Theorem: Continuity and Uniform Continuity

• If f is uniformly continuous in a domain D, then 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( ) | must be small when | x - | is small.

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 in the compact set D there exists a (possibly different) ( ) > 0 such that

• | f( ) - f(x) | < if | - x | < 1/2 ( ) (here depends on ).

Define

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

Then U is an open cover of D, so by compactness can be reduced to a finite subcover. Suppose that the sets , , ..., cover the set D: Let

• = 1/2 min{ ( ), ( ), ... ( )}

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 D we know that s is contained in, say, . How far away from the center of is t then ? By the choice of we know that

• | t - | < | t - s | + | s - | < + 1/2 ( ) < 1/2 ( ( ) + ( )) = ( )

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

• | f(t) - f(s) | < | f(t) - f( ) | + | f( ) - f(s) |

The first difference on the right is less than epsilon because | t - | < ( ). The second one is also less than epsilon because s is contained in the set . 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

• | f(t) - f(s) | < whenever | t - s | < in D

That finishes the proof.