f RPRSolutionFirst suppose that f is continuous Note that c

f: R^P->R.

First suppose that f is continuous.

Note that (, c) and (c, ) are open subsets of R.

Hence {x : f(x) < c} = f ¹ (, c) and {x : f(x) > c} = f ¹ (c, ) are open in RP

Conversely, suppose the sets {x : f(x) < c} and {x : f(x) > c} are open in RP for every c R.

Any open subset U of R can be written as the union of open balls U = A(a, b), where A is an arbitrary indexing set.

Note (a, b) = (, b) (a, ) and f ¹ (a, b) = f ¹ (, b) f ¹ (a, ) = {x : f(x) < b} {x : f(x) > a}

Since the intersection of any two open sets is open, each set f ¹ (a, b) is open.

Since the arbitrary union of open sets is open, the set f ¹ (U) = Af ¹ (a, b) is open.

Hence f is continuous on RP.