Homepage › Solution manuals › Walter Rudin › Principles of Mathematical Analysis › Exercise 4.3
Exercise 4.3
Exercise 3: Let be a continuous real function on a metric space . Let (the zero set of ) be the set of all at which . Prove that is closed.
Answers
Note that is closed, so that is closed by the continuity of .
Comments
Proof. Let be the set of all with . only contains 0, since that is how we defined above. Now consider . This set is open since it is the union of two open sets, i.e. . Since is continuous, the set which maps onto is open in . Since , we know that is closed. □