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Homework Help: [0,1) onto [0,infinity) , continuous surjection?

  1. Oct 5, 2008 #1
    1. The problem statement, all variables and given/known data

    Find a continuous surjection from [0,1) onto [0, infinity)

    2. Relevant equations


    3. The attempt at a solution

    I have only been able to come up with one mapping but then I realized it did not work. Any help would be appreciated.
     
  2. jcsd
  3. Oct 5, 2008 #2
    You should be able to get what you need by manipulating the tan(x) function. Just set it up so that f(0) = 0 and f(x) -> Inf as x -> 1.
     
  4. Oct 5, 2008 #3
    I got it!! So what about a cont. surjection from [0,infinity) --> Reals? I was close to saying ln(x) but that doesn't include 0.
     
  5. Oct 5, 2008 #4

    morphism

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    What about something that oscillates, with a greater and greater 'peak' as we move along the x-axis?
     
  6. Oct 6, 2008 #5
    All I could think of was f(x)=sin(x)ln(x), or f(x)=xsin(x)ln(x).

    Is this what (or similar, or close to) you are hinting at?
     
  7. Oct 6, 2008 #6

    morphism

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    You're sort of close - using sin(x) is a good idea. How about you multiply it by a positive, increasing function?
     
  8. Oct 7, 2008 #7
    How about sin(x)x since x is a positive increasing function? Would this work?
     
  9. Oct 7, 2008 #8
    xsin(x) takes in negative values though...
     
  10. Oct 7, 2008 #9

    Dick

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    The square of a negative is positive.
     
  11. Oct 7, 2008 #10

    HallsofIvy

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    I would be inclined to use f(x)= x/(something that goes to 0 as x goes to 1)
     
  12. Oct 7, 2008 #11

    Dick

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    The question was switched in post 3. You are out of touch on issues concerning the economy. Big debate tonight. Had to throw in a jab.
     
  13. Oct 7, 2008 #12

    Dick

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    Hey, wait a minute didn't you say in post 3 that you want the domain to be the reals? What's wrong with negative?? I think your two questions are getting all tangled up.
     
  14. Oct 7, 2008 #13
    f(x)=tan(pi x/2)
     
  15. Oct 7, 2008 #14

    Dick

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    At this point you should probably say which problem that's a solution to. The one in post 1 or the one in post 3?
     
  16. Oct 7, 2008 #15
    oo that was 4 post 1


    wat abt this??
    f(x)=tan(x*pi-pi/2)
    this is for post 3
     
  17. Oct 7, 2008 #16

    Dick

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    Thought so. The first suggestion was good for post 1. This was actually the suggestion of post 2. The new one is not so good for post 3. It's undefined at x=1. And a lot of other places. The domain is supposed to be [0,infinity).
     
    Last edited: Oct 7, 2008
  18. Oct 7, 2008 #17
    oo sorry...i thought it was from [0,1).....lemme think abt it
     
  19. Oct 8, 2008 #18

    Dick

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    Sure, can't hurt. There is a correct answer in one of the previous posts.
     
  20. Nov 21, 2008 #19
    It has been awhile and I am just browsing my old threads. I never got this problem though. Could anybody come up with the continuous surjection from

    [0,1) onto R?

    EDIT: I'm not even making sense, sorry.
     
    Last edited: Nov 21, 2008
  21. Nov 21, 2008 #20
    The hints were leaning toward letting f(x) = (x^2) * sin(x), I think.

    Not sure how to prove it's a surjection though. Any element, b, of the codomain is certainly mapped to, in fact its mapped to an infinite number of times. But the equation b = x^2 * sin(x) can't be solved explicitly for x. Or if it can, I certainly don't know how. But if you graph this function you should see that it works.

    Sorry I'm not sure how to prove it.

    EDIT: Ah, crap, I see now you wanted [0,1) to R. I was responding to the part of [0,Infinity) to R. The first surjection you needed was already answered with the tan function, sorry.
     
    Last edited: Nov 21, 2008
  22. Nov 21, 2008 #21
    Do you have one for [0,1) to Reals?
     
  23. Nov 21, 2008 #22

    Dick

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    This thread was started, like a month ago, Unassuming. morphism did the relevant observation in post #6. Make the function oscillating and unbounded. That's all. There's a gazillion ways to do this. Pick one.
     
  24. Nov 22, 2008 #23
    I have used his suggestion many times. I tried to visualize it and construct a function. I know that I needed an oscillating function and I can get something that passes through 0 and starts increasing wildly but how in the world do I constrict it enough so that it oscillates to infinity at it approaches 1?

    I've tried dividing by (1-x). On my calculator it seems that it does the trick on making it undefined at 1, but it still doesn't approach infinity before that.
     
  25. Nov 22, 2008 #24

    Dick

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    Let g(x)=1/(1-x). Your suggestion. g(0)=1 g(x)->inf as x->1. Now let's make it 'oscillate wildly'. Define f(x)=sin(g(x))*g(x). That maps [0,1) onto the reals. It's onto because there are values of x such the f(x) is arbitrarily large or arbitrarily small (negative). f(x) is continuous so it also assumes every value in between. That's ALL of R. Whether your calculator can give you a clear enough picture of it's behavior, I don't know.
     
  26. Nov 22, 2008 #25
    Ah, I was close. Thank you for the help on that. My calculator definetely mislead me becaues it doesn't look like it oscillates "wildly" but it works if you hit trace and then let x=.9, x=.99, x=.999, etc.
     
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