The Real And Complex Number Systems

1. x S and S is open.
Bx, d S for some d 0.
Bx, r S Bx, r if r d.
Bx, r S Bx, d if r d.
and
2. x clT
Bx, r T for any r 0.
From 1 and 2, we know
Bx, r S T Bx, r S T Bx, r T if r d.
Bx, r S T Bx, r S T Bx, d T if r d.
So, it means that x is an adherent point of S T. Thatis,x clS T. Hence,
S clT clS T.
Remark: It is not necessary that clS T clS clT. For example, S Q and
T Q c , then clS T and clS clT R 1 .
Note. The statements in Exercises 3.9 through 3.13 are true in any metric space.
3.14 AsetSinR n is called convex if, for every pair of points x and y in S and every
real satisfying 0 1, we have x 1 y S. Interpret this statement
geometrically (in R 2 and R 3 and prove that
(a) Every n ball in R n is convex.
Proof: Given an n ball Bp, r, and let x, y Bp, r. Consider x 1 y, where
0 1.
Then
x 1 y p x p 1 y p
x p 1 y p
r 1 r
r.
So, we have x 1 y Bp, r for 0 1. Hence, by the definition of convex,
we know that every n ball in R n is convex.
(b) Every n dimensional open interval is convex.
Proof: Given an n dimensional open interval I a 1 , b 1 ...a n , b n .Letx, y I,
and thus write x x 1 , x 2 ,...,x n and y y 1 , y 2 ,...y n . Consider
x 1 y x 1 1 y 1 , x 2 1 y 2 ,...,x n 1 y n where 0 1.
Then
a i x i 1 y i b i ,wherei 1, 2, . . , n.
So, we have x 1 y I for 0 1. Hence, by the definition of convex, we
know that every n dimensional open interval is convex.
(c) The interior of a convex is convex.
Proof: Given a convex set S, and let x, y intS. Then there exists r 0 such that
Bx, r S, andBy, r S. Consider x 1 y : p S, where 0 1, since S
is convex.