Thread: 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.

PLEASE CHECK, THANK YOU!!!

KK

Please review the statement of the problem you posted. It seems to me there are some internal inconsistencies there. In what you posted as a proof, you wrote “||Uk|| > Uk”. That has no possible meaning. Here is why: ||Uk|| is a number, but Uk is a point on a metric space R^m.

I will say that I have not seen any similar problem. That makes me suspicious of the exact statement.

Actually, there is no reason for me to even have ||Uk|| > Uk, because once I have M>=||Uk||, I have already proven Uk is bounded, so it is deleted now.