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Thread: 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; Nov 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: $\displaystyle \mathbb{Q}=\{\alpha_n:n\in\mathbb{Z}^+\}$.
    Define a function $\displaystyle f(x) = \sum\limits_{\alpha _j < x} {2^{ - j} } .$

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

    What happens if $\displaystyle t\notin\mathbb{Q}~?$
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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 $\displaystyle \varepsilon>0$, there will only be a finite number of points at which the function takes values greater than $\displaystyle \varepsilon$.

    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 $\displaystyle \varepsilon>0$, 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 $\displaystyle \varepsilon$. This means that the function gets closer and closer to 0 as you approach the point c, so that $\displaystyle 0 = f(c) = \lim_{x\to c}f(x)$, which is the definition of f being continuous at c.
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