1. 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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2. $\displaystyle f(c) > 0\quad \Rightarrow \quad \exists \delta > 0\left[ {\left| {x - c} \right| < \delta } \right]\quad \Rightarrow \quad \left| {f(x) - f(c)} \right| < \frac{{f(c)}} {2}$

Expand the last inequality. Add f(c) to all three parts. See what you get.

3. Originally Posted by Plato
$\displaystyle f(c) > 0\quad \Rightarrow \quad \exists \delta > 0\left[ {\left| {x - c} \right| < \delta } \right]\quad \Rightarrow \quad \left| {f(x) - f(c)} \right| < \frac{{f(c)}} {2}$

Expand the last inequality. Add f(c) to all three parts. See what you get.

$\displaystyle f(x) - f(c) < \frac{f(c)}{2}$ 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?

4. 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.

5. Excellent. Thanks so much, Plato.

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