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Unit 1: Topological Spaces




                                                                                                Notes
                 Example 19: Let X = {1, 2, 3, 4, 5} be a non-empty set and T = {, X, {3}, {3, 4}, {2, 3}, {2, 3, 4}}
          is a topology defined on X. Suppose a subset A = {1, 3, 5}  X. The closed set are:
          X, , {1, 2, 4, 5}, {1, 2, 5}, {1, 4, 5}, {1, 5}.

          So, we have  A  = X. Since A is finite and dense in X. So X is a separable space.
          Theorem 10: Show that the cofinite topological space (X, T) is separable.
          Solution: Let (X, T) be a cofinite topological space.

          (i)  When  X is countable.
               Then X  X and  X  =  X
               This shows that X is separable.

          (ii)  Let A  X s.t. A is finite.
               By definition of cofinite topological space A = X – A is open so that A is closed.
                every finite set A is T-closed and so  A  = X.

               Now  A  = X, A is countable.
               This shows that (X, T) is separable.


                 Example 20: A discrete space X is separable iff X is countable.

          Solution: As we know that every subset of a discrete space (X, T) is both open and closed. Also, A
          is said to be everywhere dense in X if  A = X.
          Also, X is separable if  A  X s.t.  A = X and A is countable.

          So, the only everywhere dense subset of X is X itself.
           X can have a countable dense subset iff X is countable.
          Hence, X is separable iff X is countable.

          1.6.2 Limit Point or Accumulation Point or Cluster Point


          Let (X, T) be a topological space and A  X. A point x  X is said to be the limit point or accumulation
          point or cluster point of A if each open set containing ‘x’ contains at least one point of A different
          from x.
          Thus, it is clear from the above definition that the limit point of a set A may or may not be the
          point of A.


                 Example 21: Let X = {a, b, c} with topology

          T = {, {a, b}, {c}, X} and A = {a}, then b is the only limit point of A, because the open sets containing
          b namely {a, b} and X also contains a point of A.
          Whereas, ‘a’ and ‘b’ are not limit point of C = {c}, because the open set {a, b} containing these
          points do not contain any point of C. The point c is also not a limit point of C, since then open set
          {c} containing ‘c’ does not contain any other point of C different from c. Thus, the set C = {c} has
          no limit points.




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