The Real And Complex Number Systems

A B which contradicts (*)-3. Hence, we have prove that S is connected.
Remark: We given another proof by the method of two valued function as follows. Let
f be a two valued function defined on S, and choose any two points a, b S. If we can
show that fa fb, we have proved that f is a constant which implies that S is
connected. Since f is a continuous function defined on a compact set S, then f is uniformly
on S. Thus, given 1 0, there exists a 0 such that as x y , x, y S, we
have |fx fy| 1 fx fy. Hence, for this , there exists a finite set of
points x 0 , x 1 ,...,x n in S with x 0 a and x n b such that
x k x k1 for k 1,2,..,n.
So, we have fa fx 0 fx 1 ... fx n fb.
4.43 Prove that a metric space S is connected if, and only if, every nonempty proper
subset of S has a nonempty boundary.
Proof: () Suppose that S is connected, and if there exists a nonempty proper subset U
of S such that U , thenletB clS U, wehave(defineclU A
1. A . B since S U ,
2. A B clU clS U U S U S
S A B,
3. A B clU clS U U ,
and
4. Both A and B are closed in S Both A and B are open in S.
Hence, S is disconnected. That is, if S is connected, then every nonempty proper subset of
S has a nonempty boundary.
() Suppose that every nonempty proper subset of S has a nonempty boundary. If S is
disconnected, then there exist two subsets A and B such that
1. A, B are closed in S, 2.A and B , 3.A B , and 4. A B S.
Then for this A, A is a nonempty proper subset of S with (clA A, andclB B)
A clA clS A clA clB A B
which contradicts the hypothesis that every nonempty proper subset of S has a nonempty
boundary. So, S is connected.
4.44. Prove that every convex subset of R n is connected.
Proof: Given a convex subset S of R n , and since for any pair of points a, b, the set
1 a b :0 1 : T S, i.e., g : 0, 1 T by g 1 a b is a
continuous function such that g0 a, andg1 b. So,S is path-connected. It implies
that S is connected.
Remark: 1. In the exercise, it tells us that every n ball is connected. (In fact, every
n ball is path-connected.) In particular, as n 1, any interval (open, closed, half-open, or
infinite) in R is connected. For n 2, any disk (open, closed, or not) in R 2 is connected.
2. Here is a good exercise on the fact that a path-connected set is connected. Given
0, 1 0, 1 : S, andifT is a countable subset of S. Prove that S T is connected. (In
fact, S T is path-connected.)
Proof: Given any two points a and b in S T, then consider the vertical line L passing
through the middle point a b/2. Let A x : x L S, and consider the lines form
a to A, and from b to A. Note that A is uncountable, and two such lines (form a to A, and