Define F: R -> R by f(x) = 5x if x is rational and f(x)= x^2 + 6 if x is irrational. Prove that f is discontinuous at 1 and continuous at 2. Are there any other points besides 2 at which f is continuous?
In any neighbourhood of 1 there are rational and irrational points, so there
are points arbitrarily close to 1 where f(x) is arbitrarily close to 5, and other
points arbitrarily close where f(x) is close to 7, so f must be discontinuous
at x=1. So if you want you can find a sequence x_n of rational points
converging to 1 and for this sequence f(x_n), similarly you can find a
sequence of irrational points y_n which converges to 1 and for this sequence
f(y_n) converges to 7, thus proving that f is not continuous at x=1.
All sequences x_n points all converging to 2 have f(x_n) converging to 10, so
f(x) is continuous at x=2.
Similarly the function is continuous whenever 5x = x^2 + 6, so it is
continuous at x=2 and x=3.
The above should be convertible into whatever form you use to demonstrate
continuity without too much difficulty.
RonL