There is a theorem about the integral of a continuous function which requires the notion of
uniform continuity. This is discussed in this section. Consider the function f

(x)

=

1
x

for
x ∈

(0,1)

. This is a continuous function because, by Theorem 6.0.6, it is continuous at every
point of

(0,1)

. However, for a given ε > 0, the δ needed in the ε,δ definition of continuity
becomes very small as x gets close to 0. The notion of uniform continuity involves being able
to choose a single δ which works on the whole domain of f. Here is the definition.

Definition 6.7.1Let f be a function. Then f is uniformlycontinuousif forevery ε > 0, there exists a δ depending only on ε such that if

|x− y|

< δ then

|f (x) − f (y)|

< ε.

It is an amazing fact that under certain conditions continuity implies uniform continuity.

Theorem 6.7.2Let f : K → F be continuouswhere K is a sequentiallycompact set in F. Then f is uniformly continuous on K.

Proof: If this is not true, there exists ε > 0 such that for every δ > 0 there exists a pair of
points, x_{δ} and y_{δ} such that even though

|xδ − yδ|

< δ,

|f (xδ) − f (yδ)|

≥ ε. Taking a
succession of values for δ equal to 1,1∕2,1∕3,

⋅⋅⋅

, and letting the exceptional pair of points for
δ = 1∕n be denoted by x_{n} and y_{n},

1-
|xn − yn| < n ,|f (xn) − f (yn)| ≥ ε.

Now since K is sequentially compact, there exists a subsequence,

{xnk}

such that
x_{nk}→ z ∈ K. Now n_{k}≥ k and so

|xn − yn | < 1.
k k k

Consequently, y_{nk}→ z also. ( x_{nk} is like a person walking toward a certain point and y_{nk} is
like a dog on a leash which is constantly getting shorter. Obviously y_{nk} must also move
toward the point also. You should give a precise proof of what is needed here.) By continuity
of f and Theorem 6.1.2,

0 = |f (z)− f (z)| = lkim→∞ |f (xnk)− f (ynk)| ≥ ε,

an obvious contradiction. Therefore, the theorem must be true. ■

The following corollary follows from this theorem and Theorem 4.7.2.