nice overview is contained in
https://terrytao.wordpress.com/tag/invariance-of-domain/
answered yesterday
Paul Frost
3,553420
answered yesterday
Paul Frost
3,553420
answered yesterday
Paul Frost
3,553420
answered yesterday
Paul Frost
3,553420
3,553420
add a comment |Â
add a comment |Â
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3
Hint: Any open set in $mathbbR^n$ contains an open ball, which is itself homeomorphic to $mathbbR^n$.
– Daniel Mroz
Jul 31 at 17:32
@DanielMroz but this only shows that $mathbb R^n$ is homeomorphic to an open subset of $mathbb R^m$ which in turn may not be homeomorphic to $mathbb R^m$, do I miss a point?
– Selflearner
Jul 31 at 18:11
@selflearner: the open subset of $Bbb R^m$ contains a ball homeomorphic to $Bbb R^m$. You can look at restrictions to concoct a homeomorphism between $Bbb R^n$ and $Bbb R^m$.
– Clayton
Jul 31 at 18:25
@Clayton restrictions to open balls in the domain lead to images that may not be open balls themselves
– Selflearner
Jul 31 at 18:38