A subset of [tex] A \subseteq R [/tex] of real numbers is called dense if [tex] \forall \delta > 0 , \forall x \in R , \exists a \in A: |x-a| < \delta [/tex] .
Suppose [tex]A \subseteq R [/tex] is dense. Prove that if g is a continuous function with g(x) = 0 for all [tex] x \in A [/tex], then g = 0
The Attempt at a Solution
From the continuity of g, we know that [tex] \forall \epsilon > 0, \exists \delta > 0 : |g(x) - g(x_0)| < \epsilon [/tex] whenever [tex]0 < |x-x_0| < \delta [/tex]. 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, [tex] \forall \epsilon > 0, \exists \delta > 0 : g(x_0) - \epsilon < g(x) < g(x_0) + \epsilon [/tex] whenever [tex] 0 < |x-x_0| < \delta [/tex]. Putting [tex] \epsilon = g(x_0)/2 [/tex], we have [tex] g(x_0) - g(x_0)/2 = g(x_0)/2 < g(x) [/tex]. Now g(x_0)/2 is positive and as a consequence, for all x in the interval [tex] x_0 - \delta < x < x_0 + \delta [/tex], 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?