R^2 homeomorphic to R^n

  • Thread starter gop
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In summary, to prove that R^2 and R^n are not homeomorphic if n\neq2, consider the complement of a point in R^2 or R^n. In R^n, the restriction of a continuous bijection from R to R^n is also a continuous bijection. However, since R\{0} is not connected but R^n\{f(0)} is if n\neq 1, there is a contradiction. To show that R^2 is not homeomorphic to R^n, consider a closed loop containing a point in R2\{0} and in Rn. In R2\{0}, it cannot be contracted to a point, but in Rn, it can
  • #1
gop
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Homework Statement



Prove that [tex]R^2[/tex] and [tex]R^n[/tex] are not homeomorphic if [tex]n\neq2[/tex] (Hint: Consider the complement of a point in [tex]R^2[/tex] or [tex]R^n[/tex]).

Homework Equations





The Attempt at a Solution




The proof that [tex]R^n[/tex] is not homeomorphic to [tex]R[/tex] is done by considering that if they are homeomorphic i.e. there exists a continuous bijection with continuous inverse [tex]f:R\rightarrow R^n[/tex]. Then the restriction [tex]f:R\backslash {0} \rightarrow R^n\backslash f(0)[/tex] is also a continuous bijection. Since [tex]R\backslash {0}[/tex] is not connected but [tex]R^n\backslash f(0)[/tex] is if [tex]n\neq 1[/tex] we have a contradiction.

However, to show that [tex]R^2[/tex] is not homeomorphic to [tex]R^n[/tex] this doesn't work. There is also no other topological invariant I could detect. Both [tex]R^2[/tex] and [tex]R^n[/tex] are contractible to a point and thus have the same fundamental group, for example.

I would appreciate any idea
thx
 
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  • #2
Instead of taking out a point, try taking out something bigger :)
 
  • #3
gop said:

Homework Statement



Prove that [tex]R^2[/tex] and [tex]R^n[/tex] are not homeomorphic if [tex]n\neq2[/tex] (Hint: Consider the complement of a point in [tex]R^2[/tex] or [tex]R^n[/tex]).

Homework Equations





The Attempt at a Solution




The proof that [tex]R^n[/tex] is not homeomorphic to [tex]R[/tex] is done by considering that if they are homeomorphic i.e. there exists a continuous bijection with continuous inverse [tex]f:R\rightarrow R^n[/tex]. Then the restriction [tex]f:R\backslash {0} \rightarrow R^n\backslash f(0)[/tex] is also a continuous bijection. Since [tex]R\backslash {0}[/tex] is not connected but [tex]R^n\backslash f(0)[/tex] is if [tex]n\neq 1[/tex] we have a contradiction.

However, to show that [tex]R^2[/tex] is not homeomorphic to [tex]R^n[/tex] this doesn't work. There is also no other topological invariant I could detect. Both [tex]R^2[/tex] and [tex]R^n[/tex] are contractible to a point and thus have the same fundamental group, for example.

I would appreciate any idea
thx
You can continue that idea. In R2\{0}, a closed loop containing a point p cannot be contracted to a point. In Rn, for n larger than 2, it can.
 

What does it mean for R^2 to be homeomorphic to R^n?

In mathematics, homeomorphism is a term used to describe the relationship between two topological spaces that have the same underlying structure. In this case, R^2 and R^n both refer to Euclidean spaces of different dimensions, but they can be transformed into one another through a homeomorphism.

Why is this concept important in mathematics?

Homeomorphism is a fundamental concept in topology, which is a branch of mathematics that studies the properties of spaces that are preserved under continuous deformations. It allows us to compare different spaces and identify their common features, which can be useful in solving problems and proving theorems.

How can we determine if R^2 is homeomorphic to R^n?

To determine if R^2 is homeomorphic to R^n, we can look for a continuous and bijective function between the two spaces. This function should also have a continuous inverse, meaning that it can be easily transformed back to its original form. If such a function exists, then R^2 is homeomorphic to R^n.

What are some examples of spaces that are homeomorphic to R^2?

Some examples of spaces that are homeomorphic to R^2 include the surface of a sphere, a torus, and a Mobius strip. These spaces have different shapes and dimensions, but they can be continuously deformed into R^2, making them homeomorphic to this space.

How does the concept of homeomorphism relate to real-world applications?

Homeomorphism has many practical applications, especially in fields like physics, engineering, and computer science. For example, it can be used to model the deformation of materials, analyze the behavior of complex systems, and develop efficient algorithms for solving problems. It also has applications in data visualization and image processing.

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