Adequate proof? Spivak's Calculus ; Dense sets

[Derek Hart]

Homework Statement

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

**A dense set is defined, in the book, as a set which contains a point in every open interval, such as the set of all irrational or all rational numbers.

Homework Equations

N/A

The Attempt at a Solution

I looked in the answer book, and spivak's proof is much different than mine. I want to know whether mine is adequate.
I began by considering a point (α, f(α)) such that f(α)<0. I then stated that there would have to be some σ so that f(x)<0 for all x such that α-σ < x <α+σ. This, however, contradicts the initial statement that A is a dense set (since f<0 for all x in (α-σ, α+σ)).
I had an almost identical argument for f>0, which i wish not to type. So does this prove that f(x)=0 for all x?

 

[LCKurtz]

Homework Statement

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

**A dense set is defined, in the book, as a set which contains a point in every open interval, such as the set of all irrational or all rational numbers.

Homework Equations

N/A

The Attempt at a Solution

I looked in the answer book, and spivak's proof is much different than mine. I want to know whether mine is adequate.
I began by considering a point (α, f(α)) such that f(α)<0. I then stated that there would have to be some σ so that f(x)<0 for all x such that α-σ < x <α+σ. This, however, contradicts the initial statement that A is a dense set (since f<0 for all x in (α-σ, α+σ)).
I had an almost identical argument for f>0, which i wish not to type. So does this prove that f(x)=0 for all x?

You didn't state it carefully, but what you are showing is that if f is continuous at $\alpha$ and $f(\alpha) < 0$ then you get a contradiction to A being dense. That is an indirect argument which is a reasonable way to prove it.

 

[Derek Hart]

You didn't state it carefully, but what you are showing is that if f is continuous at $\alpha$ and $f(\alpha) < 0$ then you get a contradiction to A being dense. That is an indirect argument which is a reasonable way to prove it.

Alright, so to correct it I should just explicitly state that f is continuous at α?

 

[LCKurtz]

You could write: Suppose $f$ is continuous at $\alpha$ and $f(\alpha)< 0$. Then blahdy blah blah...which is a contradiction to $A$ being dense. Therefore $f(x)$ is identically zero. But instead of assuming $f(\alpha) < 0$ I would start with $f(\alpha) \ne 0$ and use absolute value signs in your argument so you don't have to do two cases.

 

[Derek Hart]

That does seem simpler. Thanks