Page 10 - DMTH503_TOPOLOGY

Topology

Notes Cofinite Topology

Let X be a non-empty set, and let T be a collection of subsets of X whose complements are finite
along with , forms a topology on X and is called cofinite topology.

Example 5: Let X = {l, m, n} with topology
T = {, {l}, {m}, {n}, {l, m}, {m, n}, {l, n}, X}
is a cofinite topology since the compliments of all the subsets of X are finite.

Note If X is finite, then topology T is discrete.

Theorem 1: Let X be an infinite set and T be the collection of subsets of X consisting of empty set
 and all those whose complements are finite. Show that T is a topology on X.
Proof:

(i) Since X = , which is finite, so X  T.
Also   T (by definition of T)
(ii) Let G , G , T
1 2
 G , G are finite
1 2
 G  G is finite
1 2
 (G  G ) is finite (by De-Morgan’s law (G  G = (G  G ))
1 2 1 2 1 2
 G  G  T
1 2
(iii) If {G :   } is an arbitrary collection of sets in T, then

G is finite    

  {G :   } is finite

 [ {G :   }] is finite (by De-Morgan’s law)

  {G :   } T

Hence T is a topology for X.

Co-countable Topology

Let X be a non-empty set. Let T be the collection of subsets of X whose complements are countable
along with , forms a topology on X and is called co-countable topology.
Theorem 2: Let X be a non-empty set. Let T be the collection of all subsets of X, whose complements
are countable together with empty set . Show that T is a topology on X.
Proof:
(i) Since X = , which is countable
so, X  T

Also, by definition,  T

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