# Prove the inverse function of a compact set is compact.

• Apr 6th 2007, 12:53 PM
Prove the inverse function of a compact set is compact.
Problem: Let f:R^n -> R^m be continuous. Let K be a compact subset of R^m. Suppose there exists r > 0 such that ||f(u)|| >= r||u|| for each u in R^n. Suppose that f^{-1} (K) is not an empty set, prove it is compact.

My proof:

Now f^{-1}(K) = {x in R^n : f(x) in K}, with f^{-1}(K) maps f(K) to R^n.

Let {Uk} be a sequence in f^{-1}(K), then there exists r > 0 such that ||f(Uk)|| >= r||Uk||.

Now, f(Uk) is in K, K is compact implies K is bounded, implies there exists a number M such that M >= ||f(Uk)|| >= r||Uk|| > ||Uk||, thus shows {Uk} is bounded, implies f^{-1}(K) is bounded.

Furthermore, suppose the sequence {Yk} in f^{-1}(K) converges to the point Y, then for each {Yk} in R^n, we have f(Yk) in K. Now f(Yk) -> f(Y) because f is continuous, and f(Y) is in K because K is closed, furthermore, f(Y) is in the set of f^{-1}(K).

Thus proves f^{-1}(K) is closed, and we conclude that it is compact since it is closed and bounded.

Q.E.D.