2
$\begingroup$

$f(x)=0$ for all $x \in\mathbb{Q}$. Prove $f(x)=0$ for all $x > \in\mathbb{R}$.

I have to use the Epsilon-Delta method. I know it starts Let $\epsilon > 0$ be arbitrary. Let y be arbitrary. Now I have to find a $\delta>0$ such that if $x \in\mathbb{R}$ and $|x−y|<\delta$ implies $|f(x)−f(y)|<\epsilon$.

But I can't figure out how to find $\delta$ and I can't figure out how the $\mathbb{Q}$ and $\mathbb{R}$ play into the whole thing because $|f(x)−f(y)|=|0−0|=0<\epsilon$ already. Please help.

$\endgroup$
2
$\begingroup$

You have to prove that $f(x) = 0$ for all $x \in \mathbb{R}$, so don't start there. Let $x \in \mathbb{R}$ be any real number. Then we want to show that $f(x) = 0$, or equivalently that $|f(x)| < \epsilon$ for every $\epsilon > 0$. Do you see a way to do this?

$\endgroup$
2
$\begingroup$

I think you're trying to answer the wrong question.

What you're (presumably) being asked to do is as follows. If $f$ is a continuous function and $f(x)=0$ for all $x \in \mathbb{Q}$ then $f(x)=0$ for all $x \in \mathbb{R}$. (You're not being asked to prove that anything is continuous.)

So suppose $f$ is continuous and $x \in \mathbb{R}$. Fix any $\varepsilon > 0$. You know that there is a $\delta > 0$ such that $|f(x)-f(y)| < \varepsilon$ whenever $|x-y|<\delta$.

By cunning choice of $y$ you can force $|f(x)|<\varepsilon$.

...but then what does this tell you about the value of $f(x)$?

$\endgroup$
  • 1
    $\begingroup$ You have switched around $\delta$ and $\epsilon$. $\endgroup$ – Arthur Oct 24 '13 at 3:16
  • 2
    $\begingroup$ Yes he has, but he's absolutely right about my mix up with the question. Thank you for that. I would have just continued to struggle. So |f(x)|<ε, but ε>0 is arbitrary, so ε can get incredibly small. Then the only thing |f(x)| can equal is 0 because |f(x)| is greater than or equal to 0 (it has to be positive) and |f(x)| is less than ε. So f(x)=0 for all x∈R. Does that work? $\endgroup$ – Maddy Oct 24 '13 at 3:23
  • $\begingroup$ @Maddy: Yup :)${}$ $\endgroup$ – Clive Newstead Oct 24 '13 at 4:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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