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tt2348
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Homework Statement
Find three disjoint open sets in the real line that have the same nonempty
boundary.
Homework Equations
Connectedness on open intervals of \mathbb{R}.
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 \mathbb{R}. with a shared boundary is
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) \forall n \in \mathbb{Z}
But I'm fairly certain these are clopen sets, since the compliment is of the form \left(a,b\right]
I went with a proof by contradiction to say that it's not possible. I'd really appreciate any help.
Proof: Suppose \exists U,V,W \in \mathbb{R} _{.}\ni_{.}<br /> \partial U \cap \partial V = \partial V \cap \partial W = \partial W \cap \partial U = \emptyset ,<br /> \partial U=\partial W=\partial V =D And U,V,W are open on \mathbb{R}.
This means that U=\bigcup _{i \in I} X_{i},V=\bigcup _{j \in J} Y_{j},W=\bigcup_{k\in K} Z_{k} ,where I,J,K are indexing sets.
Since X_{i},Y_{j},Z_{k} are open sets on \mathbb{R} with the usual topology, we can express each X_{i},Y_{j},Z_{k}
as B_{r}(p)=\left\{q \in \mathbb{R}:\rho (p,q)=\left\|p-q\right\|< r \right\} where each open ball is a connected subspace of \mathbb{R}
Now \sphericalangle d_{0}\in D ,
\exists ! (i_{0},j_{0},k_{0}) \in I\times J\times K .\ni. d_{0}\in \partial X_{i_{0}} \cap<br /> \partial Y_{j_{0}} \cap \partial Z_{k_{0}} and X_{i_{0}},Y_{k_{0}},Z_{j_{0}} are pairwise disjoint connected subspaces of \mathbb{R} .
\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 , and X_{i_{0}} \cap Y_{j_{0}} \cap Z_{k_{0}}=\emptyset \Leftrightarrow<br /> 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
Without Loss of generality , assume that \inf X_{i_{0}}=d_{0}, then B_{\epsilon}(d_{0}) \bigcap X_{i_{0}}= \left\{x \in X_{i_{0}}: x>d_{0} \wedge \rho (x,d_{0}) < \epsilon \right\}
This means that Y_{j_{0}},Z_{k_{0}}\subset B_{\epsilon}(d_{0}) \bigcap (B_{\epsilon}(d_{0}) \bigcap X_{i_{0}})^{\boldsymbol{c}}.
Since Both remaining sets are connected subspaces of \mathbb{R}, by trichotomy of \mathbb{R} , every point on Z_{k_{0}} will be greater than every point on Y_{j_{0}} or every point on Z_{k_{0}} will be less than every point on Y_{j_{0}}.
Choose \sup (Y_{j_{0}})\leq \inf (Z_{k_{0}})<\sup (Z_{k_{0}})=d_{0}
Thus d_{0}\notin Y_{j_{0}}
Hence a contradiction, there is no way to construct three open disjoint sets with the same boundary on \mathbb{R}
