Suppose that f:[0,1] -> R and f(a) = lim{x -> a} f(x) for all a in [0, 1]. Prove that f(q) = 0 for all q in Q intersection[0,1] implies that f(x) = 0 for all x in [0, 1]. Is the statement still true if f:[0, 1] -> R and f(a) = lim{x->a} f(x) for all a in Q intersection[0, 1]?
R denotes the set of Real numbers
Q denotes the set of Rational numbers
Please explain thouroughly how to come up with the solution. Thank You.
A proof involvng limits and functions is offered.