let f:R(n)->R(n) be a cont. funct...compact???

Question

let f:R(n)->R(n) be a cont. funct. and B[a,r]={x∈R(n):d(a,x)<=r} be a closed bal in R(n).Provethat the set B[a,r]intersection f^(-1)(B[a,r]) is compact..?

Answer 1

f inverse of B[a,r] is closed because f is continuous and th eintersection with B[a,r] is bounded (because B[a,r] is) so f inverse is a closed and bounded subset of Rn, hence compact.

Answer 2

You should be able to find this proof in a real analysis textbook (maybe Rudin's?) for the one dimensional case and then generalize from there. The theorem says that a map is continuous iff f(-1)(G) is open whenever G is open. However, the same is true for closed sets (because of complements of a set; open if its complement is closed).