Show R^2 \{(0,0)} and {(x,y) | 1 < sqrt(x^2+y^2) < 3} are homeomorphic

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In summary: Apply a scaling and translation to it to map (1,3) to (0,\infty). In summary, the student is asking for help in understanding how to start a problem involving homeomorphic functions. They are using a book called Basic Topology by M.A. Armstrong and are struggling with the density of the material. The problem involves finding a function that maps a punctured plane and a circular disk, and the student is looking for hints or advice on how to approach this problem. Several suggestions have been given, including using polar coordinates and the tangent function, as well as scaling and translating the arctanh function. The end goal is to find a continuous bijection with a continuous inverse between the two sets.
  • #1
Mikaelochi
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TL;DR Summary
Basically I know that for two things to be homeomorphic the function that maps the two things have to be bijective and continuous but how do I even start this problem?
As I said in the summary, I don't really know how to even figure out which function would be appropriate to map the two sets that I described in the title. I'm using the book called Basic Topology by M.A. Armstrong. The book can sometimes be really dense. I am having a really hard time knowing where to start with this problem because it's not as simple as just proving a particular function is bijective and continuous. I know the first thing is the plane minus the origin and the second thing is a circular disk. Any help or hints or whatever would really help! This course just has me so lost sometimes.
 
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  • #2
I would probably begin by showing that ##\mathbb{R} \setminus \{0\}## and ##(-3,-1) \cup (1,3)## are homeomorphic. Next, I would use that the sets in the plane have radial symmetry.

(There are some theorems that can give faster conclusions, but I do not know if you have already proved those. Also, when you are starting out, it may not be bad to actually construct the homeomorphism.)
 
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  • #3
so for the example you just gave, would I need to find a function that maps R \ {0} to (-3, -1) U (1,3)? I'm not very well versed in homeomorphism proofs
 
  • #4
Mikaelochi said:
Summary:: Basically I know that for two things to be homeomorphic the function that maps the two things have to be bijective and continuous but how do I even start this problem?

As I said in the summary, I don't really know how to even figure out which function would be appropriate to map the two sets that I described in the title. I'm using the book called Basic Topology by M.A. Armstrong. The book can sometimes be really dense. I am having a really hard time knowing where to start with this problem because it's not as simple as just proving a particular function is bijective and continuous. I know the first thing is the plane minus the origin and the second thing is a circular disk. Any help or hints or whatever would really help! This course just has me so lost sometimes.
You have a punctured plane and a torus. Draw a circle ##\sqrt{x^2+y^2}=2##. This circle will be fixed points. All points ##1<\sqrt{x^2+y^2}<2 ## will be stretched to ##0<\sqrt{x^2+y^2}<2## and all points ##2<\sqrt{x^2+y^2}<3## will be stretched to ##2<\sqrt{x^2+y^2}< \infty .##

Now all you have to do is to figure out the transformations and prove that they are a homeomorphism. Polar coordinates might be easier to handle in this case.
 
  • #5
fresh_42 said:
You have a punctured plane and a torus.
Technically, the latter is an annulus. The sets in the problem are in ##\mathbb R^2##, not ##\mathbb R^3##.
 
  • #6
@Mikaelochi , to help you get started with @S.G. Janssens's hint, the tangent function (positive half of the principal branch) maps the interval ##(0, \pi/2)## to the interval ##(0, \infty)##. If you can transform the tangent function by compressing it a bit, followed by a translation, you should be able to map the interval ##(1, 3)## to the interval ##(0, \infty)##. If you follow that, finding a map from the other half, ##(-3, -1)## shouldn't be too hard.
 
  • #7
I think these approaches with the interval are way too complicated. If we work with rays, we have only one coordinate to bother, the radius. Furthermore, stretchings are linear maps, i.e. determined by two points. This immediately yields continuity and bijection. And if we set ##r':=r-2## we have an origin at ##r'=0.##
 
  • #8
Mikaelochi said:
Summary:: Basically I know that for two things to be homeomorphic the function that maps the two things have to be bijective and continuous but how do I even start this problem?

As I said in the summary, I don't really know how to even figure out which function would be appropriate to map the two sets that I described in the title. I'm using the book called Basic Topology by M.A. Armstrong. The book can sometimes be really dense. I am having a really hard time knowing where to start with this problem because it's not as simple as just proving a particular function is bijective and continuous. I know the first thing is the plane minus the origin and the second thing is a circular disk. Any help or hints or whatever would really help! This course just has me so lost sometimes.
Don't mean to nitpick, but this is an important detail: the homeomorphism is a continuous bijection...with a continuous inverse. Not every continuous bijection has a continuous inverse. I think this is an important detail.
 
  • #9
Mark44 said:
@Mikaelochi , to help you get started with @S.G. Janssens's hint, the tangent function (positive half of the principal branch) maps the interval ##(0, \pi/2)## to the interval ##(0, \infty)##. If you can transform the tangent function by compressing it a bit, followed by a translation, you should be able to map the interval ##(1, 3)## to the interval ##(0, \infty)##. If you follow that, finding a map from the other half, ##(-3, -1)## shouldn't be too hard.

Alternatively, [itex]\operatorname{arctanh}[/itex] maps [itex](0,1)[/itex] to [itex](0,\infty)[/itex].
 

What is the definition of homeomorphism?

A homeomorphism is a continuous function between two topological spaces that has a continuous inverse function.

What is the topological space of R^2?

The topological space of R^2 is the set of all ordered pairs of real numbers (x,y) with the standard topology.

How can R^2 \{(0,0)} and {(x,y) | 1 < sqrt(x^2+y^2) < 3} be homeomorphic?

R^2 \{(0,0)} and {(x,y) | 1 < sqrt(x^2+y^2) < 3} are homeomorphic because there exists a continuous function between them that has a continuous inverse function. This function maps points in R^2 \{(0,0)} to points in {(x,y) | 1 < sqrt(x^2+y^2) < 3} and vice versa while preserving the topology of both spaces.

How does the removal of (0,0) in R^2 affect the homeomorphism?

The removal of (0,0) in R^2 does not affect the homeomorphism between R^2 \{(0,0)} and {(x,y) | 1 < sqrt(x^2+y^2) < 3} as long as the function mapping points between the two spaces is still continuous and has a continuous inverse function.

What are the implications of R^2 \{(0,0)} and {(x,y) | 1 < sqrt(x^2+y^2) < 3} being homeomorphic?

The fact that R^2 \{(0,0)} and {(x,y) | 1 < sqrt(x^2+y^2) < 3} are homeomorphic means that they have the same topological properties. This implies that any topological property that holds for one space will also hold for the other, allowing for easier analysis and generalization of results.

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