Page 62 - DMTH503_TOPOLOGY

Topology

Notes Theorem 1: A subspace of a topological space is itself a topological space.
Proof:
(i)   T and   Y =     T ,
Y
X  T and X  Y = Y  Y  T ,
Y
(ii) Let {H :   } be any family of sets in T .
 Y
Then      a set G  T such that H = G  Y
  
 {H :   } = {G  Y :   }
 
= [ {G :   }]  Y  T
 Y
since {G :   }  

(iii) Let H and H be any two sets in T .
1 2 Y
Then H = G  Y and H = G  Y for some G , G  T.
1 1 2 2 1 2
 H  H = (G  Y) (G Y)
1 2 1 2
= (G  G )  Y  T , since G  G  T
1 2 Y 1 2
Hence, T is a topology for Y.
Y
Example 3: Let (Y, V) be a subspace of a topological space (X, T) and let (Z, W) be a
subspace of (Y, V). Then prove that (Z, W) is a subspace of (X, T).
Solution: Given that (Y, V)  (X, T) …(1)
and (Z, W)  (Y, V) …(2)
We are to prove that (Z, W)  (X, T)
From (1) and (2), we get

Z  Y  X …(3)
From (1), V = {G Y : G  T} …(4)
and (2), W = {H  Z : H  V} …(5)
From (4) and (5), we get H = G  Y
 H  Z = (G  Y)  Z
= G  (Y  Z)
= G  Z [Using (3)]

so, H  Z = G  Z …(6)
Using (6) in (5), we get
W = {G  Z : G  T}
 (Z, W)  (X, T)
Hence, (Z, W) is a subspace of (X, T).

Example 4: If T is usual topology on , then find relative topology on   .
Solution: Every open interval on  is T-open set.

æ 1 1 ö
Let G = ç è n - 2 , n + ÷ , n  .
2 ø

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