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