$Bbb R cong Bbb {R^n}$ iff $n=1$Closure of Topological SpacesRing with spectrum homeomorphic to a given...
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$Bbb R cong Bbb {R^n}$ iff $n=1$
Closure of Topological SpacesRing with spectrum homeomorphic to a given topological spaceIs there a general way to tell whether two topological spaces are homeomorphic?Do topological spaces deserve their name?What topological properties are invariant under diffeomorphism?Quotient spaces not homeomorphicProof that $S^1$ isn't homeomorphic to $[0, 1]$Homeomorphism of union of infinitely many topological spacesSets homeomorphic to convex setsHomeomorphic Topological Spaces with the Subspace Topology
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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
$endgroup$
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
$endgroup$
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
$endgroup$
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
edited 5 mins ago
user21820
40.5k544163
40.5k544163
asked 3 hours ago
Frank SambeFrank Sambe
174
174
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
A more elementary way than what Chris Custer suggests is this: if we remove a point from $mathbb{R}$, we get a disconnected set. But if $n>1$, $mathbb{R}^n$ minus one point remains connected (it is easy to prove this). For the other part, note that $mathbb{S}^n$ minus one point is homeomorphic to $mathbb{R}^n$.
$endgroup$
1
$begingroup$
You're right. Connectedness is a little more elementary.
$endgroup$
– Chris Custer
2 hours ago
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 $Bbb S^nsetminus{a}congBbb R^n$ via stereographic projection for any chosen $ainBbb S^n$.
$endgroup$
1
$begingroup$
$[a,b]$ is compact.
$endgroup$
– Chris Custer
2 hours ago
$begingroup$
@ChrisCuster right, good catch. Fixed
$endgroup$
– Masacroso
2 hours ago
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.
$endgroup$
$begingroup$
Considering "I am new to this topology stuff", this is probably too much although the quickest.
$endgroup$
– IAmNoOne
2 hours ago
$begingroup$
hahhaha yes, it is too much to me!
$endgroup$
– Frank Sambe
2 hours ago
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A more elementary way than what Chris Custer suggests is this: if we remove a point from $mathbb{R}$, we get a disconnected set. But if $n>1$, $mathbb{R}^n$ minus one point remains connected (it is easy to prove this). For the other part, note that $mathbb{S}^n$ minus one point is homeomorphic to $mathbb{R}^n$.
$endgroup$
1
$begingroup$
You're right. Connectedness is a little more elementary.
$endgroup$
– Chris Custer
2 hours ago
add a comment |
$begingroup$
A more elementary way than what Chris Custer suggests is this: if we remove a point from $mathbb{R}$, we get a disconnected set. But if $n>1$, $mathbb{R}^n$ minus one point remains connected (it is easy to prove this). For the other part, note that $mathbb{S}^n$ minus one point is homeomorphic to $mathbb{R}^n$.
$endgroup$
1
$begingroup$
You're right. Connectedness is a little more elementary.
$endgroup$
– Chris Custer
2 hours ago
add a comment |
$begingroup$
A more elementary way than what Chris Custer suggests is this: if we remove a point from $mathbb{R}$, we get a disconnected set. But if $n>1$, $mathbb{R}^n$ minus one point remains connected (it is easy to prove this). For the other part, note that $mathbb{S}^n$ minus one point is homeomorphic to $mathbb{R}^n$.
$endgroup$
A more elementary way than what Chris Custer suggests is this: if we remove a point from $mathbb{R}$, we get a disconnected set. But if $n>1$, $mathbb{R}^n$ minus one point remains connected (it is easy to prove this). For the other part, note that $mathbb{S}^n$ minus one point is homeomorphic to $mathbb{R}^n$.
answered 2 hours ago
JustDroppedInJustDroppedIn
2,467420
2,467420
1
$begingroup$
You're right. Connectedness is a little more elementary.
$endgroup$
– Chris Custer
2 hours ago
add a comment |
1
$begingroup$
You're right. Connectedness is a little more elementary.
$endgroup$
– Chris Custer
2 hours ago
1
1
$begingroup$
You're right. Connectedness is a little more elementary.
$endgroup$
– Chris Custer
2 hours ago
$begingroup$
You're right. Connectedness is a little more elementary.
$endgroup$
– Chris Custer
2 hours ago
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 $Bbb S^nsetminus{a}congBbb R^n$ via stereographic projection for any chosen $ainBbb S^n$.
$endgroup$
1
$begingroup$
$[a,b]$ is compact.
$endgroup$
– Chris Custer
2 hours ago
$begingroup$
@ChrisCuster right, good catch. Fixed
$endgroup$
– Masacroso
2 hours ago
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 $Bbb S^nsetminus{a}congBbb R^n$ via stereographic projection for any chosen $ainBbb S^n$.
$endgroup$
1
$begingroup$
$[a,b]$ is compact.
$endgroup$
– Chris Custer
2 hours ago
$begingroup$
@ChrisCuster right, good catch. Fixed
$endgroup$
– Masacroso
2 hours ago
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 $Bbb S^nsetminus{a}congBbb R^n$ via stereographic projection for any chosen $ainBbb S^n$.
$endgroup$
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 $Bbb S^nsetminus{a}congBbb R^n$ via stereographic projection for any chosen $ainBbb S^n$.
edited 2 hours ago
answered 2 hours ago
MasacrosoMasacroso
13.3k41749
13.3k41749
1
$begingroup$
$[a,b]$ is compact.
$endgroup$
– Chris Custer
2 hours ago
$begingroup$
@ChrisCuster right, good catch. Fixed
$endgroup$
– Masacroso
2 hours ago
add a comment |
1
$begingroup$
$[a,b]$ is compact.
$endgroup$
– Chris Custer
2 hours ago
$begingroup$
@ChrisCuster right, good catch. Fixed
$endgroup$
– Masacroso
2 hours ago
1
1
$begingroup$
$[a,b]$ is compact.
$endgroup$
– Chris Custer
2 hours ago
$begingroup$
$[a,b]$ is compact.
$endgroup$
– Chris Custer
2 hours ago
$begingroup$
@ChrisCuster right, good catch. Fixed
$endgroup$
– Masacroso
2 hours ago
$begingroup$
@ChrisCuster right, good catch. Fixed
$endgroup$
– Masacroso
2 hours ago
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.
$endgroup$
$begingroup$
Considering "I am new to this topology stuff", this is probably too much although the quickest.
$endgroup$
– IAmNoOne
2 hours ago
$begingroup$
hahhaha yes, it is too much to me!
$endgroup$
– Frank Sambe
2 hours ago
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.
$endgroup$
$begingroup$
Considering "I am new to this topology stuff", this is probably too much although the quickest.
$endgroup$
– IAmNoOne
2 hours ago
$begingroup$
hahhaha yes, it is too much to me!
$endgroup$
– Frank Sambe
2 hours ago
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.
$endgroup$
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.
answered 3 hours ago
Chris CusterChris Custer
14.9k3827
14.9k3827
$begingroup$
Considering "I am new to this topology stuff", this is probably too much although the quickest.
$endgroup$
– IAmNoOne
2 hours ago
$begingroup$
hahhaha yes, it is too much to me!
$endgroup$
– Frank Sambe
2 hours ago
add a comment |
$begingroup$
Considering "I am new to this topology stuff", this is probably too much although the quickest.
$endgroup$
– IAmNoOne
2 hours ago
$begingroup$
hahhaha yes, it is too much to me!
$endgroup$
– Frank Sambe
2 hours ago
$begingroup$
Considering "I am new to this topology stuff", this is probably too much although the quickest.
$endgroup$
– IAmNoOne
2 hours ago
$begingroup$
Considering "I am new to this topology stuff", this is probably too much although the quickest.
$endgroup$
– IAmNoOne
2 hours ago
$begingroup$
hahhaha yes, it is too much to me!
$endgroup$
– Frank Sambe
2 hours ago
$begingroup$
hahhaha yes, it is too much to me!
$endgroup$
– Frank Sambe
2 hours ago
add a comment |
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