# FA1

Let and be metric spaces, and be a mapping.

(i) Prove that if is open whenever is open, then is continuous.

(ii) Prove that is continuous if and only if is closed whenever is closed.

## (i)

Suppose is open whenever is open. Let and . Since the ball is open, so is . Since . Since is open there exists a such that . This implies making continuous at . Since was arbitrary, is continuous.

## (ii)

Suppose is closed whenever is closed. Let be open. Then is closed. By the hypothesis, is closed.

If , then and . Similarly, if then , and . Therefore .

Now, is closed, so is open. So is open whenever is open. From part (i), is continuous.

Suppose is continuous. Let be closed. Let converge to . Since is continuous converges to . Since is closed, and so which means is closed (since this is the limit of the arbitrary sequence chosen above). Therefore is closed whenever is closed.