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Math Help - discont

  1. #1
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    discont

    what does it mean when a function from real no to real no is continuous at infinitely many pts but between every two points of continuity, there is a point of discontinuity and vice versa?

    i dont get how the function will look..


    discont-photo.jpg

    also, in the proof, i do not understand the purpose behind saying that q> ε and then q>1/ε...isit contradicting?

    thanks
    Last edited by alexandrabel90; November 19th 2010 at 12:51 PM. Reason: add picture
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  2. #2
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    There is a sequential listing of the rational numbers: \mathbb{Q}=\{\alpha_n:n\in\mathbb{Z}^+\}.
    Define a function f(x) = \sum\limits_{\alpha _j  < x} {2^{ - j} } .

    If t\in\mathbb{Q} then f(t+)-f(t-)=~?

    What happens if t\notin\mathbb{Q}~?
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  3. #3
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    im not really sure ..

    but i guess that if t is a quotient then f(t+)-f(t-) <0 and

    im not sure what happens to f(t+)-f(t-) if t is not a quotient..i assume that it will still give <0 becos for a^j which are less than t, there will still be values that are quotient.and that f(t+) <1 while f(t-)>1
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    Quote Originally Posted by alexandrabel90 View Post
    im not really sure ..but i guess that if t is a quotient then f(t+)-f(t-) <0 and im not sure what happens to f(t+)-f(t-) if t is not a quotient..i assume that it will still give <0 becos for a^j which are less than t, there will still be values that are quotient.and that f(t+) <1 while f(t-)>1
    I can assure you that I mean you not ill by the following remark .
    From the above reply, one can infer that you have no idea what is going on here.
    You need to seek a sit down, face to face, tutorial.
    This sort of forum has no pretense to do that.
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  5. #5
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    ya i think i should do that. its just that im just starting this module and so i thought my foundations for it aint that good as of yet..

    btw, are you giving an example for the case where: a function from real no to real no is continuous at infinitely many pts but between every two points of continuity, there is a point of discontinuity and vice versa or trying to explain the photo that i have attached?
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  6. #6
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    Quote Originally Posted by alexandrabel90 View Post
    what does it mean when a function from real no to real no is continuous at infinitely many pts but between every two points of continuity, there is a point of discontinuity and vice versa?

    i dont get how the function will look..


    Click image for larger version. 

Name:	photo.jpg 
Views:	10 
Size:	248.9 KB 
ID:	19769

    also, in the proof, i do not understand the purpose behind saying that q> ε and then q>1/ε...isit contradicting?

    thanks
    That attachment is describing a function f(x) which takes the value 0 at every irrational point. Its values at rational points are like this: f(1/2) = 1/2, f(1/3) = f(2/3) = 1/3, f(1/4) = f(3/4) = 1/4, f(1/5) = f(2/5) = f(3/5) = f(4/5) = 1/5, and so on. If x = p/q is a fraction in its lowest terms, then f(p/q) = 1/q. Notice that as the denominators get larger, the values of the function get smaller. In fact, if you are given \varepsilon>0, there will only be a finite number of points at which the function takes values greater than \varepsilon.

    This function is discontinuous at every rational point x = p/q. The reason is that as you approach p/q along a sequence of irrational numbers, the value of the function jumps from 0 to 1/q when you reach the point p/q.

    Less obviously, the function is continuous at every irrational point c. The reason for that is that given \varepsilon>0, you can choose a neighbourhood of c that is small enough to avoid all the (finitely many) rational points at which the function takes values greater than \varepsilon. This means that the function gets closer and closer to 0 as you approach the point c, so that 0 = f(c) = \lim_{x\to c}f(x), which is the definition of f being continuous at c.
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