\documentclass{article}
\title{Connected Sets}
\author{Matt McCutchen}
\date{April 10, 2005}
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\begin{document}
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\bigskip
Note: In this document, a juxtaposition of two points denotes the distance between them.
\section{Kinds of connectedness}
\textbf{Unsplittable.} Two point sets $T_1$ and $T_2$ \term{split} a point set $S$ if the following conditions all hold:
\begin{itemize}
\item $T_1$ and $T_2$ are disjoint open sets.
\item $T_1$ and $T_2$ both intersect $S$ ($T_1 \cap S \not= \emptyset$ and $T_2 \cap S \not= \emptyset$).
\item $T_1$ and $T_2$ together cover $S$ ($S \subseteq T_1 \cup T_2$).
\end{itemize}
A point set $S$ is \term{unsplittable} if there do not exist sets $T_1$ and $T_2$ that split it. According to several sources, when mathematicians say ``connected'', they usually mean this definition.
\bigskip\noindent
\textbf{Arcwise connected.} A point set $S$ is \term{arcwise connected} if, for all $a, b \in S$, there exists a continuous curve lying entirely in $S$ that connects $a$ and $b$. That is, there is a continuous function $f : [0,1] \to S$ with $f(0) = a$ and $f(1) = b$.
\bigskip\noindent
\textbf{Capacitor-connected.} A point set $S$ is \term{capacitor-connected} if, for all $a, b \in S$ and all positive real numbers $\delta$, there exists a sequence of points $a_0, \ldots, a_n \in S$ with $a_0 = a$, $a_n = b$, and $a_i a_{i+1} < \delta$ for $i = 0, \ldots, n-1$.
\section{All unsplittable sets are capacitor-connected}
Let $S$ be an unsplittable point set, let $a \in S$, and let $\delta$ be a positive real number. Generate a sequence of point sets $A_0, A_1, \ldots$ as follows: $A_0$ = $\{a\}$, and for all $i \ge 0$,
\[A_{i+1} = S \cap \bigcup_{x \in A_i} Ball_\delta(x).\]
Let $A^* = \bigcup_{i \ge 0} A_i$.
Now, let
\[U = \bigcup_{x \in A^*} Ball_{\delta/2}(x)\quad\textrm{and}\quad V = \bigcup_{x \in S \setminus A^*} Ball_{\delta/2}(x).\]
Some observations about the sets $U$ and $V$:
\begin{itemize}
\item $U$ and $V$ are both open because they are both unions of open balls.
\item Together they cover $S$ because $A^* \subseteq U$ and $S \setminus A^* \subseteq V$.
\item They are disjoint. To prove this, suppose there were a point $x$ in both sets. By the definition of $U$, there must be a point $x_U \in A^*$ with $xx_U < \delta/2$, and since $x_U \in A^*$, $x_U \in A_n$ for some $n$. By the definition of $V$, there must also be a point $x_V \in S \setminus A^*$ with $xx_V < \delta/2$. But
\[x_Ux_V \le xx_U + xx_V < \delta/2 + \delta/2 = \delta,\]
so $x_V \in Ball_\delta(x_U) \subseteq A_{n+1}$ by the definition of the $A_i$. But then $x_V \in A^*$, contradicting the fact that $x_V \in S \setminus A^*$.
\item $U \cap S$ is nonempty (it contains $a$).
\end{itemize}
If $V \cap S$ were also nonempty, $U$ and $V$ would split $S$, but we know $S$ is unsplittable, so $V$ doesn't intersect $S$. This means $S \setminus A^* = \emptyset$, because if there existed $x \in S \setminus A^*$, we'd have $x \in V \cap S$. Thus $A^* = S$.
By the definition of $A^*$, any point $b \in S$ must be in $A_n$ for some $n$. Generate a sequence of points $b_n, \ldots, b_0$ as follows: $b_n = b$, and $b_{i-1} \in A_{i-1}$ with $b_i b_{i-1} < \delta$ for $i = n, \ldots, 1$. A $b_i$ must exist at every step because of the way the $A_i$ were constructed, and $b_0 = a$ because $a$ is the only point in $A_0$.
The points $a\!=\!b_0, \ b_1, \ldots, \ b_n\!=\!b$ form a sequence of hops in $S$ from $a$ to $b$ of lengths less than $\delta$. The foregoing argument can be used to create such a sequence for all $a, b \in S$ and all positive real numbers $\delta$, so $S$ is capacitor-connected. $\blacksquare$
\section{All compact, capacitor-connected sets are unsplittable}
Let $S$ be a compact, capacitor-connected set split by sets $T_1$ and $T_2$; we'll prove the section title by reaching a contradiction.
\bigskip
\textbf{Lemma.} \textsl{Let $O$ and $C$ be open and compact point sets, respectively. If $O \supseteq C$, there exists a positive real number $\delta$ such that for all points $x \in C$ and $y \notin O$, $xy \ge \delta$.}
\medskip
\textsl{Proof:} For $i = 0, 1, \ldots$, let $O_i$ be the set of all points $x$ such that $Ball_{2^{-i}}(x) \subseteq O$. Clearly $O_i \subseteq O_j$ when $i \le j$. If $x$ is any point in $O$, since $O$ is open, there exists a positive real number $r$ such that $Ball_r(x) \subseteq O$, so $x \in O_i$ where $i = max(0, \lceil -log_2 r \rceil)$; thus $\bigcup_i O_i = O$.
The $O_i$ form an infinite set of open sets that together cover $O$ and therefore $C$, so by the Heine-Borel Theorem, there also exists a finite set $U$ of $O_i$ whose members together cover $C$. Let $n$ be the highest index of any of the sets $O_i$ in $U$. Since $O_n \supseteq O_i$ for all $i \le n$, $O_n = \bigcup U$ and therefore $O_n \supseteq C$.
Let $\delta = 2^{-n}$. Let $x$ be any point in $C$. Since $C \subseteq O_n$, the definition of $O_n$ tells us that $Ball_{2^{-n}}(x) \subseteq O$. Let $y$ be any point not in $O$. Since $Ball_{2^{-n}}(x) \subseteq O$, $y \notin Ball_{2^{-n}}(x)$, so $xy \ge 2^{-n} = \delta$. $\Box$
\bigskip
Applying Lemma 1 with $O \mapsto T_1 \cup T_2$ and $C \mapsto S$, we find that any point $x$ in $S$ is always at least a fixed distance $\delta$ from any point $y$ not in either $T_i$. I claim now that any point $x$ in $S \cap T_1$ is further than $\delta$ from any point $y \notin T_1$; it is enough to prove this for $y \in T_2$.
Let $f : xy \to \mathbb{R}^n$ parametrize the line segment connecting $x$ and $y$ in the obvious fashion with $f(0) = x$ and $f(xy) = y$. Clearly $f(t_1)f(t_2) = |t_1 - t_2|$ for all $t_1, t_2 \in [0,xy]$. Let $P$ be the set of all $t \in [0,xy]$ for which $f(t) \in T_1$.
$P$ is nonempty (it contains $0$) and has an upper bound ($xy$), so by the completeness axiom, it has a supremum $s$. That is, $t \notin P$ for all $t > s$, and there exists no $s' < s$ for which $t \notin P$ for all $t > s'$. The second part implies that, for all $s' < s$, there exists $t \in (s',s]$ with $t \in P$. I now claim that $f(s)$ is a point in neither $T_1$ nor $T_2$.
If $f(s)$ were in $T_1$, then, since $T_1$ is open, there would exist a positive real number $\epsilon$ such that $Ball_\epsilon(f(s)) \subseteq T_1$. In particular, $f(s+\epsilon/2)$ would be in $T_1$, so $s + \epsilon/2 \in P$, contradicting the last paragraph. If $f(s)$ were in $T_2$, there would again exist a positive real number $\epsilon$ such that $Ball_\epsilon(f(s)) \subseteq T_2$. Setting $s' = s - \epsilon$, we find that there is a number $t \in (s',s]$ with $t \in P$, which means $f(t) \in T_1$. But $f(s)f(t) < \epsilon$, so $f(t) \in T_2$, contradicting the fact that $T_1$ and $T_2$ are disjoint.
We know that $x \in S$ and $f(s) \in \mathbb{R}^n \setminus (T_1 \cup T_2)$, so Lemma 1 tells us that $xf(s) \ge \delta$. $f(s)$ is strictly between $x$ and $y$ on a line segment, so $xy > xf(s) \ge \delta$, proving the claim that $xy \ge \delta$ for all $x \in S \cap T_1$ and $y \notin T_1$.
Since $T_1$ and $T_2$ split $S$, there exist points $a \in S \cap T_1$ and $b \in S \cap T_2$. Since $S$ is capacitor-connected, there must exist points $a_0, \ldots, a_k \in S$ with $a_0 = a$, $a_k = b$, and $a_i a_{i+1} < \delta$ for $i = 0, \ldots, n-1$. Let $j$ be the smallest index for which $a_j \in T_1$ but $a_{j+1} \notin T_1$; some such $j$ must exist because $a_0 \in T_1$ and $a_k \notin T_1$. Applying the claim, we find that $a_j a_{j+1} > \delta$, contradicting the validity of the sequence of hops $a_i$. $\blacksquare$
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