Continuity proof

**Chief65** · April 6th 2009, 06:51 PM

Let $f<img src=$ \rightarrow \Re" alt="f

\rightarrow \Re" /> be continuous at $c \in D$ and suppose that $f(c) > 0$ . Prove that there exists an $a > 0$ and a neighborhood U of c such that $f(x) > a$ for all $x \in U \cap D$ .

**TheEmptySet** · April 6th 2009, 07:03 PM

Originally Posted by Chief65

Let $f<img src=$ \rightarrow \Re" alt="f

\rightarrow \Re" /> be continuous at $c \in D$ and suppose that $f(c) > 0$ . Prove that there exists an $a > 0$ and a neighborhood U of c such that $f(x) > a$ for all $x \in U \cap D$ .

Let $\epsilon =\frac{f(c)}{2}$ since f is continous there exists a $\delta > 0$ such that when

$|x-c|< \delta \implies |f(x)-f(c)|< \epsilon$

Now working with the 2nd statement

$|f(x)-f(c)|< \frac{f(c)}{2} \iff -\frac{f(c)}{2}< f(x)-f(c) <\frac{f(c)}{2}$

Now add f(c) to all sides of the inequalilty to get

$\frac{f(c)}{2} < f(x)$ but we know that $f(c) > 0$

So now for all $x \in (c-\delta,c+\delta)$

$f(x) > \frac{f(c)}{2} > 0$

YAY!

**Chief65** · April 6th 2009, 07:51 PM

I understand your proof, but where does the $a > 0$ come in? And I can see what you did with the neighborhood but the question confused me because it says an $x \in U \cap D$ I don't understand the intersection and\or what $U \cap D$ means. Why must $U$ be in the domain $D$ ?

**TheEmptySet** · April 6th 2009, 07:57 PM

Originally Posted by Chief65

I understand your proof, but where does the $a > 0$ come in? And I can see what you did with the neighborhood but the question confused me because it says an $x \in U \cap D$ I don't understand the intersection and\or what $U \cap D$ means. Why must $U$ be in the domain $D$ ?

The a is the value of epsilon that I used. Note it is not unique I could have used any fractional part of f(c). The interscetion is to make sure that my neighborhood stays inside the domain of the function. Otherwise it doesn't make sense to talk about function values.

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