The Real And Complex Number Systems

Claim that Bp, r S as follows.
Let q Bp, r, We want to find two special points x Bx, r, andy By, r such
that q x 1 y.
Since the three n balls Bx, r, By, r, andBp, r have the same radius. By
parallelogram principle, we let x q x p, andy q y p, then
x x q p r, andy y q p r.
It implies that x Bx, r, andy By, r. In addition,
x 1 y
q x p 1 q y p
q.
Since x, y S, andS is convex, then q x 1 y S. It implies that Bp, r S
since q is arbitrary. So, we have proved the claim. That is, for 0 1,
x 1 y p intS if x, y intS, andS is convex. Hence, by the definition of
convex, we know that the interior of a convex is convex.
(d) The closure of a convex is convex.
Proof: Given a convex set S, and let x, y S. Consider x 1 y : p, where
0 1, and claim that p S, i.e., we want to show that Bp, r S .
Suppose NOT, there exists r 0 such that
Bp, r S . *
Since x, y S, then Bx, r S and By, r S . And let x Bx, r S and
2 2 2
y By, r S. Consider
2
x 1 y p x 1 y x 1 y
x x 1 y 1 y
x x x x
1 y 1 y 1 y 1 y
x x 1 y y | |x y
2 r | |x y
r
if we choose a suitable number , where 0 1.
Hence, we have the point x 1 y Bp, r. Note that x, y S and S is convex,
we have x 1 y S. It leads us to get a contradiction by (*). Hence, we have proved
the claim. That is, for 0 1, x 1 y p S if x, y S. Hence, by the
definition of convex, we know that the closure of a convex is convex.
3.15 Let F be a collection of sets in R n , and let S AF A and T AF A. For each
of the following statements, either give a proof or exhibit a counterexample.
F.
(a) If x is an accumulation point of T, then x is an accumulation point of each set A in
Proof: Let x be an accumulation point of T, then Bx, r x T for any
r 0. Note that for any A F, wehaveT A. Hence Bx, r x A for any
r 0. That is, x is an accumulation point of A for any A F.
The conclusion is that If x is an accumulation point of T AF A, then x is an
accumulation point of each set A in F.