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Complex analysis - open sets

  1. Oct 4, 2009 #1
    1. The problem statement, all variables and given/known data
    An open set in the complex plane is, by definition, one which contains a disc of positive radius about each of its points. Prove that:
    (a) the intersection of two open sets is an open set
    (b) the union of arbitrarily many open sets is an open set


    2. Relevant equations
    If A, B are two sets, A = {a, b, c} and B = {c, d, e}, then:
    AUB = {a, b, c, d, e}
    A intersect B = {b}



    3. The attempt at a solution
    Here's what I've done:

    (a) Let U, V be two non-empty open sets in the complex plane. Then, by definition, all the points of U and V have a positive radius about them.
    Then, clearly, all the points in the intersection of U and V will be points that have a positive radius about them.
    Thus the intersection of U and V is, by definition, an open set.

    (b) Let U_1, U_2, ... , U_n be any n non-empty open sets in the complex plane. Then, by definition, all the points of U_1, U_2, ... , U_n have a positive radius about them.
    Thus, clearly, all the points in the union of U_1, U_2, ... , U_n have a positive radius about them.
    Thus the union of U_1, U_2, ... , U_n is, by definition, an open set.

    ---

    I know heuristic proofs don't generally cut it in analysis, but I really don't see what else there is to say on this particular question. Am I missing something fundamental in my argument?
    Thanks for any help
     
  2. jcsd
  3. Oct 4, 2009 #2

    jbunniii

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    I think you need to be a lot more explicit about the positive radius.

    Here's how I would start (a):

    Let [itex]U[/itex] and [itex]V[/itex] be two open sets in [itex]\mathbb{C}[/itex]. If [itex]U \cap V[/itex] is empty, then the claim is obviously true. So suppose that it is not empty, and choose [itex]z \in U \cap V[/itex].

    Then [itex]z \in U[/itex], so there exists a radius [itex]r_u > 0[/itex] such that [itex]u \in U[/itex] whenever [itex]|z - u| < r_u[/itex].

    Similarly, [itex]z \in V[/itex], so there exists a radius [itex]r_v > 0[/itex] such that [itex]v \in V[/itex] whenever [itex]|z - v| < r_v[/itex].

    I now wish to show that [itex]U \cap V[/itex] contains a disc of positive radius around the point [itex]z[/itex], which means that I must find a radius [itex]r > 0[/itex] such that [itex]x \in U \cap V[/itex] whenever [itex]|z - x| < r[/itex].

    Your job is now to find such an [itex]r[/itex]. Hint: will one of [itex]r_u[/itex] or [itex]r_v[/itex] work? If so, which one? And you must prove that it works!
     
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