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.

To prove that f^-1(A) is closed in X for any closed set A in Y, we can use the fact that f(closure(A)) is a subset of closure(f(A)).

Here's a step-by-step proof:

1. Let A be a closed subset of Y.
2. We want to show that f^-1(A) is closed in X.
3. Take an arbitrary point x in the complement of f^-1(A), denoted by f^-1(A)^c.
4. Since x is not in f^-1(A), it means that f(x) is not in A.
5. Since A is closed, its complement, Y - A, is open.
6. Since f(x) is not in A, it belongs to Y - A, which is open in Y.
7. By the definition of continuity, for every open set V in Y, the preimage of V under f, denoted by f^-1(V), is open in X.
8. Since Y - A is open in Y, f^-1(Y - A) is open in X.
9. By De Morgan's Law, f^-1(Y - A) = f^-1(Y) - f^-1(A).
10. Since Y is the whole space, f^-1(Y) is the whole space X.
11. Therefore, f^-1(X - A) = X - f^-1(A) is open in X.
12. This implies that f^-1(A) is closed in X.

Hence, by establishing that f^-1(A) is closed for a closed set A in Y, we have shown that f is continuous.

In summary, to prove that f is continuous, it is necessary to show that the preimage of any closed set A in Y, denoted by f^-1(A), is a closed set in X.