Sierpiński's Theorem/Lemma 1
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Theorem
Let $\struct {S, \tau}$ be a compact connected Hausdorff space.
Let $A$ be a closed, non-empty proper subset of $S$.
Let $C$ be a component of $A$.
Then:
- $C \cap \partial A \ne \O$
where $\partial A$ denotes the boundary of $A$.
Proof
Let $p \in C$.
Let $\VV$ be the set of all subsets of $A$ containing $p$ that are clopen relative to $A$..
By Quasicomponents and Components are Equal in Compact Hausdorff Space and Quasicomponent is Intersection of Clopen Sets:
- $C$ is the intersection of $\VV$.
Aiming for a contradiction, suppose:
- $C \cap \partial A = \O$
By Boundary of Set is Closed, $K \cap \partial A$ is closed for each $K \in \VV$.
Thus by Compact Space satisfies Finite Intersection Axiom, there exists a finite set $\VV' \subseteq \VV$ such that $\partial A \cap \bigcap \VV' = \O$.
But then:
- $\ds K = \bigcap \VV' \in \VV$
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Therefore there exists a $K \in \VV$ such that $K \cap \partial A = \O$.
Since $A$ is closed in $S$, and $K$ is clopen in $A$, $K$ is closed in $S$.
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We have that:
- $\partial A = A^- \setminus \map {\operatorname {Int} } A$
where $A^-$ is the closure of $A$ and $\map {\operatorname {Int} } A$ is the interior of $A$
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Hence as $K \subseteq A$, it follows that $K \subseteq \map {\operatorname {Int} } A$.
Since $K$ is open relative to $A$, it is open relative to $\map {\operatorname {Int} } A$.
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We have that $\map {\operatorname {Int} } A$ is open in $S$.
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Therefore $K$ is open in $S$.
Thus $K$ is clopen in $S$.
We have that $p \in K \subseteq A \subsetneqq S$.
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Therefore $S$ is not connected.
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From this contradiction it follows that:
- $C \cap \partial A \ne \O$
$\blacksquare$