math
posted by gemma .
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.
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