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Unit 16: The Countability Axioms




          16.1.5 Theorems Related to Metric Spaces                                              Notes

          Theorem 9: A metric space is second countable iff it is separable.
          Proof:
          (i)  Let (x, ) be a metric space. Let T be the metric topology on X corresponding to the metric
               . Let (X, T) be second countable. To prove that X  is separable.

               Here write the complete proof of the theorem (6).
          (ii)  Conversely, suppose  that (x,  ) is  a metric  space  and  T  is  a  metric  topology  on  X
               corresponding to the metric . Also, suppose that X is separable, so that

                 A   X s.t. A   X and A is countable.

               A is countable  A is expressible as
               A = {a  : n N}
                    n
               To prove that X is second countable.
               We know that each open sphere forms an open set.
               Let a  A be arbitrary.
                   n
                                  +
               Write  = {S  (a ) : r Q , n  N}.
                         r  n
               Q is an enumerable set
                +
              Q  is an enumerable set
              Q  Q
                +
               Then  is a countable base for the topology T on X.
              X is second countable.

               Let G  T be arbitrary s.t. x G.
               x being an arbitrary point of X.
               By definition of open set in a metric space,
                a positive real number  s.t. S   G                            ...(1)
                                         (x, )
               Since A  is dense in X and so there will exist a point a A s.t.

                                
                          (a,x)                                                  ...(2)
                                3
               Since Q is dense in R for the usual topology on R and hence its subset Q  is also dense in R
                                                                        +
                                           +
               with usual topology so that r  Q  s.t.
                              2 
                             r 
                         3     3
               Aim:      S r(a)    S  (x)    G.
               Also let  y S  (a,r)   be arbitrary so that  (y,a) r             ...(3)

                                  (x,y)   (x,a)   (a,y)
                                                 2 
                                        <     r     ,                        from (3)
                                           3     3  3
                 (x,y)    y S  (x, )
                                                



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