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Math Help - Continuity - A basic question

  1. #1
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    Continuity - A basic question

    I conjecture that if a function is continuous at a discreet point (say p), it is continuous at all points in some delta-neighborhood of p.

    Am I correct? If yes - any pointers to prove this? If not - any counterexample.

    For those who know the subject deeper - is this kind of formulation even important?
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi,
    Quote Originally Posted by aman_cc View Post
    Am I correct?
    No.
    Quote Originally Posted by aman_cc View Post
    If not - any counterexample.
    Consider the function f defined on \mathbb{R} by f:x\mapsto \begin{cases}x&\text{if}\ x\in\mathbb{Q}\\0&\text{if}\ x\not\in\mathbb{Q}\end{cases}. The only point where f is continuous is 0.
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  3. #3
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    Thanks. But it sounds conter-intutive to me.

    In layman's way, being continous at 'a' means 'very' close to 'a' value of the function varies 'very' little. So the same will apply to any point which is very very close to 'a' and thus I say it will be continous to at all points in some delta-neighbourhood of a.

    I know I am talking an all vague language but it helps get a feel of definitons. Please excuse me if all this is not important/relevant.
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by aman_cc View Post
    In layman's way, being continous at 'a' means 'very' close to 'a' value of the function varies 'very' little. So the same will apply to any point which is very very close to 'a' and thus I say it will be continous to at all points in some delta-neighbourhood of a.
    Let a\neq 0 be a rational number. Let's define the sequence (a_n)_{n\in\mathbb{N}} by a_n = a+\frac{\pi}{n} (note that this sequence converges to a). For all integer n, a_n is an irrationnal number hence \lim_{n\to\infty}f(a_n)=\lim_{n\to\infty}0=0. However f(a)=a\neq 0 so f(\lim_{n\to\infty}a_n)\neq \lim_{n\to\infty}f(a_n) and f is not continuous at a. In other words, if n is 'very large' then a_n is 'very close to' a but f(a_n)=0 is 'not close to' f(a)=a. (this is true for all a\in\mathbb{Q}\setminus\{0\} so it can be assumed that a is 'very very close' to 0)

    Using the same method it can be show that this result also holds if a\in\mathbb{R}\setminus\mathbb{Q}.
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  5. #5
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    Thanks for you post. I guess I need to think more about your post and my question.
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