Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?
share|improve this question
2  
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
add comment

2 Answers

up vote 4 down vote accepted

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 $x \in (x_0 - \delta, x_0 + \delta)$

share|improve this answer
    
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
show 1 more comment

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 $f(x_n) \leq 0$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.