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Unit 5: The Subspace Topology




          Let G    be arbitrary, then G  T, by (1).                                           Notes
          We can write G = H  Y for some T-open set H.
          Now, G = H  Y, G  T  H  Y  T

          Again H  Y  T, H  T and Y  X  Y  T
          Conversely, let (Y,  )  (X, T) and let Y  T for any G     G  T
                              G      A  T s.t.  G = A  Y

          Again           A  T, Y  T  A  Y  T  G  T
          Finally, any G      G  T.

          5.1.2 Basis for the Subspace Topology



                 Example 8: Consider the subset Y = [0, 1] of the real line , in the subspace topology. The
          subspace topology has as basis all sets of the form (a, b)  Y, where (a, b) is an open interval in
          , such a set is of one of the following types:
                                ì (a,b)  if a and b are in Y,
                                ï
                                ï [0,b)  if only b is in Y,
                      (a, b)  Y = í
                                ï (a,1]  if only a is in Y,
                                ï Y or   if neither a nor b is in Y.
                                î
          By definition, each of these sets is open in Y. But sets of the second and third types are not open
          in the larger space .
          Note that these sets form a basis for the order topology on Y. Thus, we see that in the case of the
          set Y = [0, 1], its subspace topology (as a subspace of ) and its order topology are the same.


                 Example 9: Let Y be the subset [0, 1)  {2} of . In the subspace topology on Y the one-
                                                                 5
                                                               3
          point set {2} is open, because it is the intersection of the open set  ( , )  with Y. But in the order
                                                               2  2
          topology on Y, the set {2} is not open. Any basis element for the order topology on Y  that
          contains 2 is of the form
                 {x  x  Y and a < x  2}
          for some a  Y; such a set necessarily contains points of Y less than 2.
          Lemma 1: If  is a basis for the topology of X, then the collection   = {B  Y : B  } is a basis for
                                                              Y
          the subspace topology on Y.
          Proof: Given  -open in X and given y     Y, we can choose an element B of  such that
          y  B   . Then y  B  Y     Y
          Now as we know

          “If X is a topological space and  is a collection of open sets of X such that for each open set   of
          X and each x in  , there is an element c of  such that x     . The  is a basis for the topology
          of X.”
          Thus, we can say that   is a basis for the subspace topology on Y.
                             Y





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