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Math Help - 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:

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

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

    So  f(x) > \frac{3}{2} f(c) and  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,  \forall \epsilon > 0 , there exists a  \delta > 0 s.t.  \forall x \in \Re such that  |x-c|< \delta ,  |f(x) - f(c)|< \epsilon .

    As far as I can see, you've let  \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 \varepsilon  = \frac{{f(c)}}{2}> 0.
    We know that \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 \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
    \begin{gathered}<br />
  \left| {f(x) - f(c)} \right| < \frac{{f(c)}}<br />
{2} \hfill \\<br />
   - \frac{{f(c)}}<br />
{2} < f(x) - f(c) < \frac{{f(c)}}<br />
{2} \hfill \\<br />
  0 < \frac{{f(c)}}<br />
{2} < f(x) < \frac{{3f(c)}}<br />
{2} \hfill \\ <br />
\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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