Page 63 - DMTH503_TOPOLOGY

Unit 5: The Subspace Topology

Then G  T. Now = {G   : G  T} Notes

æ 1 1 ö
If G = ç n - , n + ÷
è 2 2 ø

æ 1 1 ö
then G  = n - 2 , n + ÷  
ç
2 ø
è
= {n}
Or = {{n} : n  }
Every singleton set of  is -open set.
As an arbitrary subset of  is an arbitrary union of singleton sets and so every subset of  is
-open.
Consequently,  is a discrete topology on .

Example 5: Define relative topology. Consider the topology : T = {, {a}, {b, c}, {a, b, c}, X}
on X = {a, b, c, d}. If Y = {b, c, d} is a subset of X, then find relative topology on Y.
Solution: If  is relative topology on , then

 = {G  Y, G  T}
  = {  Y, {a}  Y, {b, c}  Y, {a, b, c}  Y, X  Y}
  = { ,, {b, c}, {b, c}, Y}
  = {, Y, {b, c}}

Example 6: Let X be a topological space and let Y and Z be subspaces of X such that
Y  Z. Show that the topology which Y has a subspace of X is the same as that which it has as a
subspace of Z.

Solution: Let (X, T) be a topological space and Y, Z be subspaces of X such that
Y  Z  X.
Further assume (Y, T )  (Z, T )  (X, T) …(1)
1 2
(Y, T )  (X, T) …(2)
3
We are to show that T = T
1 3
By definition (1) declares that
T = {G  Y : G  T } …(3)
1 2
T = {H  Z : H  T} …(4)
2
T = {P  Z : P  T} …(5)
3
Using (4) in (3), we get
G  Y = (H  Z)  Y = H  (Y  Z) = H  Y
Now, (3) becomes
T = {H  Y : H  T} …(6)
1
From (5) and (6), we get T = T .
1 3

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