Page 115 - DMTH503_TOPOLOGY
P. 115

Unit 11: Connected Spaces, Connected Subspaces of Real Line




          Now from (1) and (2), we get                                                          Notes
          H = ,
          which contradicts the given fact that H is non-empty.
          Hence  A  is also a connected set.

          Theorem 3: If every two points of a set E are contained in same connected subset of E, then E is
          connected.
          Proof: Let us suppose that E is not connected.
          Then, it must a separation E = A|B
          i.e. E is the union of non-empty separated sets A and B.

          Since A and B are non-empty, let a  A and b  B.
          Then, A and B being disjoint
           a, b are two distinct points of E.
          So, by given hypothesis there exists a connected subset C of E such that a, b  c
          But, C being a connected subset of a disconnected set E with the separation E = A|B,
          we have C  A or C  B.
          This is not possible, since A and B are disjoint and C contains at least one point of A and one that
          of B, which leads to a contradiction.
          Hence E is connected.
          Theorem 4: A topological space (X, T) is connected iff the only non-empty subset of X which is
          open and closed is X itself.
          Proof: Let (X, T) be a connected space.
                                                                  c
          Let A be a non-empty subset of X that is both open and closed. Then A  is both open and closed.
                          C
              A  = A and  A  = A C
                   C
          Thus A   A  = 
                               C
                  C
            A    A  =  and A    A  = 
          Also X = A   A C
                         C
          Therefore A and A  are two separated sets whose union in X.
                          C
                                                         C
          Now if A   and A   , then we have separation X = A|A , which leads to the contradiction as
          X is connected.
          So either A =  or A  = 
                          C
                      C
          But A =  or A  = 
          But A  
          So A  = 
              C
              X = A   A  = A    = A
                       C
          This shows that the only non-empty subset of X that is both open and closed is X itself. Conversely,
          let the only subset of X which is both open and closed be X itself.
          Then, there exists no non-empty proper subset of X which is both open and closed.
          Hence (X, T) is not disconnected and therefore, it is connected.



                                           LOVELY PROFESSIONAL UNIVERSITY                                   109
   110   111   112   113   114   115   116   117   118   119   120