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Thread: Continuity proof

  1. #1
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    Continuity proof

    Hi, I'm stuck on the following question. Any help will be fantastic, thanks.

    Question:

    $\displaystyle f:\Re \to \Re $ is a continuous function. Prove that if for some $\displaystyle c \in \Re $, $\displaystyle f(c)>0 $, then there exists a $\displaystyle \delta > 0 $ such that $\displaystyle \forall x \in ( c - \delta, c + \delta) $, $\displaystyle f(x) > 0 $

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    I tried a contradiction proof but didn't really get anywhere. Please help! Thank you.
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  2. #2
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    $\displaystyle f(c) > 0\quad \Rightarrow \quad \exists \delta > 0\left[ {\left| {x - c} \right| < \delta } \right]\quad \Rightarrow \quad \left| {f(x) - f(c)} \right| < \frac{{f(c)}}
    {2}$

    Expand the last inequality. Add f(c) to all three parts. See what you get.
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  3. #3
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    Quote Originally Posted by Plato View Post
    $\displaystyle f(c) > 0\quad \Rightarrow \quad \exists \delta > 0\left[ {\left| {x - c} \right| < \delta } \right]\quad \Rightarrow \quad \left| {f(x) - f(c)} \right| < \frac{{f(c)}}
    {2}$

    Expand the last inequality. Add f(c) to all three parts. See what you get.
    Thanks for the reply, Plato!

    Following your help, I get...

    $\displaystyle f(x) - f(c) < \frac{f(c)}{2} $ if $\displaystyle f(x)-f(c) > 0 $
    and
    $\displaystyle f(c) - f(x) < \frac{f(c)}{2} $ if $\displaystyle f(x)-f(c) < 0 $

    So $\displaystyle f(x) > \frac{3}{2} f(c) $ and $\displaystyle f(x) > \frac{1}{2} f(c) $ ...each of which is greater than 0 since f(c)>0

    Is this what you meant? I don't see where the third inequality comes from so I don't think I've followed your advice properly.

    Also, if a function is continuous at c, by definition, $\displaystyle \forall \epsilon > 0 $, there exists a $\displaystyle \delta > 0 $ s.t. $\displaystyle \forall x \in \Re $ such that $\displaystyle |x-c|< \delta $, $\displaystyle |f(x) - f(c)|< \epsilon $.

    As far as I can see, you've let $\displaystyle \epsilon = \frac{f(c)}{2} $ which is fixed. So I don't see how this covers the 'for all epsilon' bit. I'm probably being very dull here though.


    EDIT: Thinking about this again, I think you might've been suggesting a proof by contradiction - which I totally missed. If you let f(x) < 0, then we have a contradiction. Is that what you wanted me to do?
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  4. #4
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    First of all, there is absolutely no reason why we should for “all epsilon”.
    The problem says that f is positive is some neighborhood of c.
    Use the definition of continuity with $\displaystyle \varepsilon = \frac{{f(c)}}{2}> 0$.
    We know that $\displaystyle \exists \delta > 0\left[ {\left| {x - c} \right| < \delta } \right]\quad \Rightarrow \quad \left| {f(x) - f(c)} \right| < \frac{{f(c)} }{2} $
    We also know that $\displaystyle \left| {x - c} \right| < \delta \quad \Leftrightarrow \quad x \in (c - \delta ,c + \delta )$ which is a neighborhood of c. For any x there we have
    $\displaystyle \begin{gathered}
    \left| {f(x) - f(c)} \right| < \frac{{f(c)}}
    {2} \hfill \\
    - \frac{{f(c)}}
    {2} < f(x) - f(c) < \frac{{f(c)}}
    {2} \hfill \\
    0 < \frac{{f(c)}}
    {2} < f(x) < \frac{{3f(c)}}
    {2} \hfill \\
    \end{gathered} $.

    So the function is positive in that neighborhood of c.
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  5. #5
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    Excellent. Thanks so much, Plato.

    I was focusing on the definition of continuity more than the question, me thinks.
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