The Real And Complex Number Systems

Section 4.18.)
Proof: Let f be a two valued function defined on S. SinceA, andB are connected in S,
then we have
fA a, andfB b, wherea, b 0, 1.
Given a sequence x n A with x n 0, 0, then we have
a lim n
fx n f limx n by continuity of f at 0
n0
f0, 0
b.
So, we have f is a constant. That is, S is connected.
Assume that S is arcwise connected, then there exists a continuous function
g : 0, 1 S such that g0 0, 0 and g1 1, sin 1. Given 1/2, there exists a
0 such that as |t| , wehave
gt g0 gt 1/2. *
1
Let N be a positive integer so that
2N
Define two subsets U and V as follows:
U
V
1
, thus let ,0 : p and 1
,0 : q.
2N 2N1
x, y : x p q
2
x, y : x p q
2
gq, p
gq, p
Then we have
(1). U V gq, p, (2). U , sincep U and V , sinceq V,
(3). U V by the given set A, and (*)
Since x, y : x pq
2
gq, p. So, we have
and x, y : x
pq
2 are open in R2 , then U and V are open in
(4). U is open in gq, p and V is open in gq, p.
From (1)-(4), we have gq, p is disconnected which is absurb since a connected subset
under a continuous function is connected. So, such g cannot exist. It means that S is not
arcwise connected.
Remark: This exercise gives us an example to say that connectedness does not imply
path-connectedness. And it is important example which is worth keeping in mind.
4.47 Let F F 1 , F 2 ,... be a countable collection of connected compact sets in R n
such that F k1 F k for each k 1. Prove that the intersection
k1 F k is connected and
closed.
Proof: Since F k is compact for each k 1, F k is closed for each k 1. Hence,
k1 F k : F is closed. Note that by Theorem 3.39, we know that F is compact. Assume
that F is not connected. Then there are two subsets A and B with
1.A , B . 2.A B . 3.A B F. 4.A, B are closed in F.
Note that A, B are closed and disjoint in R n . By exercise 4.57, there exist U and V which
are open and disjoint in R n such that A U, andB V. Claim that there exists F k such
that F k U V. Suppose NOT, thenthereexistsx k F k U V. Without loss of
generality, we may assume that x k F k1 . So, we have a sequence x k F 1 which
implies that there exists a convergent subsequence x kn , say lim kn x kn x. Itis
clear that x F k for all k since x is an accumulation point of each F k . So, we have