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Homogeneous topological space

  1. Nov 12, 2009 #1
    1. The problem statement, all variables and given/known data

    For any [itex] a \in \left( -1,1 \right) [/itex] construct a homeomorphism [itex] f_a: \left( -1,1 \right) \longrightarrow \left( -1,1 \right) [/itex] such that [itex] f_a\left( a \right) = 0 [/itex]. Deduce that [itex] \left( -1,1 \right) [/itex] is homogeneous.


    2. Relevant equations

    Definition of a homogeneous topological space, ie that the exists a homeomorphism for each pair of points x,y which maps x to y.

    3. The attempt at a solution

    I can't find a set a functions which map an arbitrary point to zero and is surjective. My attemps include f = x - a, f = |x - a|, f = sin (x-a) but these are not homeomorphic for arbitrary a.
     
    Last edited by a moderator: Nov 12, 2009
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  3. Nov 12, 2009 #2

    George Jones

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    There are lots of such homeomorphisms. Examples:

    1) join two straight lines that have different slopes (not necessarily equal to one);

    2) use the points (-1, -1), (a, 0), and (1, 1) to pin down a parabola (I think this works, haven't worked the details).
     
  4. Nov 12, 2009 #3
    I'd be careful with the parabola idea, since if it actually has a peak in that interval, then it won't take open sets to open sets, but will take any open set that contains the x-value for the axis of symmetry to a half-open set.
     
  5. Nov 12, 2009 #4

    George Jones

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    Yes, that's why I said that I think it works. I drew some sketches that make it appear that the max/mins occur outside the interval, but I didn't solve any equations. My sketches could be misleading me.

    Obvious choices of line segments in my example 1) give obvious homeomorphisms.
     
  6. Nov 12, 2009 #5
    Can you explain what you mean by joining lines? Are you talking about intersecting lines in [tex] R^2 [/tex]?
     
  7. Nov 12, 2009 #6

    George Jones

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    Yes.

    What are [itex]A[/itex] and [itex]B[/itex] such that [itex]f_a \left( x \right) = Ax + B[/itex] has

    [tex]\lim_{x \rightarrow -1} f_a \left( x \right) = -1[/tex]

    and [itex]f_a \left( a \right) = 0[/itex]?
     
    Last edited: Nov 12, 2009
  8. Nov 12, 2009 #7
    Right so you have a line starting at the point (-1,-1) and intersecting at (a,0). Unfortunately it does not finish at (1,1), and so it cannot be a bijection from the interval
    (-1, 1) to (-1,1). Is this not right?
     
  9. Nov 12, 2009 #8
    It can be a bijection so long as the lines are both increasing or both decreasing.
     
  10. Nov 12, 2009 #9
    Sorry I must be confused here. I can only see one line other than the interval (-1, 1). I thought you were talking about a line crossing the real axis at the point a.
     
  11. Nov 12, 2009 #10

    George Jones

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    So pick another line from (a,0) to (1,1), and turn f_a into a "schizophrenic" function.
     
  12. Nov 12, 2009 #11
    ok so I have for x < a [tex] f_a \left( x \right) = \frac{x-a}{1+a} [/tex]

    and for x > a [tex] f_a \left( x \right) = \frac{x-a}{1-a} [/tex]

    Which coincide at x = a. Looks good to me.
     
  13. Nov 12, 2009 #12

    George Jones

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    Looks good if an equality is included for at least one of x < a, x > a.

    The question is not done yet, though. You still need to show the deduction that gives that (-1, 1) is a homogeneous space.
     
  14. Nov 12, 2009 #13
    I know I was just thinking about that. I need to show that for any x,y in the domain, there is a homeomorphism mapping x to y. Presumably I can somehow use the function I just made, but simple addition of y onto this function doesn't make a homeomorphism.
     
  15. Nov 12, 2009 #14

    George Jones

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    Unfortunately, I know only how to give a very small hint, or a very large hint (writing down the answer). Small hint: since both x and y are arbitrary elements of (-1 , 1), you should try things that have both x and y as indices.

    Maybe someone else knows a better hint that doesn't give the whole answer away.
     
  16. Nov 12, 2009 #15
    How about this:

    [tex] f_b \left( b \right) = 0, f_a^{-1} \left( 0 \right) = a, f_a^{-1} \left( f_b \left( b \right)\right) = a [/tex]?

    So that for any a,b we have a homeomorphism (i think) mapping b to a.
     
  17. Nov 12, 2009 #16

    George Jones

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    Yes, that's it. :smile:

    I might switch the and b around in one of the places, i.e., either "So that for any b,a we have a homeomorphism," or [itex]f_b^{-1} \left( f_a \left( a \right)\right) = b
    [/itex].
     
  18. Nov 12, 2009 #17
    Ok thanks that's the problem solved!
     
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