(adsbygoogle = window.adsbygoogle || []).push({}); 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]

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

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