1549901369-Elements_of_Real_Analysis__Denlinger_

5.4 Uniform Continuity 259

N otes:

(1) The conclusion of Theorem 5.4.3 cannot be strengthened to say that

f : V(f) -->IR is continuous on A. The Dirichlet function (Example 5.1.10) is

uniformly continuous on Q but is not continuous on Q.

(2) There is no difference between the statements "f : V(f) --> IR is uni-

formly continuous on a set A ~ V(f)" and "f : A --> IR is uniformly continu-

ous." The statement "f is uniformly continuous on A " will be used to cover

both.

(3) The converse of Theorem 5.4.3 is not true. Continuity on A does not

imply uniform continuity on A. The following example demonstrates that.

1

Example 5.4.4 The function f(x) = - is continuous on (0, 1), but is not

x

uniformly continuous there. The proof is contained in the following discussion.

We already know that this function is continuous on (0, 1), by Theorem

5.1.13. What does it mean to say that it is not uniformly continuous there? In

Figure 5. 10 below, imagine that a fixed€ > 0 is given. The figure on the left

shows that when x = 2 , a fairly large 8 is satisfactory. The figure on the right

shows that as x approaches 0, the value of 8 must get quite small; and in fact

8 must --> 0 as x --> 0. It is apparent that no single 8 > 0 is small enough to

work for all x > 0. (This is a convincing argument, but is not a proof.)

y

~

8-nbd.

y

e-nbd.0 -:

--r

' '

' '

x

Figure 5.10

:

' I I

' '

' ' I

I

J_,

8-nbd.

x

In order to prove that a function is not uniformly continuous on a set A ,
we need a criterion to use. If we negate Definition 5.4.1, we get the following
criterion: