The Real And Complex Number Systems

if x Q, andgx 0ifx Q c .
Remark: It is the same rusult for minf 1 ,...f m since maxa, b mina, b a b.
4.21 Let f : S R be continuous on an open set in R n , assume that p S, and assume
that fp 0. Prove that there is an n ball Bp; r such that fx 0 for every x in the
ball.
Proof: Since p S is an open set in R n , there exists a 1 0 such that Bp, 1 S.
Since fp 0, given fp 0, then there exists an n ball Bp;
2 2 such that as
x Bp; 2 S, wehave
fp
fp fx fp 3fp .
2
2
Let min 1 , 2 , then as x Bp; , wehave
fx fp 0.
2
Remark: The exercise tells us that under the assumption of continuity at p, we roughly
have the same sign in a neighborhood of p, iffp 0 or fp 0.
4.22 Let f be defined and continuous on a closed set S in R. Let
A x : x S and fx 0 .
Prove that A is a closed subset of R.
Proof: Since A f 1 0, andf is continous on S, wehaveA is closed in S. And
since S is closed in R, we finally have A is closed in R.
Remark: 1. Roughly speaking, the property of being closed has Transitivity. Thatis,
in M, d let S T M, ifS is closed in T, andT is closed in M, then S is closed in M.
Proof: Let x be an adherent point of S in M, then B M x, r S for every r 0.
Hence, B M x, r T for every r 0. It means that x is also an adherent point of T in
M. SinceT is closed in M, we find that x T. Note that since B M x, r S for every
r 0, we have (S T)
B T x, r S B M x, r T S B M x, r S T B M x, r S .
So, we have x is an adherent point of S in T. And since S is closed in T, wehavex S.
Hence, we have proved that if x is an adherent point of S in M, then x S. Thatis,S is
closed in M.
Note: (1) Another proof of remark 1, since S is closed in T, there exists a closed subset
U in Msuch that S U T, and since T is closed in M, wehaveS is closed in M.
(2) There is a similar result, in M, d let S T M, ifS is open in T, andT is open
in M, then S is open in M. (Leave to the reader.)
2. Here is another statement like the exercise, but we should be cautioned. We write it
as follows. Let f and g be continuous on S, d 1 into T, d 2 .LetA x : fx gx,
show that A is closed in S.
Proof: Let x be an accumulation point of A, then there exists a sequence x n A
such that x n x. So, we have fx n gx n for all n. Hence, by continuity of f and g, we
have
fx f lim n
x n lim n
fx n lim n
gx n g lim n
x n gx.
Hence, x A. Thatis,A contains its all adherent point. So, A is closed.