# Continuity using Epsilon- Delta

$f(x)=0$ 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.

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?

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 $f(x)$?

• You have switched around $\delta$ and $\epsilon$. – Arthur Oct 24 '13 at 3:16
• 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? – Maddy Oct 24 '13 at 3:23
• @Maddy: Yup :)${}$ – Clive Newstead Oct 24 '13 at 4:17