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Continuous onto function

  1. Dec 10, 2008 #1
    1.Is it true that if f is continuous onto function on a closed interval then f(x) must also be a closed interval. How about the other way around. f is continuous and onto on a open bounded interval and f(x) is a closed interval

    2. Relevant equations
    f:[0,1]-->(0,1)
    f:(0,1)-->[0,1]


    3. The attempt at a solution
    There is a theorem that says that if f is continuous on a closed and bounded interval then set of f(x) is a closed and bounded interval.
     
  2. jcsd
  3. Dec 10, 2008 #2

    Dick

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    True, that is a theorem. f continuous on a closed interval then the image of f is a closed interval. If f is continous on an open interval then the range can be closed, open or neither. Can you find an example of each?
     
  4. Dec 10, 2008 #3
    Does this mean that f:[0,1]-->(0,1) does not exist. If both are closed intervals there are several examples, including the trivial f(x)=x

    I don't have an example for continuous onto f:(0,1)-->[0,1]
     
  5. Dec 10, 2008 #4

    Dick

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    Yes, there is no continuous function f mapping [0,1]->(0,1). You cited the theorem. For mapping (0,1) into [0,1], can't you think of a function where f(0)=1/2, f(1/3)=1, f(2/3)=0 and f(1)=1/2? Surely you can draw one.
     
  6. Dec 10, 2008 #5
    It has to be an onto function. I am not aware of one as you have described.
     
  7. Dec 10, 2008 #6

    Dick

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    Onto what? It's certainly onto [0,1].
     
  8. Dec 10, 2008 #7
    Yes. But I don't have an example as you have described. Sorry. I am always bad with examples.
     
  9. Dec 10, 2008 #8

    Dick

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    Can't you linearly interpolate between the values I gave? You could also fit it to a polynomial. You could scale a sine function. Any number of things you could do to get an explicit example. You did sketch one, right? That's all I care about.
     
  10. Dec 11, 2008 #9
    sin(4x) works

    Thx
     
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