The Real And Complex Number Systems

lim x
1 1 2x 1/r 1
x 1/r , 0
0
lim 1/r
x 2 x 1 r 1
1
2x
1 1 r 1
by L-Hospital Rule.
0.
Hence x n 0asn .
6. Here is a useful criterion for a function which is NOT uniformly continuous defined
a subset A in a metric space. We say a function f is not uniformly continuous on a subset A
in a metric space if, and only if, there exists 0 0, and two sequences x n and y n
such that as
lim n
x n y n 0
which implies that
|fx n fy n | 0 for n is large enough.
The criterion is directly from the definition on uniform continuity. So, we omit the
proof.
4.52 Assume that f is uniformly continuous on a bounded set S in R n . Prove that f
must be bounded on S.
Proof: Since f is uniformly continuous on a bounded set S in R n ,given 1, then
there exists a 0 such that as x y , x, y S, wehave
dfx, fy 1.
Consider the closure of S, clS is closed and bounded. Hence clS is compact. Then for
any open covering of clS, there is a finite subcover. That is,
clS xclS Bx; /2,
clS kn k1 Bx k ; /2, wherex k clS,
S kn k1 Bx k ; /2, wherex k clS.
Note that if Bx k ; /2 S for some k, then we remove this ball. So, we choose
y k Bx k ; /2 S, 1 k n and thus we have
Bx k ; /2 By k ; for 1 k n,
since let z Bx k ; /2,
z y k z x k x k y k /2 /2 .
Hence, we have
S kn k1 By k ; , wherey k S.
Given x S, then there exists By k ; for some k such that x By k ; . So,
dfx, fx k 1 fx Bfy k ;1
Note that kn k1 Bfy k ;1 is bounded since every Bfy k ;1 is bounded. So, let B be a
bounded ball so that kn k1 Bfy k ;1 B. Hence, we have every x S, fx B. Thatis,
f is bounded.
Remark: If we know that the codomain is complete, then we can reduce the above
proof. See Exercise 4.55.
4.53 Let f be a function defined on a set S in R n and assume that fS R m .Letg be
defined on fS with value in R k , and let h denote the composite function defined by