# Proving there is an interval where $f(x)$ is positive

Let $f(x)$ be a continuous real function s.t $f(x_0) > 0$

Prove: There is some interval of the form $(x_0 -\delta, x_0 + \delta)$ where $f$ is positive.

Proof:

Since $f$ is continuous: $\forall \,{\epsilon > 0}\,\, \exists \,{\delta>0}$ s.t. $|x- x_0|<\delta \implies |f(x) - f(x_0)| < \epsilon$

By contradiction suppose there is no interval $(x_0 - \delta, x_0 + \delta)$ where $f(x)$ is positive. This means that $f(x_0) - \epsilon < f(x) < f(x_0) + \epsilon < 0$. Hence we have a contradiction since $\epsilon$ and $f(x_0)$ are both greater than zero.

1. Is this correct?
2. Could someone provide a non-contradiction proof?
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You need to quantify your variables, because right now your proof doesn't make sense. –  wj32 Oct 27 '12 at 21:45
It is not a proof. For a while you might try to use fewer logical symbols. The idea is simple. We have $f(x_0)=b\gt 0$. If $x$ is close enough to $x_0$, then $f(x)$ is very close to $b$ and therefore positive. More formally, pick $\epsilon=b/2$, say. Then there is a $\delta$ such that if $|x-x_0|\lt \delta$, then $|f(x)-b|\lt b/2$, and therefore by Triangle Inequality $f(x)\gt b-b/2\gt 0$. Note how the formal stuff comes from the geometry. –  André Nicolas Oct 27 '12 at 21:55

Your proof by contradiction is incorrect. Specifically, the following statements are incorrect.

This means that $f(x_0) - \epsilon < f(x) < f(x_0) + \epsilon < 0$. Hence we have a contradiction since $\epsilon$ and $f(x_0)$ are both greater than zero.

You can argue by contradiction but what you have is not the right proof.

A direct proof is simple for this case. Choose $\epsilon = f(x_0)$ in your continuity criterion to get your $\delta$.

Now $f(x) > 0$ for $x \in (x_0 - \delta, x_0 + \delta)$

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Is my proof correct? –  CodeKingPlusPlus Oct 27 '12 at 21:38
@CodeKingPlusPlus: No. Where does your $epsilon$ come from? And from $f(x)<0$ and $f(x)<f(x_0)+\epsilon$ you cannot infer $f(x_0)+\epsilon<0$. –  Hagen von Eitzen Oct 27 '12 at 21:41
my $\epsilon$ is any arbitrary number greater than $0$ –  CodeKingPlusPlus Oct 27 '12 at 21:42
@CodeKingPlusPlus Why does that mean $f(x_0) + \epsilon < 0$? –  user17762 Oct 27 '12 at 21:43
@CodeKingPlusPlus ... and for which $x$? –  Hagen von Eitzen Oct 27 '12 at 21:44

Choose $\epsilon = \frac{f(x_0)}{2}> 0$. Then there exists a $\delta>0$ such that for $|x-x_0| < \delta$, $|f(x)-f(x_0)| < \epsilon = \frac{f(x_0)}{2}$. Then $-\frac{f(x_0}{2} < f(x)-f(x_0)$ from which we get $0 < \frac{f(x_0)}{2} < f(x)$ for all $x$ such that $|x-x_0| < \delta$.

Alternatively, a proof by contradiction is straightforward as well:

Suppose on every interval of the form $I_\delta = (x_0-\delta, x_0+\delta)$, there is some $x \in I_\delta$ such that $f(x) \leq 0$. Then choose $\delta = \frac{1}{n}$ and let $x_n$ be the corresponding $x \in I_{\frac{1}{n}}$. Then clearly $x_n \to x_0$, and since $f$ is continuous, $f(x_n) \to f(x_0) >0$ which contradicts $f(x_n) \leq 0$.

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