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Unit 6: Closed Sets and Limit Point




          Theorem 1: Let X be a topological space. Then the following conditions hold:          Notes
          (a)   and X are closed.
          (b)  Arbitrary intersections of closed sets are closed.
          (c)  Finite unions of closed sets are closed.
          Proof:

          (a)   and X are closed because they are the complements of the open sets X and , respectively.
                                          A
          (b)  Given a collection of closed sets    ,  we apply De Morgan’s law,
                                             J
                                       X   A    X A  .
                                                      
                                           J   J
               Since the sets  X A     are open by definition, the right side of this equation represents an
               arbitrary union of open sets, and is thus open. Therefore, A  is closed.
                                                                 
          (c)  Similarly, if A  is closed for i = 1, …, n, consider the equation
                          i
                                           n     n
                                        X   A    X A .  
                                              i        i
                                                 
                                           
                                           i 1  i 1
               The set on the right side of this equation is a finite intersection of open sets and is therefore
               open. Hence  A  is closed.
                             i
          Theorem 2: Let Y be a subspace of X. Then a set A is closed in Y if and only if it equals the
          intersection of a closed set of X with Y.
          Proof: Assume that A = C   Y, where C is closed  in X.  Then  X    C  is open in  X,  so that
          (X  C)  Y is open in Y, by definition of the subspace topology. But (X  C)  Y = Y  A. Hence
          Y  A is open in Y, so that A is closed in Y. Conversely, assume that A is closed in Y. Then Y  A
          is open in Y, so that by definition it equals the intersection of an open set U of X with Y. The set
          X  U is closed in X and A = Y  (X  U), so that A equals the intersection of a closed set of X with
          Y, as desired.


                 Example 3: Let (Y, U)   (X, T) and A  Y.
          Then A is U-closed iff A = F  Y for some T closed set F.
          or

          A is U-closed iff A is the intersection of Y and a T-closed F.
          Solution: Let (Y, U)  (X, T) and A  Y, i.e. (Y, U) is subspace of (X, T).
          To prove that A is U-closed iff
          A = F  Y for some T-closed set F.
          A is U-closed  Y  A is U-open.

          Then Y  A can be expressed as:
                           Y  A = G  Y for some T-open set G.
          From which          A = Y  G  Y = X  Y  G  Y
                                = (X  G)  Y
                                = F  Y, where F = X  G is a T-closed set.
          This completes the proof.




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