Page 37 - DMTH503_TOPOLOGY

Unit 2: Basis for Topology

So,  is a basis and generates a topology T; we must show T = T. Notes
Take  T; by hypothesis, there is a set B   with x  B  ; this is the definition of being an
open set in topology T.

Conversely, take V open in topology T. Then by the previous lemma, V equals a union of
elements of sets in .
By hypothesis, each set in B is open in topology T; thus V is a union of open sets from T, so it is
open in T.
Lemma 3: Let  and  be basis for the topologies T and T, respectively, on X. Then the following
are equivalent:
1. T is finer than T.
2. For each x  X and each basis element B   containing x, there is a basis element B  
such that x  B  B.
Proof: (2)  (1)
Given any element of T,
We are to show that  T.

Let x  .
Since  generates T, there is an element B   such that x  B  .
Condition (2) tells us  an element B   such that x  B  B. Then
x  B,  ,

so,  T, by definition
(1)  (2)
Given x  X and B  , with x  B.
Now B belongs to T by definition

and T T by condition (1)
 B  T.
Since T is generated by ,
there is an element B   such that x  B  B.

2.1.2 A Characterisation of a Base for a Topology

Theorem 1: Let (X, T) be a topological space. A sub-collection  of T is a base for T iff every T-open
set can be expressed as union of members of .
or
If T be a topology on X and   T, show that following conditions are equivalent:

(i) Each G  T is the union of members of .
(ii) For any x belonging to an open set G,  B   with x  B  G.
Proof: Let  be a base for the topological space (X, T) so that x  G  T.

  B   s.t. x  B   s.t. x  B  G …(1)

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