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Why is this set Open and Connected?

  1. Feb 2, 2012 #1
    1. The problem statement, all variables and given/known data
    This isn't a homework problem but just a fundamental concept that my entire complex analysis course is based on.

    The set D = ℂ \ non-positive real axis. Why is it considered open and connected (domain)?


    2. Relevant equations



    3. The attempt at a solution
    When it says non-positive real axis, is the origin included? I can see why D is connected. But why is it considered open? The set is open when it contains all of its boundaries. But the set D clearly does not contain any point on the non-positive real axis.
     
  2. jcsd
  3. Feb 2, 2012 #2

    HallsofIvy

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    Yes, 0 is not a positive number so it is part the "non-positive real axis".

    Your error is the statement "The set is open when it contains all of its boundaries." That is exactly backwards. A set is open if and only if it contains none of its boundary. A set is closed if and only if it contains all of its boundary.
     
  4. Feb 2, 2012 #3
    An open set does not contain all of it's boundaries. Moreover the boundary of a set is defined as it's closure without it's interior. An open set is equal to it's interior since the interior is defined as the 'biggest' open set that is a subset of the set it is the interior of. (i.e. it's the union of all open sets in the set it is the interior of).

    all you really have to ask yourself in this case since you're probably using the Euclidian topology on the complex plane is: Does every point in the subset have sphere (disk) around it that is also in the subset. Which is in this case is clearly true (if you are not on the non-positive real axis you can allways get closer to it, which is why actually zero is also taken oout of the plane).
     
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