Homeomorphism between [0,1]/~ and the Hawaiian Earring
0
$begingroup$
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
asked Dec 9 '18 at 19:11
111111
275
$endgroup$
add a comment |
0
$begingroup$
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
asked Dec 9 '18 at 19:11
111111
275
$endgroup$
$begingroup$
What have you tried so far?
$endgroup$
– Algebear
Dec 9 '18 at 19:15
$begingroup$
I'm having trouble figuring out how to send a point x from [0,1] to a circle of radius x.
$endgroup$
– 111
Dec 9 '18 at 19:17
$begingroup$
Your question is not related to algebraic topology!
$endgroup$
– Paul Frost
Dec 9 '18 at 23:08
$begingroup$
Sorry I'm new to topology and I just wanted to make sure that any answers did not employ algebraic techniques, sorry
$endgroup$
– 111
Dec 9 '18 at 23:10
1
$begingroup$
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.
$endgroup$
– Paul Frost
Dec 9 '18 at 23:13
add a comment |
0
0
0
0
$begingroup$
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
asked Dec 9 '18 at 19:11
111111
275
$endgroup$
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
asked Dec 9 '18 at 19:11
111111
275
asked Dec 9 '18 at 19:11
111111
275
asked Dec 9 '18 at 19:11
111111
275
asked Dec 9 '18 at 19:11
111111
275
asked Dec 9 '18 at 19:11
111111
275
275
$begingroup$
What have you tried so far?
$endgroup$
– Algebear
Dec 9 '18 at 19:15
$begingroup$
I'm having trouble figuring out how to send a point x from [0,1] to a circle of radius x.
$endgroup$
– 111
Dec 9 '18 at 19:17
$begingroup$
Your question is not related to algebraic topology!
$endgroup$
– Paul Frost
Dec 9 '18 at 23:08
$begingroup$
Sorry I'm new to topology and I just wanted to make sure that any answers did not employ algebraic techniques, sorry
$endgroup$
– 111
Dec 9 '18 at 23:10
1
$begingroup$
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.
$endgroup$
– Paul Frost
Dec 9 '18 at 23:13
add a comment |
$begingroup$
What have you tried so far?
$endgroup$
– Algebear
Dec 9 '18 at 19:15
$begingroup$
I'm having trouble figuring out how to send a point x from [0,1] to a circle of radius x.
$endgroup$
– 111
Dec 9 '18 at 19:17
$begingroup$
Your question is not related to algebraic topology!
$endgroup$
– Paul Frost
Dec 9 '18 at 23:08
$begingroup$
Sorry I'm new to topology and I just wanted to make sure that any answers did not employ algebraic techniques, sorry
$endgroup$
– 111
Dec 9 '18 at 23:10
1
$begingroup$
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.
$endgroup$
– Paul Frost
Dec 9 '18 at 23:13
$begingroup$
What have you tried so far?
$endgroup$
– Algebear
Dec 9 '18 at 19:15
$begingroup$
What have you tried so far?
$endgroup$
– Algebear
Dec 9 '18 at 19:15
$begingroup$
I'm having trouble figuring out how to send a point x from [0,1] to a circle of radius x.
$endgroup$
– 111
Dec 9 '18 at 19:17
$begingroup$
I'm having trouble figuring out how to send a point x from [0,1] to a circle of radius x.
$endgroup$
– 111
Dec 9 '18 at 19:17
$begingroup$
Your question is not related to algebraic topology!
$endgroup$
– Paul Frost
Dec 9 '18 at 23:08
$begingroup$
Your question is not related to algebraic topology!
$endgroup$
– Paul Frost
Dec 9 '18 at 23:08
$begingroup$
Sorry I'm new to topology and I just wanted to make sure that any answers did not employ algebraic techniques, sorry
$endgroup$
– 111
Dec 9 '18 at 23:10
$begingroup$
Sorry I'm new to topology and I just wanted to make sure that any answers did not employ algebraic techniques, sorry
$endgroup$
– 111
Dec 9 '18 at 23:10
1
1
$begingroup$
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.
$endgroup$
– Paul Frost
Dec 9 '18 at 23:13
$begingroup$
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.
$endgroup$
– Paul Frost
Dec 9 '18 at 23:13
add a comment |
1 Answer
1
active
oldest
votes
1
$begingroup$
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, $f(0)=(0,0)$.
answered Dec 9 '18 at 19:19
José Carlos SantosJosé Carlos Santos
161k22127232
$endgroup$
$begingroup$
What is the motivation for this formula?
$endgroup$
– 111
Dec 9 '18 at 22:09
$begingroup$
That's a strange question. That's what occured to me to solve the problem.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 22:16
1
$begingroup$
You are right: you stated it from the start. I shall edit my answer then.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 22:40
1
$begingroup$
I've edited my answer. I hope that everything is clear now.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 22:45
1
$begingroup$
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.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 23:25
|
show 8 more comments
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
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, $f(0)=(0,0)$.
answered Dec 9 '18 at 19:19
José Carlos SantosJosé Carlos Santos
161k22127232
$endgroup$
$begingroup$
What is the motivation for this formula?
$endgroup$
– 111
Dec 9 '18 at 22:09
$begingroup$
That's a strange question. That's what occured to me to solve the problem.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 22:16
1
$begingroup$
You are right: you stated it from the start. I shall edit my answer then.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 22:40
1
$begingroup$
I've edited my answer. I hope that everything is clear now.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 22:45
1
$begingroup$
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.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 23:25
|
show 8 more comments
1
$begingroup$
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, $f(0)=(0,0)$.
answered Dec 9 '18 at 19:19
José Carlos SantosJosé Carlos Santos
161k22127232
$endgroup$
$begingroup$
What is the motivation for this formula?
$endgroup$
– 111
Dec 9 '18 at 22:09
$begingroup$
That's a strange question. That's what occured to me to solve the problem.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 22:16
1
$begingroup$
You are right: you stated it from the start. I shall edit my answer then.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 22:40
1
$begingroup$
I've edited my answer. I hope that everything is clear now.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 22:45
1
$begingroup$
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.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 23:25
|
show 8 more comments
1
1
1
$begingroup$
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, $f(0)=(0,0)$.
answered Dec 9 '18 at 19:19
José Carlos SantosJosé Carlos Santos
161k22127232
$endgroup$
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, $f(0)=(0,0)$.
answered Dec 9 '18 at 19:19
José Carlos SantosJosé Carlos Santos
161k22127232
edited Dec 9 '18 at 22:45
answered Dec 9 '18 at 19:19
José Carlos SantosJosé Carlos Santos
161k22127232
answered Dec 9 '18 at 19:19
José Carlos SantosJosé Carlos Santos
161k22127232
answered Dec 9 '18 at 19:19
José Carlos SantosJosé Carlos Santos
161k22127232
161k22127232
$begingroup$
What is the motivation for this formula?
$endgroup$
– 111
Dec 9 '18 at 22:09
$begingroup$
That's a strange question. That's what occured to me to solve the problem.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 22:16
1
$begingroup$
You are right: you stated it from the start. I shall edit my answer then.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 22:40
1
$begingroup$
I've edited my answer. I hope that everything is clear now.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 22:45
1
$begingroup$
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.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 23:25
|
show 8 more comments
$begingroup$
What is the motivation for this formula?
$endgroup$
– 111
Dec 9 '18 at 22:09
$begingroup$
That's a strange question. That's what occured to me to solve the problem.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 22:16
1
$begingroup$
You are right: you stated it from the start. I shall edit my answer then.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 22:40
1
$begingroup$
I've edited my answer. I hope that everything is clear now.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 22:45
1
$begingroup$
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.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 23:25
$begingroup$
What is the motivation for this formula?
$endgroup$
– 111
Dec 9 '18 at 22:09
$begingroup$
What is the motivation for this formula?
$endgroup$
– 111
Dec 9 '18 at 22:09
$begingroup$
That's a strange question. That's what occured to me to solve the problem.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 22:16
$begingroup$
That's a strange question. That's what occured to me to solve the problem.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 22:16
1
1
$begingroup$
You are right: you stated it from the start. I shall edit my answer then.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 22:40
$begingroup$
You are right: you stated it from the start. I shall edit my answer then.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 22:40
1
1
$begingroup$
I've edited my answer. I hope that everything is clear now.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 22:45
$begingroup$
I've edited my answer. I hope that everything is clear now.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 22:45
1
1
$begingroup$
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.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 23:25
$begingroup$
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.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 23:25
|
show 8 more comments
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$begingroup$
What have you tried so far?
$endgroup$
– Algebear
Dec 9 '18 at 19:15
$begingroup$
I'm having trouble figuring out how to send a point x from [0,1] to a circle of radius x.
$endgroup$
– 111
Dec 9 '18 at 19:17
$begingroup$
Your question is not related to algebraic topology!
$endgroup$
– Paul Frost
Dec 9 '18 at 23:08
$begingroup$
Sorry I'm new to topology and I just wanted to make sure that any answers did not employ algebraic techniques, sorry
$endgroup$
– 111
Dec 9 '18 at 23:10
1
$begingroup$
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.
$endgroup$
– Paul Frost
Dec 9 '18 at 23:13