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R^2 homeomorphic to R^n

  1. Nov 9, 2008 #1


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    1. The problem statement, all variables and given/known data

    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]).

    2. Relevant equations

    3. 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
  2. jcsd
  3. Nov 9, 2008 #2


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    Science Advisor
    Homework Helper

    Instead of taking out a point, try taking out something bigger :)
  4. Nov 9, 2008 #3


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    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.
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