Prove: Dense Set A for Continuous f=0 for all x

[ejionheara]

Homework Statement

A set A of real numbers is said to be dense if every open interval contains a point of A

Prove that if f is continuous and f(x) = 0 for all numbers x in a dense set A, then f(x) = 0 for all x.

Homework Equations

none

The Attempt at a Solution

The way I understand that a set of numbers are dense would be to use an example such as the rational numbers are dense in the reals because no matter what open interval one could take on the reals, it will contain an element of the set of the rationals.

Applying that reasoning to this question, I think it would follow that because
A is dense, no matter where the interval was taken, one would find an element a in A where f(a) is zero. Since f is continuous, there exists a neighborhood in which f will intersect A. However since it intersects A, it must be a part of A, then f(x) = 0.

The problem I'm having is I don't really know if I can assert that, and if I can't I was thinking about a proof by contradiction.

Any help would be very much appreciated.

 

[micromass]

Why is this true? "However since it intersects A, it must be a part of A"?

Do you know the following theorems?
1) A is dense in R iff for every point x in R, there exist a sequence (xn) in A which converges to x.
2) f is continuous in x iff (xn--> x implies f(xn)--> f(x))

It would be handy if you knew these two things...

 

[ejionheara]

No, I do not know those theorems.

The class I'm currently in uses Spivak's Calculus (4th) and we just got through the Least Upper Bound chapter. So we really haven't discussed convergence of a sequence at all.

 

[micromass]

Tp bad.

No worries tho, an epsilon-delta proof should be able to do the trick to.

Assume that f(a) is not zero. Let \epsilon=|f(a)|. This should be in contradiction with the continuity of f, which says that

\exists \delta: \forall x\in X:~|x-a|<\delta~\Rightarrow~|f(x)-f(a)|<\epsilon

Say something if you need more hints...

 

[ejionheara]

If I'm correct, if one were to set epsilon = |f(a)|, then wouldn't |f(x)| be strictly between 0 and -2|(f(a)|, (by the reverse triangle inequality). If this is true, then that would be a contradiction because f(x) = 0. Then f(a) = 0, which means a is actually in the set A?

 

[micromass]

"then that would be a contradiction because f(x) = 0."

That's a bit to fast. f(x)=0 is only true for elements x in A. You'll first have to show that there exists an x in A such that |x-a|<delta. That's were density comes in...


"which means a is actually in the set A?"
No, a is certainly not in a in general. The contradiction is that f(a)=0, which is not what we assumed.

 

[ejionheara]

Okay, so we proved that there was a contradiction in asserting that f(a) != 0, so f(a) = 0. So then I need to show that there exists an x in A such |x-a| < delta. But A is dense, meaning every open interval has to contain a point of A.

Is there a formal way of stating density? Right now it seems as if it's just going by definition.

 

[micromass]

Yes, it is just by definition

 

[ejionheara]

Oh, wow, I'm so used to assuming that going by the definition is wrong. If it's not complicated, it's obviously false...

Anyway, now that we've shown that there exists an x in A such |x-a|< delta. Then A intersects this delta neighborhood meaning that f(x) = 0. Then no matter what delta is taken as f(x) = 0?

 

[micromass]

Yes. Since A always intersects the delta-neighbourhood, this implies that there always exists and x such that |x-a|<delta. This implies that f(x)=0, and thus (by choice of epsilon), that f(a)=0. Which is a contradiction.

 

[ejionheara]

Thank you very much for the help!