## Solutions to Homework 7

Rudin Chapter 4.

Problem 1.
The condition

does not imply continuity ofr at It certainly holds if is continuous, so consider a function which is not continuous at one point, such as

Certainly the condition holds for by continuity. For it also holds, since the definition of limit excludes the value thus for

Problem 4.
If is continuous and is dense, then for every there is a sequence such that in as Then, by the continuity of in which shows that is dense in Now, if and are both continuoues and for all where is dense it follows that that is for all Indeed, by the density of in there exists a sequence in with for all Then

Problem 15.
Let be continuous and open. Thus if is open then both and are open. If in then assumes its maximum, and minimum, on by the continuity of If there is an interior point at which assumes its maximum or minumum then cannot be open, since the maximum, or minimum is not an interior point of it. Since is connected, is an interval and In particular since otherwise these cannot be the endpoints. Thus is continuous and injective. Such a map is necessarily monotonic. To see this explicitly, suppose (otherwise replace by Then if since if not, contains point on both sides of and and so would contain violating the injetivity. Similarly, if then The same argument now shows that given implies and similarly for Thus is monotonic.

Richard B. Melrose 2004-05-18