Exercise 9.  Consider the following sets.   Sketch each set.   State, with reasons, which of the following terms apply to the above sets: open; connected; domain; region; closed region; bounded.  

9 (iii).  [Graphics:Images/ComplexPlaneTopologyModHome_gr_372.gif].

Solution 9 (iii).

See text and/or instructor's solution manual.

    First, observe that

        [Graphics:../Images/ComplexPlaneTopologyModHome_gr_373.gif]

The set  [Graphics:../Images/ComplexPlaneTopologyModHome_gr_374.gif][Graphics:../Images/ComplexPlaneTopologyModHome_gr_375.gif]  is the closed disk of radius 2 centered at the point [Graphics:../Images/ComplexPlaneTopologyModHome_gr_376.gif].

                It is a closed connected region that is bounded.

                It is not open, hence it is not a domain.

[Graphics:../Images/ComplexPlaneTopologyModHome_gr_377.gif]

        

         [Graphics:../Images/ComplexPlaneTopologyModHome_gr_378.gif]

  

        The boundary circle  [Graphics:../Images/ComplexPlaneTopologyModHome_gr_379.gif]  is included in the set  [Graphics:../Images/ComplexPlaneTopologyModHome_gr_380.gif].

We are done.   

For each  [Graphics:../Images/ComplexPlaneTopologyModHome_gr_381.gif]  satisfying  [Graphics:../Images/ComplexPlaneTopologyModHome_gr_382.gif]  there exists an  [Graphics:../Images/ComplexPlaneTopologyModHome_gr_383.gif]  so that  [Graphics:../Images/ComplexPlaneTopologyModHome_gr_384.gif]  (choose  [Graphics:../Images/ComplexPlaneTopologyModHome_gr_385.gif]).  These are the points interior to  S.  

If  [Graphics:../Images/ComplexPlaneTopologyModHome_gr_386.gif]  satisfies  [Graphics:../Images/ComplexPlaneTopologyModHome_gr_387.gif]  then for any  [Graphics:../Images/ComplexPlaneTopologyModHome_gr_388.gif]  the epsilon neighborhood  [Graphics:../Images/ComplexPlaneTopologyModHome_gr_389.gif]  is not entirely contained in  S.  This argument establishes that S is not an open  set.   

Every pair of points in S can be connected with a straight line.  This argument establishes that S is a connected set.

Every point in the boundary of S is also in S.  This argument establishes that S is a closed set.

[Graphics:../Images/ComplexPlaneTopologyModHome_gr_390.gif].  This establishes that S is a bounded set.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(c) 2008 John H. Mathews, Russell W. Howell