Homeomorphism between [0,1]/~ and the Hawaiian Earring












0














Let $X$ be the quotient space [0,1]/~ where 0 ~ 1 ~ 1/2 ~ $cdots$ ~ 1/n ~ $cdots$



Let $H$ (the Hawaiian Earring) be the subspace of $mathbb{R}^2$ consisting of the union of circles of radius 1/n and centered at (0,1/n) for all n=1,2,3,...



I need to find a homeomorphism between X and H.



Note: I am studying general topology, so I can't use algebraic topology.






general-topology quotient-spaces







share|cite|improve this question






















  • What have you tried so far?
    – Guus Palmer
    Dec 9 at 19:15










  • I'm having trouble figuring out how to send a point x from [0,1] to a circle of radius x.
    – 111
    Dec 9 at 19:17










  • Your question is not related to algebraic topology!
    – Paul Frost
    Dec 9 at 23:08










  • Sorry I'm new to topology and I just wanted to make sure that any answers did not employ algebraic techniques, sorry
    – 111
    Dec 9 at 23:10






  • 1




    There is no reason to be sorry, your question is fine. I just wanted to say that it is a question to be treated in general topology.
    – Paul Frost
    Dec 9 at 23:13


















0














Let $X$ be the quotient space [0,1]/~ where 0 ~ 1 ~ 1/2 ~ $cdots$ ~ 1/n ~ $cdots$



Let $H$ (the Hawaiian Earring) be the subspace of $mathbb{R}^2$ consisting of the union of circles of radius 1/n and centered at (0,1/n) for all n=1,2,3,...



I need to find a homeomorphism between X and H.



Note: I am studying general topology, so I can't use algebraic topology.






general-topology quotient-spaces







share|cite|improve this question






















  • What have you tried so far?
    – Guus Palmer
    Dec 9 at 19:15










  • I'm having trouble figuring out how to send a point x from [0,1] to a circle of radius x.
    – 111
    Dec 9 at 19:17










  • Your question is not related to algebraic topology!
    – Paul Frost
    Dec 9 at 23:08










  • Sorry I'm new to topology and I just wanted to make sure that any answers did not employ algebraic techniques, sorry
    – 111
    Dec 9 at 23:10






  • 1




    There is no reason to be sorry, your question is fine. I just wanted to say that it is a question to be treated in general topology.
    – Paul Frost
    Dec 9 at 23:13
















0












0








0


1





Let $X$ be the quotient space [0,1]/~ where 0 ~ 1 ~ 1/2 ~ $cdots$ ~ 1/n ~ $cdots$



Let $H$ (the Hawaiian Earring) be the subspace of $mathbb{R}^2$ consisting of the union of circles of radius 1/n and centered at (0,1/n) for all n=1,2,3,...



I need to find a homeomorphism between X and H.



Note: I am studying general topology, so I can't use algebraic topology.






general-topology quotient-spaces







share|cite|improve this question













Let $X$ be the quotient space [0,1]/~ where 0 ~ 1 ~ 1/2 ~ $cdots$ ~ 1/n ~ $cdots$



Let $H$ (the Hawaiian Earring) be the subspace of $mathbb{R}^2$ consisting of the union of circles of radius 1/n and centered at (0,1/n) for all n=1,2,3,...



I need to find a homeomorphism between X and H.



Note: I am studying general topology, so I can't use algebraic topology.






general-topology quotient-spaces




general-topology quotient-spaces






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 9 at 19:11









111

275




275












  • What have you tried so far?
    – Guus Palmer
    Dec 9 at 19:15










  • I'm having trouble figuring out how to send a point x from [0,1] to a circle of radius x.
    – 111
    Dec 9 at 19:17










  • Your question is not related to algebraic topology!
    – Paul Frost
    Dec 9 at 23:08










  • Sorry I'm new to topology and I just wanted to make sure that any answers did not employ algebraic techniques, sorry
    – 111
    Dec 9 at 23:10






  • 1




    There is no reason to be sorry, your question is fine. I just wanted to say that it is a question to be treated in general topology.
    – Paul Frost
    Dec 9 at 23:13




















  • What have you tried so far?
    – Guus Palmer
    Dec 9 at 19:15










  • I'm having trouble figuring out how to send a point x from [0,1] to a circle of radius x.
    – 111
    Dec 9 at 19:17










  • Your question is not related to algebraic topology!
    – Paul Frost
    Dec 9 at 23:08










  • Sorry I'm new to topology and I just wanted to make sure that any answers did not employ algebraic techniques, sorry
    – 111
    Dec 9 at 23:10






  • 1




    There is no reason to be sorry, your question is fine. I just wanted to say that it is a question to be treated in general topology.
    – Paul Frost
    Dec 9 at 23:13


















What have you tried so far?
– Guus Palmer
Dec 9 at 19:15




What have you tried so far?
– Guus Palmer
Dec 9 at 19:15












I'm having trouble figuring out how to send a point x from [0,1] to a circle of radius x.
– 111
Dec 9 at 19:17




I'm having trouble figuring out how to send a point x from [0,1] to a circle of radius x.
– 111
Dec 9 at 19:17












Your question is not related to algebraic topology!
– Paul Frost
Dec 9 at 23:08




Your question is not related to algebraic topology!
– Paul Frost
Dec 9 at 23:08












Sorry I'm new to topology and I just wanted to make sure that any answers did not employ algebraic techniques, sorry
– 111
Dec 9 at 23:10




Sorry I'm new to topology and I just wanted to make sure that any answers did not employ algebraic techniques, sorry
– 111
Dec 9 at 23:10




1




1




There is no reason to be sorry, your question is fine. I just wanted to say that it is a question to be treated in general topology.
– Paul Frost
Dec 9 at 23:13






There is no reason to be sorry, your question is fine. I just wanted to say that it is a question to be treated in general topology.
– Paul Frost
Dec 9 at 23:13












1 Answer
1






active

oldest

votes


















1














You just map $left[frac12,1right]$ onto the largest circle, $left[frac13,frac12right]$ onto the largest one and so on…



That is, if $xin(0,1]$, then $x=fraclambda{n+1}+frac{1-lambda}n$ for some $ninmathbb N$ and some $lambdain[0,1]$. Then$$f(x)=left(frac1nsin(2pilambda),frac1n+-frac1ncos(2pilambda)right).$$And, of course, $f(0)=(0,0)$.






share|cite|improve this answer























  • What is the motivation for this formula?
    – 111
    Dec 9 at 22:09










  • That's a strange question. That's what occured to me to solve the problem.
    – José Carlos Santos
    Dec 9 at 22:16






  • 1




    You are right: you stated it from the start. I shall edit my answer then.
    – José Carlos Santos
    Dec 9 at 22:40






  • 1




    I've edited my answer. I hope that everything is clear now.
    – José Carlos Santos
    Dec 9 at 22:45






  • 1




    For continuity, see $f$ as a map from $[0,1]$ onto $H$. It is clear that $f$ is continuous on each interval $left[frac1{n+1},frac1nright]$. So, it is continuous on $(0,1]$. FInally, it is clear that $f(0)=lim_{xto0}f(x)$ and that therefore $f$ is continuous on $0$ too. Then, yes, just use the fact that $[0,1]$ is compact.
    – José Carlos Santos
    Dec 9 at 23:25













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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









1














You just map $left[frac12,1right]$ onto the largest circle, $left[frac13,frac12right]$ onto the largest one and so on…



That is, if $xin(0,1]$, then $x=fraclambda{n+1}+frac{1-lambda}n$ for some $ninmathbb N$ and some $lambdain[0,1]$. Then$$f(x)=left(frac1nsin(2pilambda),frac1n+-frac1ncos(2pilambda)right).$$And, of course, $f(0)=(0,0)$.






share|cite|improve this answer























  • What is the motivation for this formula?
    – 111
    Dec 9 at 22:09










  • That's a strange question. That's what occured to me to solve the problem.
    – José Carlos Santos
    Dec 9 at 22:16






  • 1




    You are right: you stated it from the start. I shall edit my answer then.
    – José Carlos Santos
    Dec 9 at 22:40






  • 1




    I've edited my answer. I hope that everything is clear now.
    – José Carlos Santos
    Dec 9 at 22:45






  • 1




    For continuity, see $f$ as a map from $[0,1]$ onto $H$. It is clear that $f$ is continuous on each interval $left[frac1{n+1},frac1nright]$. So, it is continuous on $(0,1]$. FInally, it is clear that $f(0)=lim_{xto0}f(x)$ and that therefore $f$ is continuous on $0$ too. Then, yes, just use the fact that $[0,1]$ is compact.
    – José Carlos Santos
    Dec 9 at 23:25


















1














You just map $left[frac12,1right]$ onto the largest circle, $left[frac13,frac12right]$ onto the largest one and so on…



That is, if $xin(0,1]$, then $x=fraclambda{n+1}+frac{1-lambda}n$ for some $ninmathbb N$ and some $lambdain[0,1]$. Then$$f(x)=left(frac1nsin(2pilambda),frac1n+-frac1ncos(2pilambda)right).$$And, of course, $f(0)=(0,0)$.






share|cite|improve this answer























  • What is the motivation for this formula?
    – 111
    Dec 9 at 22:09










  • That's a strange question. That's what occured to me to solve the problem.
    – José Carlos Santos
    Dec 9 at 22:16






  • 1




    You are right: you stated it from the start. I shall edit my answer then.
    – José Carlos Santos
    Dec 9 at 22:40






  • 1




    I've edited my answer. I hope that everything is clear now.
    – José Carlos Santos
    Dec 9 at 22:45






  • 1




    For continuity, see $f$ as a map from $[0,1]$ onto $H$. It is clear that $f$ is continuous on each interval $left[frac1{n+1},frac1nright]$. So, it is continuous on $(0,1]$. FInally, it is clear that $f(0)=lim_{xto0}f(x)$ and that therefore $f$ is continuous on $0$ too. Then, yes, just use the fact that $[0,1]$ is compact.
    – José Carlos Santos
    Dec 9 at 23:25
















1












1








1






You just map $left[frac12,1right]$ onto the largest circle, $left[frac13,frac12right]$ onto the largest one and so on…



That is, if $xin(0,1]$, then $x=fraclambda{n+1}+frac{1-lambda}n$ for some $ninmathbb N$ and some $lambdain[0,1]$. Then$$f(x)=left(frac1nsin(2pilambda),frac1n+-frac1ncos(2pilambda)right).$$And, of course, $f(0)=(0,0)$.






share|cite|improve this answer














You just map $left[frac12,1right]$ onto the largest circle, $left[frac13,frac12right]$ onto the largest one and so on…



That is, if $xin(0,1]$, then $x=fraclambda{n+1}+frac{1-lambda}n$ for some $ninmathbb N$ and some $lambdain[0,1]$. Then$$f(x)=left(frac1nsin(2pilambda),frac1n+-frac1ncos(2pilambda)right).$$And, of course, $f(0)=(0,0)$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 9 at 22:45

























answered Dec 9 at 19:19









José Carlos Santos

150k22121221




150k22121221












  • What is the motivation for this formula?
    – 111
    Dec 9 at 22:09










  • That's a strange question. That's what occured to me to solve the problem.
    – José Carlos Santos
    Dec 9 at 22:16






  • 1




    You are right: you stated it from the start. I shall edit my answer then.
    – José Carlos Santos
    Dec 9 at 22:40






  • 1




    I've edited my answer. I hope that everything is clear now.
    – José Carlos Santos
    Dec 9 at 22:45






  • 1




    For continuity, see $f$ as a map from $[0,1]$ onto $H$. It is clear that $f$ is continuous on each interval $left[frac1{n+1},frac1nright]$. So, it is continuous on $(0,1]$. FInally, it is clear that $f(0)=lim_{xto0}f(x)$ and that therefore $f$ is continuous on $0$ too. Then, yes, just use the fact that $[0,1]$ is compact.
    – José Carlos Santos
    Dec 9 at 23:25




















  • What is the motivation for this formula?
    – 111
    Dec 9 at 22:09










  • That's a strange question. That's what occured to me to solve the problem.
    – José Carlos Santos
    Dec 9 at 22:16






  • 1




    You are right: you stated it from the start. I shall edit my answer then.
    – José Carlos Santos
    Dec 9 at 22:40






  • 1




    I've edited my answer. I hope that everything is clear now.
    – José Carlos Santos
    Dec 9 at 22:45






  • 1




    For continuity, see $f$ as a map from $[0,1]$ onto $H$. It is clear that $f$ is continuous on each interval $left[frac1{n+1},frac1nright]$. So, it is continuous on $(0,1]$. FInally, it is clear that $f(0)=lim_{xto0}f(x)$ and that therefore $f$ is continuous on $0$ too. Then, yes, just use the fact that $[0,1]$ is compact.
    – José Carlos Santos
    Dec 9 at 23:25


















What is the motivation for this formula?
– 111
Dec 9 at 22:09




What is the motivation for this formula?
– 111
Dec 9 at 22:09












That's a strange question. That's what occured to me to solve the problem.
– José Carlos Santos
Dec 9 at 22:16




That's a strange question. That's what occured to me to solve the problem.
– José Carlos Santos
Dec 9 at 22:16




1




1




You are right: you stated it from the start. I shall edit my answer then.
– José Carlos Santos
Dec 9 at 22:40




You are right: you stated it from the start. I shall edit my answer then.
– José Carlos Santos
Dec 9 at 22:40




1




1




I've edited my answer. I hope that everything is clear now.
– José Carlos Santos
Dec 9 at 22:45




I've edited my answer. I hope that everything is clear now.
– José Carlos Santos
Dec 9 at 22:45




1




1




For continuity, see $f$ as a map from $[0,1]$ onto $H$. It is clear that $f$ is continuous on each interval $left[frac1{n+1},frac1nright]$. So, it is continuous on $(0,1]$. FInally, it is clear that $f(0)=lim_{xto0}f(x)$ and that therefore $f$ is continuous on $0$ too. Then, yes, just use the fact that $[0,1]$ is compact.
– José Carlos Santos
Dec 9 at 23:25






For continuity, see $f$ as a map from $[0,1]$ onto $H$. It is clear that $f$ is continuous on each interval $left[frac1{n+1},frac1nright]$. So, it is continuous on $(0,1]$. FInally, it is clear that $f(0)=lim_{xto0}f(x)$ and that therefore $f$ is continuous on $0$ too. Then, yes, just use the fact that $[0,1]$ is compact.
– José Carlos Santos
Dec 9 at 23:25




















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