Continuity and Dense Subsets of the Real Numbers

[jgens]

Homework Statement

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}.

Homework Equations

N/A

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.

 

[radou]

Looks fine to me.

 

[jgens]

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

 

[Dick]

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

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?

 

[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.

 

[Dick]

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.

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.

 

[jgens]

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