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Showing that R and R^2 are not homeomorphic

  1. Aug 20, 2010 #1

    radou

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

    Prove that R and R^2 are not homeomorphic.

    3. The attempt at a solution

    So, to prove this, one needs to conclude that there is no homeomorphism between R and R^2. A homeomorphism is a continuous bijection f with a continuous inverse. (Does there exist a bijection at all between these two sets? I assume yes, since they have the same cardinality, but I don't see how to construct it.)

    Assume such mapping f exists - could we derive a contradiction here. I'm really clueless.
     
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  3. Aug 20, 2010 #2

    Office_Shredder

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    You can kind of envision a direct bijection between (0,1) and (0,1)x(0,1) by given a pair of real numbers, construct a single one by alternating their digits. So the pair (.1275, .9999) becomes .19297959.

    There are some issues with well-definedness at points with more than one decimal expansion (for example, .1 vs .099999).

    As far as proving no homeomorphism exists, the classic method involves removing the origin from both and seeing what happens to one and not the other
     
  4. Aug 21, 2010 #3

    radou

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    Could you please be a bit more specific, I'm not sure I understand what you meant.

    I assume I have to find some topological property which R and R^2 don't share, since if there would exist a homeomorphism between them, this property would be preserved. But I simply can't see it - both spaces are connected and separable, right? I should find some other property.
     
  5. Aug 21, 2010 #4

    Office_Shredder

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    If you remove a point from R, what topological property does R have that R minus that point doesn't have?
     
  6. Aug 21, 2010 #5

    radou

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    If we remove some point x from R, then R\{x} is not connected anymore, since it can be written as a union of <-∞, x> and <x, +∞>.
     
  7. Aug 21, 2010 #6

    Office_Shredder

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    Right. Now if R is homeomorphic to R2, what can we conclude about R2?
     
  8. Aug 21, 2010 #7

    radou

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    Well, if R is homeomorphic to R^2, we know that R^2 is connected, too, since continuous functions (and homeomorphisms in particulas) preserve that property. If we remove some x from R now, R\{x} isn't connected anymore. I'm not sure what's the picture. Is we look at the restriction of the mapping to R\{x}, it stays continuous, right? But I only know that it maps connected sets to connected sets. Can it happen for the domain not to be connected, and the image of the map connected?
     
  9. Aug 21, 2010 #8

    Office_Shredder

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    R\{x} is not connected, and our homeomorphism is still a homeomorphism. The space R\{x} is allegedly homeomorphic to is going to be R2\{f(x)}

    In general a domain can be disconnected and have a connected image; for example any constant map. But this can't happen for homeomorphisms (we would hope not, otherwise homeomorphic would be a terribly weak property) and is something that can be proven fairly simply
     
  10. Aug 22, 2010 #9

    radou

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    OK, we proved that R\{x} and R\{f(x)}^2 are not homeomorphic, but how does this imply that R and R^2 are not homeomorphic? Obviuosly both R\{x} and R\{f(x)}^2 are not connected, and our homeomorphism stays a homeomorphism from a disconnected set to a disconnected set. Should we prove that a homeomorphism doesn't in general map a disconnected set into a disconnected set? Or is it simply a game of logic - if we have the implication (which is true) "A is connected => f(A) is connected under a homeomorphism f", then its negation is false, i.e. "A is not connected => f(A) is not connected under a homeomorphism f"? (btw, I'm not really sure my logic is right here) Thanks for your patience :)
     
  11. Aug 22, 2010 #10

    Office_Shredder

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    Not R\{f(x)}2 (which doesn't even make sense), we're looking at R2\{f(x)}; the plane minus a point

    Suppose f is a homeomorphism from R to R2. Then here are the key points:
    1) f is also a homeomorphism from R\{x} to R2\{f(x)}
    2) R\{x} is disconnected, R2\{x} is connected
    3) A connected set cannot be homeomorphic to a disconnected set

    You'll have to fill in the details for the proof
     
  12. Aug 22, 2010 #11

    radou

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    You meant R^2\{f(x)} is connected, right? (in 2) )

    So, if f : R\{x} --> R^2\{f(x)} is a homeomorphism, then it has a continuous inverse f^-1, and since continuous function preserve connectedness, it follows that R\{x} is connected, which is a contradiction.

    I hope I got it right now, thanks for your help.
     
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