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Topology

Notes To show: G =  {B : B   and B  G} …(2)
From (1), the statement (2) at once follows.
Conversely, suppose that   T s.t. (2) holds.
Also, suppose that (X, T) is a topological space.

To prove: statement (1).
Let x  X be arbitrary and G be an open set s.t. x  G.
Then x  G  T.
Now (2) suggests that

 B   s.t. x  B  G.
Hence the result (1).

Self Assessment

1. Let X = {a, b, c, d} and A = {{a, b}, {b, c}, {d}}. Determine a base  (generated by A) for a
unique topology T on X.
2. Let  be a base for the topology T on X. Let *  T s.t.   *. Show that * is a base for the
topology T on X.

3. What is necessary and sufficient condition for a family to become a base for a topology?
4. Let B be a base for X and let Y be a subspace of X. Then if we intersect each element of B with
Y, the resulting collection of sets is a base for the subspace Y. Prove it.

2.2 Sub-base

Definition: Let (X, T) be a topological space. Let   T s.t. S  .
 is said to be sub, base or open sub-base or semi bases for the topology T on X if finite
intersections of the members of  form a base for the topology T on X i.e. the unions of the
members of  give all the members of T. The elements of  are referred to as sub-basic open sets.

Example 3: Let a, b   be arbitrary s.t. a < b. Clearly (–, b)  (a, ) = (a, b)
The open intervals (a, b) form a base for the usual topology on . Hence, by definition, the
family of infinite open intervals forms a sub-base for the usual topology on .

Theorem 2: Let  be a non-empty collection of subsets of a non empty set X. Then  is a sub-base
for a unique topology T for X, i.e., finite intersections of members of  form a base for T.
Proof: Let  be the collection of all finite intersections of members of . Then we have to show
that  is a base for a unique topology on X.
For this, we have to show that  satisfies conditions (1) and (2).
(1) Since X is the intersection of empty collection of members of , it follows that

X   and so X = {B : B  }.
(2) Let B , B   and x  B  B . Then B , B are finite intersections of members of . Hence,
1 2 1 2 1 2
B  B is also a finite intersection of members of S and so B  B  .
1 2 1 2
Hence,  is a base for a unique topology on X for which  is sub-base.

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