# Homework Help: Continuity and Dense Subsets of the Real Numbers

1. Sep 9, 2010

### jgens

1. The problem statement, all variables and given/known data

If $f$ is continuous and $f(x)=0$ for all $x$ in a dense subset of the real numbers, then $f(x)=0$ for all $x \in \mathbb{R}$.

2. Relevant equations

N/A

3. The attempt at a solution

Does this solution work? And if it does, can it be improved in some way?

Proof: From the continuity of $f$, for every $\varepsilon > 0$ we can find a $\delta > 0$, such that if $|x-a| < \delta$, then $|f(x)-f(a)| < \varepsilon$. Because $f(x)=0$ for all $x$ in a dense subset of the real numbers, it's clearly possible to choose a number $x_0$ from $(a-\delta,a+\delta)$ with $f(x_0)=0$. This means that $|f(x_0)-f(a)| = |f(a)| < \varepsilon$. Since this is necessarily true for any given $\varepsilon > 0$, it follows that $|f(a)| < \varepsilon$ for all $\varepsilon > 0$. By the Archimedean property of the real numbers, 0 is the only real number satisfying this criterion, so $f(a)=0$. Clearly, this holds for any number where $f$ is continuous, so $f(x)=0$ for all $x \in \mathbb{R}$, completing the proof.

2. Sep 9, 2010

Looks fine to me.

3. Sep 9, 2010

### jgens

Thanks. Is there any way that I can improve my proof then?

4. Sep 9, 2010

### Dick

Actually, your proof is assuming f is uniformly continuous, which you don't necessarily have. It's more straightforward to do a proof by contradiction. Assume f(x0)=c where c is not zero. Can you show that leads to a contradiction?

5. Sep 9, 2010

### jgens

Dick, can you explain how my proof assumes uniform continuity? And yes, I know that I can prove it that way because that was my first thought, I'm just trying to see if it's possible to do it without using a contradiction.

Last edited: Sep 9, 2010
6. Sep 9, 2010

### Dick

Actually, I take that back. The only reason I thought that is because you mentioned delta before you mentioned 'a'. I would put an "At any point a since f is continuous at a..." at the beginning. Or at least make it clear that delta depends on a. Otherwise a careless reader like me might think you are assuming otherwise.

7. Sep 9, 2010

### jgens

Thanks! I'm sorry about that. I'll be more careful in the future.