Page 29 - DMTH503_TOPOLOGY

Unit 1: Topological Spaces

By construction of T,  i(A)  T. Notes
Thus, i(A) is T-open set s.t. i(A)  A.
Let B be an open set s.t. B  A.
B  T, B  A  i(B) = (B), i(B)  i(A)

 B  i(A)
Thus i(A) contains any open set B s.t. B  A. It follows that i(A) is the largest open subset of A.
Consequently i(A) = A°.

1.7.4 Properties of Exterior

Theorem 19: Let (X, T) be a topological space and A, B  X. Then

(i) ext (X) = 
(ii) ext ) = X
(iii) ext (A)  A

(iv) ext (A) = ext [(ext (A))]
(v) A  B  ext (B)  ext (A)
(vi) A°  ext [ext (A)]
(vii) ext (A  B) = ext (A)  ext (B).
Proof:

(i) ext (X) = (X – X)° = as we know that ext (A) = (X – A)°
(ii) ext ) = (X – )° = X° = X
(iii) ext (A) = (X – A)° X – A = A or ext (A)  A for B°  B

(iv) [ext (A)] = [(X – A)°] =X – (X – A)°
or ext [{ext (A)}] = ext [X – (X – A)°]
= [X – {X – (X – A)°}]°
= [(X – A)°]° = (X – A)°°
= (X – A)° [As B°° = B°  B]

= ext (A)
 ext (A) = ext [(ext (A))]
(v) A  B  X – B  X – A

 (X – B)°  (X – A)°
 ext (B)  ext (A)
(vi) ext (A) = (X – A)°  X – A
 ext (A)  X – A
As A  B  ext (B)  ext (A), we get

ext (X – A)  ext [ext (A)] …(1)

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