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Math Help - Determine if this set of continuous functions is a group

  1. #1
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    Determine if this set of continuous functions is a group

    Problem: The set of continuous functions f:[0,1] \rightarrow [0,1] such that f(0)=0, f(1)=1, and f is increasing. Binary operation is composition of functions.

    First off, I'm a little confused about the f:[0,1] \rightarrow [0,1] part. Is that just saying the domain and range of the function is from 0 to 1?

    Now, I trying to test this for associativity.

    a*(b*c)=(a*b)*c

    But I started to run into problems. There are only two elements in this set, but I need three to test for associativity, don't I? Or do I just use one twice? But what if one works twice, but the other doesn't?

    Regardless, I tried using composition to test for associativity, and got stuck.

    a*b = a(b(x)) ??

    Any help is appreciated.
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  2. #2
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    Re: Determine if this set of continuous functions is a group

    You need to test if (f \circ (g\circ h))(x) = ((f\circ g) \circ h )(x)

    where f,g,h are in the set of functions described
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    Re: Determine if this set of continuous functions is a group

    There are more than just two elements...

    For instance, the functions
    x
    x^2
    x^3
    etc
    Are all in your supposed group.
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    Re: Determine if this set of continuous functions is a group

    I still don't understand. Where are x, x^2 etc coming from?

    Is the first [0,1] in the problem specifying the domain of the functions and the second is specifying the range? And therefore, is x, x^2 etc coming from the fact that putting 0 or 1 into any of those will result in 0 or 1?
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    Super Member TheChaz's Avatar
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    Re: Determine if this set of continuous functions is a group

    Quote Originally Posted by tangibleLime View Post
    I still don't understand. Where are x, x^2 etc coming from?

    Is the first [0,1] in the problem specifying the domain of the functions and the second is specifying the range? And therefore, is x, x^2 etc coming from the fact that putting 0 or 1 into any of those will result in 0 or 1?
    The group is a SET (see the first line of your first post). Each element in the set is a function from [0, 1] to [0, 1]

    The first is the domain, the second is the range.
    Last edited by TheChaz; October 10th 2011 at 06:37 PM. Reason: I read the conditions again!
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    Re: Determine if this set of continuous functions is a group

    Ah okay. So I'm testing for associativity like this...

    f(g(h(x))) = h(f(g(x)))

    Which is true since if x=0, they all return 0 and if x=1 they all return one, so it ends up as 0=0 or 1=1, right?

    If x=0,
    f(g(h(0))) = f(g(0)) = f(0) = 0
    h(f(g(0))) = h(f(0)) = h(0) = 0

    Is that incorrect?

    Assuming it isn't, I'm now trying to come up with the unique identity element...

    a*e=e*a=a

    In other words,
    f(g(x)) = g(f(x)) = f(x)

    Which is true if g(x)=x... or f(x)=x?? So the identity is... what? I think I'm on the wrong track.
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  7. #7
    Super Member TheChaz's Avatar
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    Re: Determine if this set of continuous functions is a group

    f(x) = x is the identity, since when you compose this with any function g, gf = g.

    Your "associativity" is wrong. It should be
    f(gh) = (fg)h. You don't want to move them around, just change the way they are grouped.
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    Re: Determine if this set of continuous functions is a group

    Quote Originally Posted by TheChaz View Post
    I'm pretty sure this set is not closed under composition.
    suppose f,g are in the set under consideration. then fg(0) = f(g(0)) = f(0) = 0, while fg(1) = f(g(1)) = f(1) = 1.

    clearly, fg is continuous if f and g are (standard proof from first-year calculus).

    so the main thing is: is fg increasing if f and g are?

    suppose 0 ≤ x < y ≤ 1.

    then 0 = g(0) ≤ g(x) < g(y) ≤ g(1) = 1, because g is increasing.

    so 0 = f(g(0)) ≤ f(g(x)) < f(g(y)) ≤ f(g(1)) = 1, because f is increasing.


    @tangibleLime: for any functions f,g,h: we can show that if the compositions are defined, then

    (f \circ g) \circ h = f \circ (g \circ h).

    ((f \circ g) \circ h)(x) = (f \circ g)(h(x)) = f(g(h(x)) = f((g \circ h)(x)) = f \circ (g \circ h)(x).
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    Super Member TheChaz's Avatar
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    Re: Determine if this set of continuous functions is a group

    Quote Originally Posted by Deveno View Post
    suppose f,g are in the set under consideration. then fg(0) = f(g(0)) = f(0) = 0, while fg(1) = f(g(1)) = f(1) = 1.

    clearly, fg is continuous if f and g are (standard proof from first-year calculus)....
    I didn't see the f(1) = 1 condition until later.
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    Re: Determine if this set of continuous functions is a group

    and apparently, you edited your post, as i was writing mine. it's all good ^^.
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  11. #11
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    Re: Determine if this set of continuous functions is a group

    You may be confusing "[0, 1]" with "{0, 1}". The first is an interval containing an infinite collection of numbers, the second is a set containing only two numbers. Saying that f:[0, 1]->[0, 1] means that f maps all numbers in the closed interval 0\le x\le 1 to a number in that interval.
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