Showing $S^1$ is not homeomorphic to $mathbb R^2$ with the subspace topology












1














So I want to say we let $f:S^1→mathbb R^2$ be a homeomorphism, and since we can find a bijection, I want to say we can open an open set in the plane using the standard product topology, and take its pre-image, which is assumed to be open by continuity of f but this is not always the case in $S^1$.



Help or hint please.






general-topology







share|cite|improve this question




















  • 2




    Do you know compactness?
    – Randall
    Nov 28 '18 at 4:49










  • very well, oh I see where you're going, use that the circle is and R2 isn't? I Wass trying to avoid that proof, its sort of the easier way out no?
    – eyeheartmath
    Nov 28 '18 at 4:50






  • 2




    Why avoid it? It’s the shortest possible proof.
    – Randall
    Nov 28 '18 at 4:51






  • 3




    If you don't want to use compactness, you could use connectedness. Removing any two points from the circle disconnects it, but this isn't true about the plane.
    – Santana Afton
    Nov 28 '18 at 4:54






  • 2




    Homeomorphism is a topological thing, so any property you use will necessarily be topological. It looks like what you might be thinking of is something similar to "Invariance of domain" though, by which the embedding of $S^1$ would have to be open (but isn't). Note though that this is harder than using compactness.
    – Carl
    Nov 28 '18 at 5:40


















1














So I want to say we let $f:S^1→mathbb R^2$ be a homeomorphism, and since we can find a bijection, I want to say we can open an open set in the plane using the standard product topology, and take its pre-image, which is assumed to be open by continuity of f but this is not always the case in $S^1$.



Help or hint please.






general-topology







share|cite|improve this question




















  • 2




    Do you know compactness?
    – Randall
    Nov 28 '18 at 4:49










  • very well, oh I see where you're going, use that the circle is and R2 isn't? I Wass trying to avoid that proof, its sort of the easier way out no?
    – eyeheartmath
    Nov 28 '18 at 4:50






  • 2




    Why avoid it? It’s the shortest possible proof.
    – Randall
    Nov 28 '18 at 4:51






  • 3




    If you don't want to use compactness, you could use connectedness. Removing any two points from the circle disconnects it, but this isn't true about the plane.
    – Santana Afton
    Nov 28 '18 at 4:54






  • 2




    Homeomorphism is a topological thing, so any property you use will necessarily be topological. It looks like what you might be thinking of is something similar to "Invariance of domain" though, by which the embedding of $S^1$ would have to be open (but isn't). Note though that this is harder than using compactness.
    – Carl
    Nov 28 '18 at 5:40
















1












1








1







So I want to say we let $f:S^1→mathbb R^2$ be a homeomorphism, and since we can find a bijection, I want to say we can open an open set in the plane using the standard product topology, and take its pre-image, which is assumed to be open by continuity of f but this is not always the case in $S^1$.



Help or hint please.






general-topology







share|cite|improve this question















So I want to say we let $f:S^1→mathbb R^2$ be a homeomorphism, and since we can find a bijection, I want to say we can open an open set in the plane using the standard product topology, and take its pre-image, which is assumed to be open by continuity of f but this is not always the case in $S^1$.



Help or hint please.






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 Nov 28 '18 at 5:02









Saad

19.7k92352




19.7k92352










asked Nov 28 '18 at 4:47









eyeheartmath

657




657








  • 2




    Do you know compactness?
    – Randall
    Nov 28 '18 at 4:49










  • very well, oh I see where you're going, use that the circle is and R2 isn't? I Wass trying to avoid that proof, its sort of the easier way out no?
    – eyeheartmath
    Nov 28 '18 at 4:50






  • 2




    Why avoid it? It’s the shortest possible proof.
    – Randall
    Nov 28 '18 at 4:51






  • 3




    If you don't want to use compactness, you could use connectedness. Removing any two points from the circle disconnects it, but this isn't true about the plane.
    – Santana Afton
    Nov 28 '18 at 4:54






  • 2




    Homeomorphism is a topological thing, so any property you use will necessarily be topological. It looks like what you might be thinking of is something similar to "Invariance of domain" though, by which the embedding of $S^1$ would have to be open (but isn't). Note though that this is harder than using compactness.
    – Carl
    Nov 28 '18 at 5:40
















  • 2




    Do you know compactness?
    – Randall
    Nov 28 '18 at 4:49










  • very well, oh I see where you're going, use that the circle is and R2 isn't? I Wass trying to avoid that proof, its sort of the easier way out no?
    – eyeheartmath
    Nov 28 '18 at 4:50






  • 2




    Why avoid it? It’s the shortest possible proof.
    – Randall
    Nov 28 '18 at 4:51






  • 3




    If you don't want to use compactness, you could use connectedness. Removing any two points from the circle disconnects it, but this isn't true about the plane.
    – Santana Afton
    Nov 28 '18 at 4:54






  • 2




    Homeomorphism is a topological thing, so any property you use will necessarily be topological. It looks like what you might be thinking of is something similar to "Invariance of domain" though, by which the embedding of $S^1$ would have to be open (but isn't). Note though that this is harder than using compactness.
    – Carl
    Nov 28 '18 at 5:40










2




2




Do you know compactness?
– Randall
Nov 28 '18 at 4:49




Do you know compactness?
– Randall
Nov 28 '18 at 4:49












very well, oh I see where you're going, use that the circle is and R2 isn't? I Wass trying to avoid that proof, its sort of the easier way out no?
– eyeheartmath
Nov 28 '18 at 4:50




very well, oh I see where you're going, use that the circle is and R2 isn't? I Wass trying to avoid that proof, its sort of the easier way out no?
– eyeheartmath
Nov 28 '18 at 4:50




2




2




Why avoid it? It’s the shortest possible proof.
– Randall
Nov 28 '18 at 4:51




Why avoid it? It’s the shortest possible proof.
– Randall
Nov 28 '18 at 4:51




3




3




If you don't want to use compactness, you could use connectedness. Removing any two points from the circle disconnects it, but this isn't true about the plane.
– Santana Afton
Nov 28 '18 at 4:54




If you don't want to use compactness, you could use connectedness. Removing any two points from the circle disconnects it, but this isn't true about the plane.
– Santana Afton
Nov 28 '18 at 4:54




2




2




Homeomorphism is a topological thing, so any property you use will necessarily be topological. It looks like what you might be thinking of is something similar to "Invariance of domain" though, by which the embedding of $S^1$ would have to be open (but isn't). Note though that this is harder than using compactness.
– Carl
Nov 28 '18 at 5:40






Homeomorphism is a topological thing, so any property you use will necessarily be topological. It looks like what you might be thinking of is something similar to "Invariance of domain" though, by which the embedding of $S^1$ would have to be open (but isn't). Note though that this is harder than using compactness.
– Carl
Nov 28 '18 at 5:40












1 Answer
1






active

oldest

votes


















2














At least two proofs come to mind.




  1. $S^1$ is compact, and $mathbb R^2$ is not, but any continuous map takes compact sets to compact sets, so $mathbb R^2$ cannot be the image under any continuous map of $S^1$ with the given topologies.


  2. $S^1$ minus any two points is disconnected, but $mathbb R^2$ minus any two points remains connected. This forms a proof as follows : consider $p,q in mathbb R^2$ ,then the set $mathbb R^2 setminus {p,q}$ is a connected (open) set, hence its image under any injective continuous map $f$ to $mathbb S^1$ should be connected, but the image is $mathbb S^1 setminus {f(p),f(q)}$ which is disconnected, since $f(p) neq f(q)$ by injectivity of $f$. Therefore, no injective continuous map from $mathbb R^2$ to $S^1$ exists in the given topologies.



Also see if something like this works out (exercise) : $S^1$ is the quotient space of $[0,1]$ under the identification $a sim b iff a=b $ or ${a,b} = {0,1}$. Therefore, any continuous map $f$ from $S^1$ to $mathbb R^2$ lifts to a continuous map $psi$ from $[0,1]$ to $mathbb R^2$, which satisfies $psi(0) = psi(1)$, via $psi(a) = f([a])$. The reverse is also true : any such continuous map $phi : [0,1] to mathbb R^2$ with $phi(0) = phi(1)$ factors to a continuous map given by $g([a]) = phi(a)$. So, we have an identification between continuous maps from $S^1$ to $mathbb R^2$, and from $[0,1]$ to $mathbb R^2$.



Using the fact that $[0,1]$ is a bounded interval, see if you can make any argument using the above identification to show that $S^1 not cong mathbb R^2$. This is a more comfortable setting to try out your "direct method" that you were attempting in the post.






share|cite|improve this answer





















  • that is exactly what I needed, my first thoughts were to use connectedness and compactness but was trying to do it without the use of these topological invariants. thanks for that proof, it looks like it works to me and I have to wrap my head round it some more. I was more or less tying to do this using basic principles of topology rather than compactness/connectedness tricks.
    – eyeheartmath
    Nov 28 '18 at 5:31






  • 1




    It is ok. Usually, you will need to be familiar with these tricks : more than the basic principles, they will come to your help as you progress.
    – астон вілла олоф мэллбэрг
    Nov 28 '18 at 5:32










  • exactly, I wanted to know all the different proofs for this one, I like the proof how it uses covering spaces which I was rusty with. Thanks so much! im working through it now
    – eyeheartmath
    Nov 28 '18 at 5:37











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1 Answer
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active

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active

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2














At least two proofs come to mind.




  1. $S^1$ is compact, and $mathbb R^2$ is not, but any continuous map takes compact sets to compact sets, so $mathbb R^2$ cannot be the image under any continuous map of $S^1$ with the given topologies.


  2. $S^1$ minus any two points is disconnected, but $mathbb R^2$ minus any two points remains connected. This forms a proof as follows : consider $p,q in mathbb R^2$ ,then the set $mathbb R^2 setminus {p,q}$ is a connected (open) set, hence its image under any injective continuous map $f$ to $mathbb S^1$ should be connected, but the image is $mathbb S^1 setminus {f(p),f(q)}$ which is disconnected, since $f(p) neq f(q)$ by injectivity of $f$. Therefore, no injective continuous map from $mathbb R^2$ to $S^1$ exists in the given topologies.



Also see if something like this works out (exercise) : $S^1$ is the quotient space of $[0,1]$ under the identification $a sim b iff a=b $ or ${a,b} = {0,1}$. Therefore, any continuous map $f$ from $S^1$ to $mathbb R^2$ lifts to a continuous map $psi$ from $[0,1]$ to $mathbb R^2$, which satisfies $psi(0) = psi(1)$, via $psi(a) = f([a])$. The reverse is also true : any such continuous map $phi : [0,1] to mathbb R^2$ with $phi(0) = phi(1)$ factors to a continuous map given by $g([a]) = phi(a)$. So, we have an identification between continuous maps from $S^1$ to $mathbb R^2$, and from $[0,1]$ to $mathbb R^2$.



Using the fact that $[0,1]$ is a bounded interval, see if you can make any argument using the above identification to show that $S^1 not cong mathbb R^2$. This is a more comfortable setting to try out your "direct method" that you were attempting in the post.






share|cite|improve this answer





















  • that is exactly what I needed, my first thoughts were to use connectedness and compactness but was trying to do it without the use of these topological invariants. thanks for that proof, it looks like it works to me and I have to wrap my head round it some more. I was more or less tying to do this using basic principles of topology rather than compactness/connectedness tricks.
    – eyeheartmath
    Nov 28 '18 at 5:31






  • 1




    It is ok. Usually, you will need to be familiar with these tricks : more than the basic principles, they will come to your help as you progress.
    – астон вілла олоф мэллбэрг
    Nov 28 '18 at 5:32










  • exactly, I wanted to know all the different proofs for this one, I like the proof how it uses covering spaces which I was rusty with. Thanks so much! im working through it now
    – eyeheartmath
    Nov 28 '18 at 5:37
















2














At least two proofs come to mind.




  1. $S^1$ is compact, and $mathbb R^2$ is not, but any continuous map takes compact sets to compact sets, so $mathbb R^2$ cannot be the image under any continuous map of $S^1$ with the given topologies.


  2. $S^1$ minus any two points is disconnected, but $mathbb R^2$ minus any two points remains connected. This forms a proof as follows : consider $p,q in mathbb R^2$ ,then the set $mathbb R^2 setminus {p,q}$ is a connected (open) set, hence its image under any injective continuous map $f$ to $mathbb S^1$ should be connected, but the image is $mathbb S^1 setminus {f(p),f(q)}$ which is disconnected, since $f(p) neq f(q)$ by injectivity of $f$. Therefore, no injective continuous map from $mathbb R^2$ to $S^1$ exists in the given topologies.



Also see if something like this works out (exercise) : $S^1$ is the quotient space of $[0,1]$ under the identification $a sim b iff a=b $ or ${a,b} = {0,1}$. Therefore, any continuous map $f$ from $S^1$ to $mathbb R^2$ lifts to a continuous map $psi$ from $[0,1]$ to $mathbb R^2$, which satisfies $psi(0) = psi(1)$, via $psi(a) = f([a])$. The reverse is also true : any such continuous map $phi : [0,1] to mathbb R^2$ with $phi(0) = phi(1)$ factors to a continuous map given by $g([a]) = phi(a)$. So, we have an identification between continuous maps from $S^1$ to $mathbb R^2$, and from $[0,1]$ to $mathbb R^2$.



Using the fact that $[0,1]$ is a bounded interval, see if you can make any argument using the above identification to show that $S^1 not cong mathbb R^2$. This is a more comfortable setting to try out your "direct method" that you were attempting in the post.






share|cite|improve this answer





















  • that is exactly what I needed, my first thoughts were to use connectedness and compactness but was trying to do it without the use of these topological invariants. thanks for that proof, it looks like it works to me and I have to wrap my head round it some more. I was more or less tying to do this using basic principles of topology rather than compactness/connectedness tricks.
    – eyeheartmath
    Nov 28 '18 at 5:31






  • 1




    It is ok. Usually, you will need to be familiar with these tricks : more than the basic principles, they will come to your help as you progress.
    – астон вілла олоф мэллбэрг
    Nov 28 '18 at 5:32










  • exactly, I wanted to know all the different proofs for this one, I like the proof how it uses covering spaces which I was rusty with. Thanks so much! im working through it now
    – eyeheartmath
    Nov 28 '18 at 5:37














2












2








2






At least two proofs come to mind.




  1. $S^1$ is compact, and $mathbb R^2$ is not, but any continuous map takes compact sets to compact sets, so $mathbb R^2$ cannot be the image under any continuous map of $S^1$ with the given topologies.


  2. $S^1$ minus any two points is disconnected, but $mathbb R^2$ minus any two points remains connected. This forms a proof as follows : consider $p,q in mathbb R^2$ ,then the set $mathbb R^2 setminus {p,q}$ is a connected (open) set, hence its image under any injective continuous map $f$ to $mathbb S^1$ should be connected, but the image is $mathbb S^1 setminus {f(p),f(q)}$ which is disconnected, since $f(p) neq f(q)$ by injectivity of $f$. Therefore, no injective continuous map from $mathbb R^2$ to $S^1$ exists in the given topologies.



Also see if something like this works out (exercise) : $S^1$ is the quotient space of $[0,1]$ under the identification $a sim b iff a=b $ or ${a,b} = {0,1}$. Therefore, any continuous map $f$ from $S^1$ to $mathbb R^2$ lifts to a continuous map $psi$ from $[0,1]$ to $mathbb R^2$, which satisfies $psi(0) = psi(1)$, via $psi(a) = f([a])$. The reverse is also true : any such continuous map $phi : [0,1] to mathbb R^2$ with $phi(0) = phi(1)$ factors to a continuous map given by $g([a]) = phi(a)$. So, we have an identification between continuous maps from $S^1$ to $mathbb R^2$, and from $[0,1]$ to $mathbb R^2$.



Using the fact that $[0,1]$ is a bounded interval, see if you can make any argument using the above identification to show that $S^1 not cong mathbb R^2$. This is a more comfortable setting to try out your "direct method" that you were attempting in the post.






share|cite|improve this answer












At least two proofs come to mind.




  1. $S^1$ is compact, and $mathbb R^2$ is not, but any continuous map takes compact sets to compact sets, so $mathbb R^2$ cannot be the image under any continuous map of $S^1$ with the given topologies.


  2. $S^1$ minus any two points is disconnected, but $mathbb R^2$ minus any two points remains connected. This forms a proof as follows : consider $p,q in mathbb R^2$ ,then the set $mathbb R^2 setminus {p,q}$ is a connected (open) set, hence its image under any injective continuous map $f$ to $mathbb S^1$ should be connected, but the image is $mathbb S^1 setminus {f(p),f(q)}$ which is disconnected, since $f(p) neq f(q)$ by injectivity of $f$. Therefore, no injective continuous map from $mathbb R^2$ to $S^1$ exists in the given topologies.



Also see if something like this works out (exercise) : $S^1$ is the quotient space of $[0,1]$ under the identification $a sim b iff a=b $ or ${a,b} = {0,1}$. Therefore, any continuous map $f$ from $S^1$ to $mathbb R^2$ lifts to a continuous map $psi$ from $[0,1]$ to $mathbb R^2$, which satisfies $psi(0) = psi(1)$, via $psi(a) = f([a])$. The reverse is also true : any such continuous map $phi : [0,1] to mathbb R^2$ with $phi(0) = phi(1)$ factors to a continuous map given by $g([a]) = phi(a)$. So, we have an identification between continuous maps from $S^1$ to $mathbb R^2$, and from $[0,1]$ to $mathbb R^2$.



Using the fact that $[0,1]$ is a bounded interval, see if you can make any argument using the above identification to show that $S^1 not cong mathbb R^2$. This is a more comfortable setting to try out your "direct method" that you were attempting in the post.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 28 '18 at 5:18









астон вілла олоф мэллбэрг

37.4k33376




37.4k33376












  • that is exactly what I needed, my first thoughts were to use connectedness and compactness but was trying to do it without the use of these topological invariants. thanks for that proof, it looks like it works to me and I have to wrap my head round it some more. I was more or less tying to do this using basic principles of topology rather than compactness/connectedness tricks.
    – eyeheartmath
    Nov 28 '18 at 5:31






  • 1




    It is ok. Usually, you will need to be familiar with these tricks : more than the basic principles, they will come to your help as you progress.
    – астон вілла олоф мэллбэрг
    Nov 28 '18 at 5:32










  • exactly, I wanted to know all the different proofs for this one, I like the proof how it uses covering spaces which I was rusty with. Thanks so much! im working through it now
    – eyeheartmath
    Nov 28 '18 at 5:37


















  • that is exactly what I needed, my first thoughts were to use connectedness and compactness but was trying to do it without the use of these topological invariants. thanks for that proof, it looks like it works to me and I have to wrap my head round it some more. I was more or less tying to do this using basic principles of topology rather than compactness/connectedness tricks.
    – eyeheartmath
    Nov 28 '18 at 5:31






  • 1




    It is ok. Usually, you will need to be familiar with these tricks : more than the basic principles, they will come to your help as you progress.
    – астон вілла олоф мэллбэрг
    Nov 28 '18 at 5:32










  • exactly, I wanted to know all the different proofs for this one, I like the proof how it uses covering spaces which I was rusty with. Thanks so much! im working through it now
    – eyeheartmath
    Nov 28 '18 at 5:37
















that is exactly what I needed, my first thoughts were to use connectedness and compactness but was trying to do it without the use of these topological invariants. thanks for that proof, it looks like it works to me and I have to wrap my head round it some more. I was more or less tying to do this using basic principles of topology rather than compactness/connectedness tricks.
– eyeheartmath
Nov 28 '18 at 5:31




that is exactly what I needed, my first thoughts were to use connectedness and compactness but was trying to do it without the use of these topological invariants. thanks for that proof, it looks like it works to me and I have to wrap my head round it some more. I was more or less tying to do this using basic principles of topology rather than compactness/connectedness tricks.
– eyeheartmath
Nov 28 '18 at 5:31




1




1




It is ok. Usually, you will need to be familiar with these tricks : more than the basic principles, they will come to your help as you progress.
– астон вілла олоф мэллбэрг
Nov 28 '18 at 5:32




It is ok. Usually, you will need to be familiar with these tricks : more than the basic principles, they will come to your help as you progress.
– астон вілла олоф мэллбэрг
Nov 28 '18 at 5:32












exactly, I wanted to know all the different proofs for this one, I like the proof how it uses covering spaces which I was rusty with. Thanks so much! im working through it now
– eyeheartmath
Nov 28 '18 at 5:37




exactly, I wanted to know all the different proofs for this one, I like the proof how it uses covering spaces which I was rusty with. Thanks so much! im working through it now
– eyeheartmath
Nov 28 '18 at 5:37


















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build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

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0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
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Mont Emei

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Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
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