$Bbb R cong Bbb R^n$ iff $n=1$ [closed]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

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$Bbb R cong Bbb R^n$ iff $n=1$ [closed]


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











$endgroup$



closed as off-topic by user21820, max_zorn, José Carlos Santos, YiFan, Cesareo Apr 30 at 8:34


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
    Apr 30 at 8:22















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











$endgroup$



closed as off-topic by user21820, max_zorn, José Carlos Santos, YiFan, Cesareo Apr 30 at 8:34


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
    Apr 30 at 8:22













0












0








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











$endgroup$




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




general-topology






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 30 at 8:09









Arnaud D.

16.4k52445




16.4k52445










asked Apr 30 at 0:12









Frank SambeFrank Sambe

154




154




closed as off-topic by user21820, max_zorn, José Carlos Santos, YiFan, Cesareo Apr 30 at 8:34


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.







closed as off-topic by user21820, max_zorn, José Carlos Santos, YiFan, Cesareo Apr 30 at 8:34


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
    Apr 30 at 8:22












  • 2




    $begingroup$
    Possible duplicate of Prove that for every $n geq 2$, $Bbb R$ is not homeomorphic to $Bbb R^n$
    $endgroup$
    – YuiTo Cheng
    Apr 30 at 8:22







2




2




$begingroup$
Possible duplicate of Prove that for every $n geq 2$, $Bbb R$ is not homeomorphic to $Bbb R^n$
$endgroup$
– YuiTo Cheng
Apr 30 at 8:22




$begingroup$
Possible duplicate of Prove that for every $n geq 2$, $Bbb R$ is not homeomorphic to $Bbb R^n$
$endgroup$
– YuiTo Cheng
Apr 30 at 8:22










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









$endgroup$








  • 1




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


















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











$endgroup$








  • 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


















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









$endgroup$












  • $begingroup$
    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

















3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






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









$endgroup$








  • 1




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















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









$endgroup$








  • 1




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













3












3








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









$endgroup$



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












share|cite|improve this answer



share|cite|improve this answer










answered Apr 30 at 0:24









JustDroppedInJustDroppedIn

2,467420




2,467420







  • 1




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












  • 1




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







1




1




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




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











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











$endgroup$








  • 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















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











$endgroup$








  • 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













2












2








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











$endgroup$



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














share|cite|improve this answer



share|cite|improve this answer








edited Apr 30 at 0:30

























answered Apr 30 at 0:21









MasacrosoMasacroso

13.4k41749




13.4k41749







  • 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












  • 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







1




1




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




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












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




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











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









$endgroup$












  • $begingroup$
    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















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









$endgroup$












  • $begingroup$
    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













0












0








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









$endgroup$



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












share|cite|improve this answer



share|cite|improve this answer










answered Apr 30 at 0:20









Chris CusterChris Custer

15.1k3827




15.1k3827











  • $begingroup$
    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
















  • $begingroup$
    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















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




$begingroup$
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




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



-

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