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Topology




                    Notes          Let n be fixed for the moment. Choose a neighbourhood U  of x  that intersects only finitely
                                                                                   n    0
                                   many elements of the collection  . This means that as B ranges over  , all but finitely many of
                                                              n                            n
                                   the functions f  are identically equal to zero on U . Because each function f  is continuous, we
                                              n,B                          n                   n,B
                                   can now choose a neighbourhood V  of x  contained in  U  on which  each of the remaining
                                                                 n   0             n
                                   functions f  for B   , varies by at most /2.
                                           n,B       n
                                                                                                1   
                                   Choose such a neighbourhood V  of x  for each n  Z . Then choose N so that     ,  and define
                                                            n   0           +                   N   2
                                   W = V    ...   V . We assert that W is the desired neighbourhood of x . Let x  W. If n  , then
                                        1      n                                          0
                                                                |f (x) – f (x )|  /2
                                                                 n,B   n,B  0
                                   because the function f  either vanishes identically or varies by at most /2 on W. If n > N, then
                                                    n,B
                                                              |f (x) – f (x )|  Y  < /2
                                                               n,B   n,B  0   n
                                                         1 
                                   because f  maps X into  0,  .  Therefore,
                                          n,B            
                                                        n 
                                                               (F(x), F(x ))  /2 < ,
                                                                       0
                                   as desired.
                                   Step II: Now we prove the converse.
                                   Assume X is metrizable. We know X is regular; let us show that X has a basis that is countably
                                   locally finite.
                                                                                                        1
                                   Choose a metric for X. Given m, let   be the covering of X by all open balls of radius   .  There
                                                                m                                      m
                                   is an open covering   of X refining   such that   is countably locally finite. Note that each
                                                    m             m         m
                                                                 2
                                   element of   has diameter at most   .  Let  be the union of the collections  , for m   .
                                             m                   m                                 m          +
                                   Because each collection   is countably locally finite, so is . We show that  is a basis for X.
                                                       m
                                   Given x   X  and given   > 0, we show that there is  an element  B of   containing x that  is
                                                                      1  
                                   contained in B(x, ). First choose m so that     .  Then, because   covers X, we can choose an
                                                                     m   2              m
                                                                                               2
                                   element B of   that contains x. Since B contains x and has diameter at most    ,  it is contained
                                              m
                                                                                               m
                                   in B(x, ), as desired.
                                   25.2 Summary

                                      A subset A of a space X is called a G  set in X if it equals the intersection of a countable
                                                                     
                                       collection of open subsets of X.
                                      Let X be a regular space with a basis  that is countably locally finite. Then X is normal,
                                       and every closed set in X is a G  set in X.
                                                                
                                      A space X is metrizable iff X is regular and has a basis that is countably locally finite.

                                   25.3 Keywords

                                   Basis: A collection of subsets B of X is called a basis for a topology if:
                                   (1)  The union of the elements of B is X.
                                   (2)  If x  B    B , B , B   B, then there exists a B  of B such that x  B   B    B .
                                             1   2  1  2                   3                3  1   2



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