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Finding topologies of sets in complex space

  1. May 12, 2010 #1
    1. The problem statement, all variables and given/known data

    Consider the following subsets of [tex]\mathbb{C}[/tex], whose
    descriptions are given in polar coordinates. (Take [tex]r \geq 0[/tex] in
    this question.)
    [tex]
    \begin{align*}
    X_1 =& \{ (r,\theta) | r = 1 \} \\
    X_2 =& \{ (r,\theta) | r < 1 \} \\
    X_3 =& \{ (r,\theta) | 0 < \theta < \pi, r > 0 \} \\
    X_4 =& \{ (r,\theta) | r = \cos 2\theta \}
    \end{align*}
    [/tex]
    Give each set the usual topology inherited from [tex]\mathcal{C}[/tex].
    Which, if any, of these sets are homeomorphic?

    2. Relevant equations



    3. The attempt at a solution

    [tex]\tau_1 = \varnothing[/tex]. [tex]\tau_2 = \{ B(z,r') \cap X_2 | r'
    > 0 \}[/tex]. [tex]\tau_3 = \{ B(z,r') \cap X_3 | r' > 0 \}[/tex]. [tex]\tau_4 =
    \varnothing[/tex].

    [tex]X_2[/tex] is homeomorphic.

    Are my answers correct? I am not sure if the topologies I wrote make sense at all.
     
  2. jcsd
  3. May 12, 2010 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Do you even understand what the words you are using mean?

    It makes no sense to say that "[itex]X_2[/itex] is homeomorphic". The word "homeomorphic" does not apply to a single set. Two sets are "homeomorphic" is one can be continuously changed into the other. That is, we can shrink, stretch, or otherwise deform a set but we cannot cut, tear, or otherwise break it. In particular, you cannot change the dimension of a set and you cannot change the number of times it intersects itself. While you can change the size of a set with a "continuous" transformation, you cannot change whether it is bounded or not.

    Do you mean [itex]\tau_1, \tau_2, \tau_3, \tau_4[/itex] to be the topologies on each of these sets? If so, [itex]\tau_1[/itex] and [itex]\tau_4[/itex] are certainly incorrect! A topology must include the entire set and the empty set at least so a toplogy is never empty. In any case, it is not necessary to write out the topologies- they are just the usual topology on [itex]R^2[/itex] restricted to the given set.

    [itex]X_1[/itex] is the circle with center at (0, 0) and radius 1. It is a one dimensional closed path.

    [itex]X_2[/itex] is the disk with center at (0, 0) and radius 1. it is a two dimensional, bounded, set.

    [itex]X_3[/itex] is the upper half plane. It is two dimsensional and unbounded.

    [itex]X_4[/itex] is a "four petal rose". It is one dimensional and intersects itself.

    Which, if any, of those sets are homeomorphic to each other?
     
  4. May 13, 2010 #3
    Thank you very much for detailed explanation about homeomorphic sets. I did not know those rules and thought the question is asking which set is homeomorphic to the complex space [tex]\mathbb{C}[/tex].

    None of these sets are homeomorphic to each other. Is it correct?
     
  5. May 13, 2010 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, that is correct.
     
  6. Jun 7, 2010 #5
    I think X2 and X3 are homeomorphic.
     
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