Principles of Mathematics in Operations Research

144 10 Basic Topology

3qx G Br(p) r\E3qi^p^ d(q,p) = n > 0.

3q 2 eBr(p)DE 9 q 2 ^p.

Then, q 2 ^qi- Since q 2 ^ p, r 2 = d(q 2 ,p) > 0.

3q 3 e Br2(p) D E 9 q 3 ^p^q 2 ^qi;---. D

Corollary 10.1.38 7/F is a/mite set, E' = 0.

Theorem 10.1.39 E is open if and only if Ec is closed.

Proof. (=>): Let E be open, Let p be a limit point of Ec. Show p G Ec.

Suppose not:

p G E =>• 3r > 0 9 -Br(p) C E [because E is open] (*)

Since p is a limit point of Ec, for every neighborhood N of p, N D Ec ^ 0. In

particular (by taking N = Br(p)), Br(p) OEc ^%, Contradiction to (*).

(<=): Assume Ec is closed. Show E is open; i.e. Vp G E, 3r > 0 9 Fr(p) C

E. Let p G E =4> p £ Fc => p is not a limit point of Fc. So 3r > 0 9 Br(p)f)Ec

does not contain any q / p (p either). =*> _Br(p) (~l Ec = 0 =>• Br(p) C E. D

Theorem 10.1.40 Let E C X, then

(a) E is closed.

(b) E-E-^Eis closed.

(c) E is the smallest closed set which contains E; i.e. if F is closed and

E C F => E CF.

Proof. EcX.
(a): (F)c is open.
Let p G (F)c => p <£ E_=> 3r > 0 B Br{p) n F = 0 => Br(p) C (F)c.
Show that Br{p) C {E)c:
If it is not true 3q G Br(p) and q $ (E)c =$• q€ Ec.
Find s > 0 3 £s(g) C £r(p). Then Bs(g) n E ^ 0 => Br(g) n F ^ 0.
Contradiction,
(b): (=>): Immediate from (a).
(«=): F is closed. Show E = E, i.e. F C F. Let p G F = F U F', if p G F,
we are done.
If p G F' =>• p G F (because F is closed),
(c): Let F be closed, E C F. Show that E C F. Let p G E = E U E', if
p G F => p G F. If p G F' we have to show that p G F':
Given r > 0, show Br(p) fl F contains a point q ^ p. Since p G F',
Sr(p) n F contains a point </ ^ p. Then, q G Fr(p) n F (because F c F).
So, p G F' =*> p G F (because F is closed). •