## Theorem 6.2.9: Continuity and Uniform Continuity

*f*is uniformly continuous in a domain

*, then*

**D***f*is continuous in

**D**.

If *f* is continuous on a compact domain * D*, then

*f*is uniformly continuous in

*.*

**D**### Proof:

The first part of the proof is obvious. If *|f(t) - f(s)|* is
small whenever *|s - t|* is small, regardless of the particular
location of *s* and *t*, then in particular
*|f(x) - f( x _{0})|* must be small when

*|x - x*is small.

_{0}|
The second part is much more complicated, and relies on the structure
of compact sets on the real line. Recall that a set is compact
if every open cover has a finite subcover. We first want to define
a suitable open cover: pick an
* > 0*. For every fixed
*x _{0}* in the compact set

*there exists a (possibly different)*

**D***(x*such that

_{0}) > 0| f( x_{0}) - f(x) | < if | x_{0}- x | < 1/2 (x_{0})

where depends on
*x _{0}*. Define

(xU_{0}) = { x: |x_{0}- x| < 1/2 (x_{0})} and= {U(xU_{0}): x_{0}is contained in}D

Then * U* is an open cover of

*, so by compactness can be reduced to a finite subcover. Suppose that the sets*

**D***U*cover the set

_{1}, U_{2}, ..., U_{n}*:*

**D**Let

= 1/2 min{(x_{1}), (x_{2}), ... (x_{n})}

Since this is a minimum over a finite set, we know that
* > 0*. Now take any two numbers
*t, s* in * D* such that

*|t - s| <*. Since the finite collection covers the compact set

*we know that*

**D***s*is contained in, say,

*U*. How far away from the center of

_{2}*U*is

_{2}*t*then ?

By the choice of we know that

|t - x_{2}| < |t - s| + |s - x_{2}| < + 1/2 (x_{2}) < 1/2 (( x_{2}) + ( x_{2})) = ( x_{2})

Now we are almost done, because if *|t - s| < *
then

|f(t) - f(s)| < |f(t) - f( x_{2})| + |f( x_{2}) - f(s)|

The first difference on the right is less than epsilon because
*|t - x _{2}| < (x_{2})*.
The second one is also less than epsilon because

*s*is contained in the set

*U*. Hence, the difference on the left is less than twice epsilon. But now it should be easy for you to modify the proof so that we can arrive at

_{2}|f(t) - f(s)| < whenever |t - s| < inD

That finishes the proof.