Is there a function f: R² → R that is not continuous but f(U) is open in R for all open sets UC R²?

Prove that if f: R→ R is continuous, then the image of a bounded set over f is bounded.

Let f: R" → R be a continuous function, and let x1, x2 E R" be distinct. Prove that there exist a continuous function g: R" → R and disjoint neighborhoods U1, U2 of x1, x2, respectively, such that g = f in U, and g = 1- f in U2.

Let f be a function from Z×Z to Z×Z (that is, f maps pairs of integers to pairs of integers) given by f (x, y) = (x + y, xy). Is f one-to-one? onto? Prove or give a counterexample