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$Bbb R cong Bbb R^n$ iff $n=1$ [on hold]

Prove that for every $n geq 2$, $Bbb R$ is not homeomorphic to $Bbb R^n$homeomorphism between the same spaces with different topologiesCan this be a way to prove that $BbbR^2$ and $BbbR^3$ are not homeomorphic?Does there exist a neighborhood around a real number on the real line whose complement is finite?Is there a topology such that $(Bbb R, +, mathcal T)$ is a compact Hausdorff topological group?Demonstrate Topological Spaces are HomeomorphicTwo graphs are homeomorphic iff they have isomorphic subdivisionsWhat topological properties are invariant under diffeomorphism?1) Is $Bbb R setminus Bbb Q$ compact? 2) Prove $[0, 1] times [0, 1]$ and $(0, 1) times (0, 1)$ are not homeomorphicIs $Bbb Q / Bbb Z$ discrete?Show homeomorphism between two quotient topologies

0

$begingroup$

I am new on this topology stuff, so came to ask for help, my intuition tells me that I need to use a topological invariant to see when they are homeomorphic but I am not quite sure how to proceed.

Prove that $Bbb R cong Bbb R^n$ if and only if $n=1$ and $Bbb S^1 cong Bbb S^n$ if and only if $n=1$

general-topology

share|cite|improve this question

edited 2 days ago

Arnaud D.

16.4k52445

asked Apr 30 at 0:12

Frank SambeFrank Sambe

154

$endgroup$

put on hold as off-topic by user21820, max_zorn, José Carlos Santos, YiFan, Cesareo 2 days ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, max_zorn, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Prove that for every $n geq 2$, $Bbb R$ is not homeomorphic to $Bbb R^n$
  $endgroup$
  – YuiTo Cheng
  2 days ago
  

  
  
  
  

add a comment | 

0

$begingroup$

I am new on this topology stuff, so came to ask for help, my intuition tells me that I need to use a topological invariant to see when they are homeomorphic but I am not quite sure how to proceed.

Prove that $Bbb R cong Bbb R^n$ if and only if $n=1$ and $Bbb S^1 cong Bbb S^n$ if and only if $n=1$

general-topology

share|cite|improve this question

edited 2 days ago

Arnaud D.

16.4k52445

asked Apr 30 at 0:12

Frank SambeFrank Sambe

154

$endgroup$

put on hold as off-topic by user21820, max_zorn, José Carlos Santos, YiFan, Cesareo 2 days ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, max_zorn, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Prove that for every $n geq 2$, $Bbb R$ is not homeomorphic to $Bbb R^n$
  $endgroup$
  – YuiTo Cheng
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

I am new on this topology stuff, so came to ask for help, my intuition tells me that I need to use a topological invariant to see when they are homeomorphic but I am not quite sure how to proceed.

Prove that $Bbb R cong Bbb R^n$ if and only if $n=1$ and $Bbb S^1 cong Bbb S^n$ if and only if $n=1$

general-topology

share|cite|improve this question

edited 2 days ago

Arnaud D.

16.4k52445

asked Apr 30 at 0:12

Frank SambeFrank Sambe

154

$endgroup$

share|cite|improve this question

edited 2 days ago

Arnaud D.

16.4k52445

asked Apr 30 at 0:12

Frank SambeFrank Sambe

154

share|cite|improve this question

edited 2 days ago

Arnaud D.

16.4k52445

asked Apr 30 at 0:12

Frank SambeFrank Sambe

154

edited 2 days ago

Arnaud D.

16.4k52445

edited 2 days ago

Arnaud D.

16.4k52445

asked Apr 30 at 0:12

Frank SambeFrank Sambe

154

asked Apr 30 at 0:12

Frank SambeFrank Sambe

154

3 Answers
3

active

oldest

votes

3

$begingroup$

A more elementary way than what Chris Custer suggests is this: if we remove a point from $mathbbR$, we get a disconnected set. But if $n>1$, $mathbbR^n$ minus one point remains connected (it is easy to prove this). For the other part, note that $mathbbS^n$ minus one point is homeomorphic to $mathbbR^n$.

share|cite|improve this answer

answered Apr 30 at 0:24

JustDroppedInJustDroppedIn

2,467420

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You're right. Connectedness is a little more elementary.
  $endgroup$
  – Chris Custer
  Apr 30 at 0:38
  

  
  
  
  

add a comment | 

2

$begingroup$

Suppose there is such homeomorphism $f:Bbb RtoBbb R^n$, then setting $A:=(-infty,0)cup(0,infty)$ it is enough to check that $Bbb R^nsetminus f(0)$ is connected for $n>1$ but $A$ is disconnected, hence $f^-1$ cannot be continuous, so it is not an homeomorphism.

For the other case you can proceed analogously noticing that $Bbb S^nsetminusacongBbb R^n$ via stereographic projection for any chosen $ainBbb S^n$.

share|cite|improve this answer

edited Apr 30 at 0:30

answered Apr 30 at 0:21

MasacrosoMasacroso

13.3k41749

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $[a,b]$ is compact.
  $endgroup$
  – Chris Custer
  Apr 30 at 0:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ChrisCuster right, good catch. Fixed
  $endgroup$
  – Masacroso
  Apr 30 at 0:26
  

  
  
  
  

add a comment | 

0

$begingroup$

One nice way is to use the invariant of homology. $H_n(S^n)congBbb Z$ and $H_m(S^n)cong0,,mneq n,0$.

$Bbb R^n$ can then be done using the one-point compactification.

share|cite|improve this answer

answered Apr 30 at 0:20

Chris CusterChris Custer

14.9k3827

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Considering "I am new to this topology stuff", this is probably too much although the quickest.
  $endgroup$
  – IAmNoOne
  Apr 30 at 0:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  hahhaha yes, it is too much to me!
  $endgroup$
  – Frank Sambe
  Apr 30 at 0:40
  

  
  
  
  

add a comment | 

3

$begingroup$

A more elementary way than what Chris Custer suggests is this: if we remove a point from $mathbbR$, we get a disconnected set. But if $n>1$, $mathbbR^n$ minus one point remains connected (it is easy to prove this). For the other part, note that $mathbbS^n$ minus one point is homeomorphic to $mathbbR^n$.

share|cite|improve this answer

answered Apr 30 at 0:24

JustDroppedInJustDroppedIn

2,467420

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You're right. Connectedness is a little more elementary.
  $endgroup$
  – Chris Custer
  Apr 30 at 0:38
  

  
  
  
  

add a comment | 

3

$begingroup$

A more elementary way than what Chris Custer suggests is this: if we remove a point from $mathbbR$, we get a disconnected set. But if $n>1$, $mathbbR^n$ minus one point remains connected (it is easy to prove this). For the other part, note that $mathbbS^n$ minus one point is homeomorphic to $mathbbR^n$.

share|cite|improve this answer

answered Apr 30 at 0:24

JustDroppedInJustDroppedIn

2,467420

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You're right. Connectedness is a little more elementary.
  $endgroup$
  – Chris Custer
  Apr 30 at 0:38
  

  
  
  
  

add a comment | 

$begingroup$

A more elementary way than what Chris Custer suggests is this: if we remove a point from $mathbbR$, we get a disconnected set. But if $n>1$, $mathbbR^n$ minus one point remains connected (it is easy to prove this). For the other part, note that $mathbbS^n$ minus one point is homeomorphic to $mathbbR^n$.

share|cite|improve this answer

answered Apr 30 at 0:24

JustDroppedInJustDroppedIn

2,467420

$endgroup$

share|cite|improve this answer

answered Apr 30 at 0:24

JustDroppedInJustDroppedIn

2,467420

answered Apr 30 at 0:24

JustDroppedInJustDroppedIn

2,467420

answered Apr 30 at 0:24

JustDroppedInJustDroppedIn

2,467420

2

$begingroup$

Suppose there is such homeomorphism $f:Bbb RtoBbb R^n$, then setting $A:=(-infty,0)cup(0,infty)$ it is enough to check that $Bbb R^nsetminus f(0)$ is connected for $n>1$ but $A$ is disconnected, hence $f^-1$ cannot be continuous, so it is not an homeomorphism.

For the other case you can proceed analogously noticing that $Bbb S^nsetminusacongBbb R^n$ via stereographic projection for any chosen $ainBbb S^n$.

share|cite|improve this answer

edited Apr 30 at 0:30

answered Apr 30 at 0:21

MasacrosoMasacroso

13.3k41749

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $[a,b]$ is compact.
  $endgroup$
  – Chris Custer
  Apr 30 at 0:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ChrisCuster right, good catch. Fixed
  $endgroup$
  – Masacroso
  Apr 30 at 0:26
  

  
  
  
  

add a comment | 

2

$begingroup$

Suppose there is such homeomorphism $f:Bbb RtoBbb R^n$, then setting $A:=(-infty,0)cup(0,infty)$ it is enough to check that $Bbb R^nsetminus f(0)$ is connected for $n>1$ but $A$ is disconnected, hence $f^-1$ cannot be continuous, so it is not an homeomorphism.

For the other case you can proceed analogously noticing that $Bbb S^nsetminusacongBbb R^n$ via stereographic projection for any chosen $ainBbb S^n$.

share|cite|improve this answer

edited Apr 30 at 0:30

answered Apr 30 at 0:21

MasacrosoMasacroso

13.3k41749

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $[a,b]$ is compact.
  $endgroup$
  – Chris Custer
  Apr 30 at 0:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ChrisCuster right, good catch. Fixed
  $endgroup$
  – Masacroso
  Apr 30 at 0:26
  

  
  
  
  

add a comment | 

$begingroup$

Suppose there is such homeomorphism $f:Bbb RtoBbb R^n$, then setting $A:=(-infty,0)cup(0,infty)$ it is enough to check that $Bbb R^nsetminus f(0)$ is connected for $n>1$ but $A$ is disconnected, hence $f^-1$ cannot be continuous, so it is not an homeomorphism.

For the other case you can proceed analogously noticing that $Bbb S^nsetminusacongBbb R^n$ via stereographic projection for any chosen $ainBbb S^n$.

share|cite|improve this answer

edited Apr 30 at 0:30

answered Apr 30 at 0:21

MasacrosoMasacroso

13.3k41749

$endgroup$

share|cite|improve this answer

edited Apr 30 at 0:30

answered Apr 30 at 0:21

MasacrosoMasacroso

13.3k41749

answered Apr 30 at 0:21

MasacrosoMasacroso

13.3k41749

answered Apr 30 at 0:21

MasacrosoMasacroso

13.3k41749

0

$begingroup$

One nice way is to use the invariant of homology. $H_n(S^n)congBbb Z$ and $H_m(S^n)cong0,,mneq n,0$.

$Bbb R^n$ can then be done using the one-point compactification.

share|cite|improve this answer

answered Apr 30 at 0:20

Chris CusterChris Custer

14.9k3827

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Considering "I am new to this topology stuff", this is probably too much although the quickest.
  $endgroup$
  – IAmNoOne
  Apr 30 at 0:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  hahhaha yes, it is too much to me!
  $endgroup$
  – Frank Sambe
  Apr 30 at 0:40
  

  
  
  
  

add a comment | 

0

$begingroup$

One nice way is to use the invariant of homology. $H_n(S^n)congBbb Z$ and $H_m(S^n)cong0,,mneq n,0$.

$Bbb R^n$ can then be done using the one-point compactification.

share|cite|improve this answer

answered Apr 30 at 0:20

Chris CusterChris Custer

14.9k3827

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Considering "I am new to this topology stuff", this is probably too much although the quickest.
  $endgroup$
  – IAmNoOne
  Apr 30 at 0:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  hahhaha yes, it is too much to me!
  $endgroup$
  – Frank Sambe
  Apr 30 at 0:40
  

  
  
  
  

add a comment | 

$begingroup$

One nice way is to use the invariant of homology. $H_n(S^n)congBbb Z$ and $H_m(S^n)cong0,,mneq n,0$.

$Bbb R^n$ can then be done using the one-point compactification.

share|cite|improve this answer

answered Apr 30 at 0:20

Chris CusterChris Custer

14.9k3827

$endgroup$

share|cite|improve this answer

answered Apr 30 at 0:20

Chris CusterChris Custer

14.9k3827

answered Apr 30 at 0:20

Chris CusterChris Custer

14.9k3827

answered Apr 30 at 0:20

Chris CusterChris Custer

14.9k3827