# Sierpiński's Theorem/Lemma 1

## 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 $A$..

$C$ 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:

$\displaystyle K = \bigcap \VV' \in \VV$

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 $S$.

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 $A$

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 $\map {\operatorname {Int} } A$.

We have that $\map {\operatorname {Int} } A$ is open in $S$.

Therefore $K$ is open in $S$.

Thus $K$ is clopen in $S$.

We have that $p \in K \subseteq A \subsetneqq S$.

Therefore $S$ is not connected.

From this contradiction it follows that:

$C \cap \partial A \ne \O$

$\blacksquare$