# Prove continuous function = 0

## Homework Statement

A subset of $$A \subseteq R$$ of real numbers is called dense if $$\forall \delta > 0 , \forall x \in R , \exists a \in A: |x-a| < \delta$$ .

Suppose $$A \subseteq R$$ is dense. Prove that if g is a continuous function with g(x) = 0 for all $$x \in A$$, then g = 0

## The Attempt at a Solution

From the continuity of g, we know that $$\forall \epsilon > 0, \exists \delta > 0 : |g(x) - g(x_0)| < \epsilon$$ whenever $$0 < |x-x_0| < \delta$$. Now let x_0 be any real number that is not an element of A. We must prove that g(x_0) = 0.

To argue by contradiction, first assume that g(x_0) > 0. Again from g's continuity, $$\forall \epsilon > 0, \exists \delta > 0 : g(x_0) - \epsilon < g(x) < g(x_0) + \epsilon$$ whenever $$0 < |x-x_0| < \delta$$. Putting $$\epsilon = g(x_0)/2$$, we have $$g(x_0) - g(x_0)/2 = g(x_0)/2 < g(x)$$. Now g(x_0)/2 is positive and as a consequence, for all x in the interval $$x_0 - \delta < x < x_0 + \delta$$, g(x) > 0.

Since A is dense, then we can find a value x = x* in our delta-interval that is an element of A. We know that g(x*) = 0, but from our assumption that g(x) > 0 we've proved that g(x*) > 0, and so this is a contradiction.

The proof for assuming g(x) < 0 is similar, and so g(x) = 0 for all x. QED

How does this proof look?

## Answers and Replies

Dick
Science Advisor
Homework Helper
That looks just fine to me.

Thanks for checking :)