Posted by
**gemma** on
.

Given: f(closure(A)) subset of closure(f(A))

Prove: For any closed set A in Y, f^-1(A) is closed in X (AKA f is continuous)

X and Y are metric spaces

f: X -> Y

f^-1 is f inverse.

closure(A) = A U limit points of A

The first line is Let C be a closed subset of Y. Then we have to show that f^-1(C) is closed in X.

I have made several attempts but nothing is working. Anyone have any ideas?

30 minutes ago - 4 days left to answer.