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 (i).  [Graphics:Images/ComplexPlaneTopologyModHome_gr_335.gif].

Solution 9 (i).

See text and/or instructor's solution manual.

    First, observe that  [Graphics:../Images/ComplexPlaneTopologyModHome_gr_336.gif]   iff   [Graphics:../Images/ComplexPlaneTopologyModHome_gr_337.gif]

The set  [Graphics:../Images/ComplexPlaneTopologyModHome_gr_338.gif]  is the right half plane  [Graphics:../Images/ComplexPlaneTopologyModHome_gr_339.gif].

                It is an open connected domain.  It is also a region.  

                It is not closed and not bounded.

[Graphics:../Images/ComplexPlaneTopologyModHome_gr_340.gif]

        

         [Graphics:../Images/ComplexPlaneTopologyModHome_gr_341.gif]

  

        The boundary line  x=1  is not included in the set  [Graphics:../Images/ComplexPlaneTopologyModHome_gr_342.gif].

We are done.   

For each  [Graphics:../Images/ComplexPlaneTopologyModHome_gr_343.gif]  there exists an  [Graphics:../Images/ComplexPlaneTopologyModHome_gr_344.gif]  so that  [Graphics:../Images/ComplexPlaneTopologyModHome_gr_345.gif]   (choose  [Graphics:../Images/ComplexPlaneTopologyModHome_gr_346.gif]).  This argument establishes that S is 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.

[Graphics:../Images/ComplexPlaneTopologyModHome_gr_347.gif].  This establishes that S is an unbounded set.

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/ComplexPlaneTopologyModHome_gr_348.gif]
[Graphics:../Images/ComplexPlaneTopologyModHome_gr_349.gif]

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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