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Continuity of a complex function defined on the union of an open and closed set

  1. Jan 5, 2013 #1
    1. The problem statement, all variables and given/known data

    (i) Let U and V be open subsets of C with a function f defined on [tex] U \cup V[/tex] suppose that both restrictions, [tex] f_u \mathrm{and} f_v [/tex] are continuous. Show that f is continuous.

    (ii) Illustrate by a specific example that this may not hold if one of the sets U, V is not open.

    2. Relevant equations


    3. The attempt at a solution

    Now I'm pretty sure I can prove the first part. As, since f is continuous on U and V then there exists a [tex] \delta_1 [/tex] and on U there will be a [tex] \delta_2 [/tex] and then taking the minimum of these will put [tex] |f(z) - f(w)| < \epsilon [/tex] for all epsilon. This works because U and V are open, as it ensures that for each w, there will be a delta disc such that z is in either U or V.

    I'm pretty stumped as to an example for the second part. I think I get why it may not hold if U or V is not open, as there will not always be a delta disc around a point in a closed set which is completely contained in that set. But I can't think of a way to use this to arrive at an example.
     
  2. jcsd
  3. Jan 5, 2013 #2

    micromass

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    Hint: complex logarithm
     
  4. Jan 5, 2013 #3
    Hmm, I'm not really sure how to use the logarithm. Is it because if you take the principal logarithm, there will be a discontinuity in the principal argument going from -π to π? So if you take the logarithm defined on [tex] \mathbb{C} \backslash {(-\pi, \pi]} [/tex] then the set isn't open. Or perhaps, take U to be the upper half plane and V to be the lower, and the principal argument will be discontinuous? I'm not sure if this is making sense.

    I'm confusing myself now. Our lecturer taught this module from a theoretical point of view but decides to make the exam this type of problem solving, which he hasn't really prepared us for. Thanks for any help.
     
  5. Jan 5, 2013 #4

    jbunniii

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    Hint: Consider what can happen if [itex]U[/itex] and [itex]V[/itex] are disjoint and [itex]f[/itex] is constant on [itex]U[/itex] and constant on [itex]V[/itex].
     
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