Page 30 - DMTH503_TOPOLOGY

Topology

Notes But ext (X – A) = ext (A) = (X – A)° = [X – (X – A)]°
= A°
Now (1) becomes A°  ext [ext (A)]
(vii) ext (A  B) = [X – (A  B)]° = [(X – A)  (X – B)]°

= (A  B)°
= (A)°  (B)°
= ext (A)  ext (B).
Theorem 20: Exterior Operator: The exterior, by definition of interior function ‘e’ on X is a
function
e : P(X)  P(X) s.t.
(i) e (X) = 

(ii) e ) = X
(iii) e (A)  A
(iv) e (A) = e [(e (A))]
(v) e (A  B) = e (A)  e (B)

For any sets A, B  X. Then there exists a unique topology T on X s.t. e (A) = T-exterior of A.
Proof: Write T = {G  X : e (G) = G}
We are to show that T is a topology on X.
(i) e () = e (X) =  by (i)

e (X) = e () = X by (ii)
Now e () = , e (X) = X  , X  T
(ii) Let G , G  T
1 2
Then e (G) = G , e (G) = G
1 1 2 2
But (G  G ) = G  G
1 2 1 2
e [(G  G )] = e (G  G)
1 2 1 2
= e (G)  e (G) by (v)
1 2
= G  G
1 2
 G  G  T
1 2
(iii) Firstly, we shall show that
A  B  e (B)  e (A) …(1)
A  B  A  B = B  e (B) = e (A  B)

= e (A)  e (B)  e (A)
 e (B)  e (A)
Let G  T    

Then e (G) = G …(2)
 

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