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1-1 mapping question

  1. Feb 3, 2012 #1
    1. The problem statement, all variables and given/known data
    Consider the open interval (0,1), and lets S be the set of points in the open unit square
    that is, S={(x,y):0<x,y<1}.
    Find a function that maps (0,1) into S. but not necessarily onto.
    3. The attempt at a solution
    so I can describe any point in my square with x and y coordinates like (x,y)
    so for all of my points ill just take so I will map the coordinate (x,y) to [itex] 2^x3^y [/itex] .
    so now each coordinate goes to one real. and (0,1) can be mapped to the whole real line so I shouldn't have problems with some of the numbers being bigger than one.
    But I guess I could just map the coordinates to [itex] 2^{-x}3^{-y} [/itex]
    and I think this function might be onto.
     
  2. jcsd
  3. Feb 3, 2012 #2

    HallsofIvy

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    You appear to be completely misunderstanding the problem. There is NO "(x, y)". The problem asks you to map (0, 1), a subset of R, onto the square in R2.

    You need to find a function that maps a single number, x, to a pair, (a, b).
     
  4. Feb 3, 2012 #3

    micromass

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    Not even that. You just need to map into the square. It doesn't even need to be surjective.
     
  5. Feb 3, 2012 #4

    HallsofIvy

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    Yes, thanks.
     
  6. Feb 3, 2012 #5
    ok , so I just need to map the points from (0,1) into the square, but not nessicarily all the points in the sqaure to the line segmenet. but does my original statement work.
     
  7. Feb 3, 2012 #6

    HallsofIvy

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    No, because there is NO "(x, y)" to begin with. If x is a number in (0, 1), what does x map to?
     
  8. Feb 3, 2012 #7
    I dont understand what you mean. why cant you map x to x on the line.
    what do you mean there is no (x,y) to begin with.
    thanks for the help by the way.
     
  9. Feb 3, 2012 #8

    Deveno

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    the domain of your map (function) is 1-dimensional (a line segment, minus its endpoints).

    the co-domain (set that contains the range) is 2-dimensional (interior of a square).

    your map should look something like this:

    f(t) = (x(t),y(t))

    i can think of a possible map that has a fairly simple form for both x(t) and y(t).
     
  10. Feb 4, 2012 #9
    so could I just map my my line into the square using f(t)=(x,1/2)
    where x varies and y is always at 1/2
     
  11. Feb 4, 2012 #10

    Deveno

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    that's closer, but how is "x" determined....?
     
  12. Feb 4, 2012 #11
    how about y=x
     
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