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Homework Help: Topology : 3 sets on the Real line with the wada property.

  1. Feb 22, 2010 #1
    1. The problem statement, all variables and given/known data
    Find three disjoint open sets in the real line that have the same nonempty
    boundary.


    2. Relevant equations
    Connectedness on open intervals of [tex]\mathbb{R}[/tex].

    3. The attempt at a solution

    If this is it all possible, the closest thing I could come up with to 3 disjoint open intervals of [tex]\mathbb{R}[/tex]. with a shared boundary is
    [tex]A_{n}=\left\{3n\right\}\cup (3n+1,3n+2) , B_{n}=(3n,3n+1)\cup \left\{3n+2\right\}, C_{n}=\left\{3n+1\right\}\cup (3n+2,3n+3)[/tex] [tex]\forall n \in \mathbb{Z} [/tex]

    But I'm fairly certain these are clopen sets, since the compliment is of the form [tex]\left(a,b\right][/tex]

    I went with a proof by contradiction to say that it's not possible. I'd really appreciate any help.



    [tex]Proof:[/tex] Suppose [tex]\exists U,V,W \in \mathbb{R}[/tex] [tex] _{.}\ni_{.}
    \partial U \cap \partial V = \partial V \cap \partial W = \partial W \cap \partial U = \emptyset ,
    \partial U=\partial W=\partial V =D [/tex] And [tex]U,V,W[/tex] are open on [tex]\mathbb{R}[/tex].

    This means that [tex]U=\bigcup _{i \in I} X_{i},V=\bigcup _{j \in J} Y_{j},W=\bigcup_{k\in K} Z_{k} [/tex] ,where [tex] I,J,K [/tex] are indexing sets.

    Since [tex] X_{i},Y_{j},Z_{k} [/tex] are open sets on [tex] \mathbb{R} [/tex] with the usual topology, we can express each [tex] X_{i},Y_{j},Z_{k} [/tex]

    as [tex] B_{r}(p)=\left\{q \in \mathbb{R}:\rho (p,q)=\left\|p-q\right\|< r \right\} [/tex] where each open ball is a connected subspace of [tex]\mathbb{R}[/tex]


    Now [tex]\sphericalangle d_{0}\in D [/tex] ,

    [tex]\exists ! (i_{0},j_{0},k_{0}) \in I\times J\times K .\ni. d_{0}\in \partial X_{i_{0}} \cap
    \partial Y_{j_{0}} \cap \partial Z_{k_{0}}[/tex] and [tex]X_{i_{0}},Y_{k_{0}},Z_{j_{0}} [/tex] are pairwise disjoint connected subspaces of [tex] \mathbb{R}[/tex] .

    [tex] \because B_{\epsilon}(d_{0}) \cap X_{i_{0}}\neq \emptyset ,B_{\epsilon}(d_{0}) \cap Y_{j_{0}}\neq \emptyset , B_{\epsilon}(d_{0}) \cap Z_{k_{0}}\neq \emptyset ,[/tex] and [tex] X_{i_{0}} \cap Y_{j_{0}} \cap Z_{k_{0}}=\emptyset \Leftrightarrow
    B_{\epsilon}(d_{0}) \cap X_{i_{0}} \bigcap B_{\epsilon}(d_{0}) \cap Y_{j_{0}} \bigcap B_{\epsilon}(d_{0}) \cap Z_{k_{0}}= \emptyset [/tex]

    Without Loss of generality , assume that [tex] \inf X_{i_{0}}=d_{0}, [/tex] then [tex] B_{\epsilon}(d_{0}) \bigcap X_{i_{0}}= \left\{x \in X_{i_{0}}: x>d_{0} \wedge \rho (x,d_{0}) < \epsilon \right\} [/tex]

    This means that [tex] Y_{j_{0}},Z_{k_{0}}\subset B_{\epsilon}(d_{0}) \bigcap (B_{\epsilon}(d_{0}) \bigcap X_{i_{0}})^{\boldsymbol{c}} [/tex].

    Since Both remaining sets are connected subspaces of [tex] \mathbb{R}, [/tex] by trichotomy of [tex] \mathbb{R} [/tex] , every point on [tex]Z_{k_{0}}[/tex] will be greater than every point on [tex] Y_{j_{0}} [/tex] or every point on [tex]Z_{k_{0}}[/tex] will be less than every point on [tex] Y_{j_{0}} [/tex].

    Choose [tex]\sup (Y_{j_{0}})\leq \inf (Z_{k_{0}})<\sup (Z_{k_{0}})=d_{0} [/tex]

    Thus [tex]d_{0}\notin Y_{j_{0}} [/tex]
    Hence a contradiction, there is no way to construct three open disjoint sets with the same boundary on [tex]\mathbb{R} [/tex]
     
  2. jcsd
  3. Feb 22, 2010 #2
    I just really need to know if I am in the right direction here, I'm not the best at topology.
     
  4. Feb 23, 2010 #3

    Dick

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    Your proof is pretty hard to read. Besides, I don't think it's correct. Think about what the complement of your three open sets would have to be like. It would have to be closed. It would have to have no isolated points or connected closed intervals. Starting to remind you of someone? Sounds like the Cantor set to me.
     
  5. Feb 23, 2010 #4
    Well the cantor set is closed on the interval it's defined, and if you pick any subset of the cantor set, you end up with a covering for a subset , and how can you determine the boundary would be the same??
     
  6. Feb 23, 2010 #5
    I don't see how the cantor set could be useful, if you could pinpoint an error perhaps in my proof..
     
  7. Feb 23, 2010 #6

    Dick

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    The complement of the Cantor is a countable union of open intervals. You just have to split them into three groups so every neighborhood of a point in the Cantor set contains points from each of the three groups.
     
  8. Feb 23, 2010 #7
    Hmm. I'm gonna see what I can do with it then, do you know of there are morphisms between sets with the wada property on various Euclidean spaces?
     
  9. Feb 23, 2010 #8

    Dick

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    Well, you haven't even really said what the "wada" property is. I think that's a pretty vague question.
     
  10. Feb 23, 2010 #9
    The wada property is if for a given space , you can find 3 disjoint open sets such that they have the same boundary.. I apologize I thought it was a relatively well known property in topology ( ie the case of the wada basin for R2)
     
  11. Feb 23, 2010 #10

    Dick

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    It's new to me. Thanks for the definition. I still think you can make an example by grouping intervals in the complement of the Cantor set.
     
  12. Feb 23, 2010 #11
    But the compliment will have an isolated point, since the only way an open set will have a specific point on it's boundary is if the point is on it's compliment and is contained on the closure . So the compliments of the sets would be an isolated point..
     
  13. Feb 23, 2010 #12

    Dick

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    The boundary of ONE open interval is two isolated points. The boundary of a A LOT of open intervals is A LOT of points which may not be isolated. The Cantor set doesn't have any isolated points.
     
    Last edited: Feb 23, 2010
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