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Unit 17: The Separation Axioms




          Finally {x}, {y} are closed sets in X.                                                Notes
          Generalising this result.
          {x} is closed    x  X.


                 Example 3: Prove that in a T -space all finite sets are closed.
                                       1
          Solution: Let (X, T) be a T -space.
                              1
          To prove that {x} is closed    x  X.
          Now write (ii) part of the proof of the theorem 1
          Let A be an arbitrary finite subset of X.

          Then A =  {{x}} : x  A}
          = finite union of closed sets = closed set.
              A is a closed set.


                 Example 4: A topological space (X, T) is a T -space iff T contains the cofinite topology
                                                    1
          on X.
          Solution: Let (X, T) be a T -space.
                              1
          To prove that T contains cofinite topology on X, we have to show that T contains subsets A of X
          s.t. X – A is finite.

          Here we shall make use of the fact that
          X is T -space  {x} is closed    x  X
               1
              X – {x} is open subset of X  X – {x}  T

          Thus X – {x}  T  X – (X – {x}) = {x} = finite set.
          This is true    x  X.
          Hence by definition T contains cofinite topology on X.

          Conversely, suppose that T contains cofinite topology on X.
          To prove that (X, T) is T -space.
                             1
          {x} is a finite subset of X.
          Also T contains cofinite topology.

          Consequently X – {x}  T so that
          {x} is closed    x  X
           (X, T) is T -space.
                    1
          Theorem 2: A topological space X is a T -space of X iff every singleton subset {x} of X is closed.
                                          1
          Proof: Let X be a T -space and x  X.
                         1
          By the T -axiom, we know that if y  x  X, than there exists an open set G  which contain y but
                 1                                                   y
          not x i.e.
          y = G   {x} c
               y



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