Page 28 - DMTH503_TOPOLOGY

Topology

Notes To prove that  unique topology T on X s.t. i(A) = A°, where A° = T-interior of A.
Write T = {A  X : i(A) = A}
(1) X  T, for i(X) = X
(2) To prove   T

i()  , by (ii)
But   i()
i()   So that   T
(3) G , G ,  T  G  G  T
1 2 1 2
For G , G ,  T  i(G ) = G , i(G ) = G
1 2 1 1 2 2
 i(G  G ) = i(G )  i(G ) by (iii)
1 2 1 2
= G  G
1 2
 i(G  G ) = G  G
1 2 1 2
 G  G  T
1 2
(4) To prove G  T       {G :   }  T
 
Firstly we shall prove that
A  B  i(A)  i(B),

where A, B,  X …(1)
A  B  A  B = A
 i(A) = i(A  B)
= i(A)  i(B), by (iii)

 I (B)
 i(A) i(B). Hence the result (1).
Let G  T     so that

i(G ) = G …(2)
 
Also let  {G :   } = G.

Then G  G  i(G )  i(G), by (1)
 
 G  i(G), by (2)

  {G :   }  i(G)

 G  i(G)
But i(G)  G, by (ii).
Consequently i(G) = G so that G  T. Hence the result (4). From (1), (2), (3) and (4), it follows that
T is a topology on X.
Remains to prove that
i(A) = A°.
By (iv), i[i(A)] = i(A)

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