Dense set and continuity

1. Feb 10, 2010

Bleys

This is probably very simple but I'm not sure if the last step is right.
Let A be a dense set in the reals and f(x)=0 for all x in A. If f is continuous, prove that f(x)=0 for all x.
Let a be in real number. By definition, for all $$\epsilon > 0$$ there exists $$\delta > 0$$ such that $$|x-a|< \delta \Rightarrow |f(x)-f(a)|< \epsilon$$.
But in any open interval there lies element of A, so in particular: for all x in A such that
$$|x-a|< \delta$$ we have $$|f(a)|< \epsilon$$.
Here's where I deduce f(a)=0 and this is the step I'm not sure about. Does this make sense?

2. Feb 10, 2010

torquil

Yes, this is correct. You are first using continuity of f, and then the fact that x can always be chosen to be in A, for any delta>0. Since epsilon>0 was arbitrary, you have shown that |f(a)| is smaller than any positive number. The only non-negative number smaller than any positive real number is 0 :-)

Torquil

3. Feb 10, 2010

Bleys

Ah ok, this is what was making me unsure. I didn't know whether choosing some other epsilon and therefore having a different delta would change things. Thank you! :)

4. Feb 10, 2010

torquil

Yeah, I was using the following:
It can be used since your open subset around x (a such that |x-a|<delta) is always a neighbourhood of x, for any delta>0.

Torquil