5.4 Uniform Continuity 263Proof. Part 1 (=>): See Theorem 5.4.8 above.
Part 2 ( <=): We shall prove the contrapositive. Suppose A is a bounded set
and the function f : A --> JR. is not uniformly continuous on A. Then as noted in
Lemma 5.4.5, 3 c > 0 3 VJ> 0, 3x,y EA 3 Ix -y\ < 6 but \f(x) -f(y)\ 2". c.
Fix this value of c throughout the remainder of this proof. Then,
Vn EN, 3xn, Yn EA 3 \xn - Yn\ < ~ and \f(xn) - f(Yn)\ 2". c. (10)
Consider the two sequences { Xn} and {Yn}:
(1) The sequence {xn} is a bounded sequence, so it has a convergent sub-
sequence, which we shall denote { Xin}. Say Xin --> L.
(2) The sequence {Yin } is a bounded sequence, so it has a convergent sub-
sequence, which we shall denote {Yjn}.
(3) Consider the sequences {xjn} and {Yin}. Since {xin } is a subsequence
of {xin}, we must have Xjn --> L.
(4) We shall prove that Yjn --> L.
Let c > 0. Then 3n 1 EN 3 ~ < :.
2. Since xfo --> L, 3n2 E N3 n 2". n2 =>
n1
c
\xin - L\ < 2· Choose no= max{ni,n 2 }. Then
1 1 c c
n -> no => -Jn < - < -- n 2 and \x Jn · - L\ < - 2
1 c c
=> \x Jn · -y Jn · \ < -Jn < - 2 and IxJn · -L\ < - 2(from (10) above)
I Ic c
=> \Yin - L\ :::; Yin - XJn + \xin - L\ < 2 + 2
=> IYin - L\ < c.
Therefore, Yjn --> L.
(5) Consider the sequence
{zn} = {xill Yjl) xh, Yi2 ,Xja, Yiai ... 'Xjnl Yjn> ... }.
We have Zn --> L , by Exercise 2.6.6. Thus, {Zn} is a Cauchy sequence in A.
(6) However, Vn E N, IJ(xjJ - f(YiJ\ 2". c from (10) above. Therefore,
{f (zn)} is not a Cauchy sequence.
Thus, if f : A --> JR. is not uniformly continuous on A , then there exists a
Cauchy sequence {zn} in A 3 {f(zn)} is not a Cauchy sequence. •
*CONTINUOUS EXTENSIONSA useful property of uniformly continuous functions f : (a, b) --> JR. is that
they can be "extended continuously" to [a, b]. That is because, for such func-
tions, lim f(x) and lim f(x) always exist. The following definition and the-
x-.a+ x-.b-
orems will make these ideas precise.