R^2 and R^2 - {(0,0)} are not homeomorphic

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In summary, the two spaces are not homeomorphic because there is no point in R that is homeomorphic to the point (0,0) in R^2.
  • #1
poet_3000
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Let R denote the real line and R^2 the real plane.

How do I show that R^2 and R^2 - {(0,0)} are not homeomorphic? I don't even know where to start since I cannot think of a topological property that is in one but not in the other. Can anyone give me a clue on what that property is?

A similar problem is this:
Consider S = the union of the x and y- axes together with the subspace topology inherited from R^2. Show that S and R are not homeomorphic.

Thanks in advance!
 
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  • #2
poet_3000 said:
Let R denote the real line and R^2 the real plane.

How do I show that R^2 and R^2 - {(0,0)} are not homeomorphic? I don't even know where to start since I cannot think of a topological property that is in one but not in the other. Can anyone give me a clue on what that property is?

Simply connectedness. Or equivalent: calculate the fundamental group of the two spaces. Of R^2 the group is trivial, but of R^2\{(0,0)} the group is Z.

A similar problem is this:
Consider S = the union of the x and y- axes together with the subspace topology inherited from R^2. Show that S and R are not homeomorphic.

Thanks in advance!

In S, there exists a point such that if you remove the point, then there are 4 connected components. But there is no such point in R: any point that you remove in R will yield 2 connected components.
 
  • #3
poet_3000 said:
Let R denote the real line and R^2 the real plane.

How do I show that R^2 and R^2 - {(0,0)} are not homeomorphic? I don't even know where to start since I cannot think of a topological property that is in one but not in the other. Can anyone give me a clue on what that property is?

A similar problem is this:
Consider S = the union of the x and y- axes together with the subspace topology inherited from R^2. Show that S and R are not homeomorphic.

Thanks in advance!

for the first one there are classic algebraic topology proofs. Do you want one of these or would you like something more elementary?
 
  • #4
Oh. Thank you for your replies! It really helped me. Thank you micromass and lavinia.

lavinia, may I know your other suggestion?
 
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  • #5
poet_3000 said:
Oh. Thank you for your replies! It really helped me. Thank you micromass and lavinia.

lavinia, may I know your other suggestion?

I never tried a simpler proof but let's give it a try.

If there were a homeomorphism from R^-0 to R^2, a Cauchy sequence that converges to the origin must be mapped to infinity. This means that the punctured disc around the origin must be mapped to a neighborhood of infinity that is bounded by a simple closed curve.

But then the neighborhood of infinity must be mapped into the inside this simple closed curve. A sequence that diverges to infinity would be mapped to a bounded sequence which must have a limit point. that is impossible.This generalizes to higher dimensions.

It might be fun to think up some others.
 
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  • #6
For the second problem, here is a more complicated proof.

If there is a homeomorphism from the union of the x and y axes then consider the image of the x axis. This is homeomorphic to R so the image of the origin is surrounded by an open interval. That does it because now the image of the y-axis can not penetrate this interval (which it must by continuity) because the map is 1-1.
 
  • #7
the way i said it for my senior thesis (i am not joking) is:

ok we have da loop, and we have da group. we loop de loop and group da loop, to da loop group.

but onoz! we have hole. so sorry, no homeomorphism for j00...

(and then i popped a balloon with a pin, and said: see?)
 
  • #8
Deveno said:
the way i said it for my senior thesis (i am not joking) is:

ok we have da loop, and we have da group. we loop de loop and group da loop, to da loop group.

but onoz! we have hole. so sorry, no homeomorphism for j00...

(and then i popped a balloon with a pin, and said: see?)

that is entertaining deveno! LOL! :p
 

1. What is R^2 and R^2 - {(0,0)}?

R^2 refers to the two-dimensional Euclidean space, also known as the Cartesian coordinate system. R^2 - {(0,0)} refers to the same space with the origin point (0,0) removed.

2. Why are R^2 and R^2 - {(0,0)} not homeomorphic?

Homeomorphism is a mathematical concept that describes a continuous function between two topological spaces that has a continuous inverse function. Since the origin point (0,0) is removed in R^2 - {(0,0)}, the space is not connected and therefore cannot have a continuous inverse function. This means that R^2 and R^2 - {(0,0)} cannot be homeomorphic.

3. How do you prove that R^2 and R^2 - {(0,0)} are not homeomorphic?

The most common way to prove that two spaces are not homeomorphic is by showing that they have different topological properties, such as connectedness, compactness, or separation axioms. In the case of R^2 and R^2 - {(0,0)}, we can show that R^2 is connected while R^2 - {(0,0)} is not, thus proving that they are not homeomorphic.

4. Can you give an example to illustrate why R^2 and R^2 - {(0,0)} are not homeomorphic?

Yes, we can use the concept of path-connectedness to illustrate why R^2 and R^2 - {(0,0)} are not homeomorphic. In R^2, any two points can be connected by a continuous path. However, in R^2 - {(0,0)}, this is not possible since the origin point is removed, and there is no way to continuously connect any point to the origin. This difference in path-connectedness is a topological property that proves these two spaces are not homeomorphic.

5. What is the significance of R^2 and R^2 - {(0,0)} not being homeomorphic?

The fact that R^2 and R^2 - {(0,0)} are not homeomorphic highlights the importance of the origin point in the two-dimensional Euclidean space. It also shows that removing a single point can drastically change the topological properties of a space. This concept is fundamental in topology and has many applications in mathematics and other fields.

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