Is A dense set in the reals and f(x)=0 for all x in A, does f(x)=0 for all x?

[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?

 

[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

 

[Bleys]

x can always be chosen to be in A, for any delta>0

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! :)

 

[torquil]

Yeah, I was using the following:

Formally, a subset A of a topological space X is dense in X if for any point x in X, any neighborhood of x contains at least one point from A.

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