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So I know that the definition of arcwise connected is that there is a continuous function $f:[0,1] -> S$ s.t. $f(0) = p$ and $f(1) = q$ when $p,q \in S$, $S$ = any connected open subset of $E^n$... can anyone push me in the right direction?

user20391
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2 Answers2

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Pick $p \in S$. Let $C_p \subset S$ be given by $C_p = \{ q \in S | \text{there is a path from }p\text{ to }q\text{ in }S \}$.

Since $S$ is open, $C_p$ is open.

Suppose $q \in S \setminus C_p$. Since $S$ is open, we have some $B(q,\epsilon) \subset S$ (with $\epsilon>0$), hence $B(q,\epsilon) \cap C_p = \emptyset$ (otherwise we get a quick contradiction). Hence $S\setminus C_p$ is open, which contradicts the connectedness of $S$. Hence $C_p =S$.

The following construction shows that we can choose the connecting path to be polygonal.

Choose $p,q \in S$, then there exists a path from $p$ to $q$, call it $\gamma$.

The set $\Gamma=\gamma([0,1])$ is compact. Since $S$ is open, for each $s \in \Gamma$, there is some $\epsilon_s>0$ such that $B(s,\epsilon_s) \subset S$, and this collection is an open cover of $\Gamma$, hence has a finite subcover.

Call the elements of this subcover $B_1,...,B_n$, where $p \in B_1$ and $q \in B_n$, and let $x_k$ be the centre of $B_k$.

Let $i_1 = 1$ and define $t_2 = \sup \{ t \in [0,1] | \gamma(\tau) \in B_{i_1} \forall \tau \in [0,t] \}$. If $t_2 = 1$ we are finished, otherwise $\gamma(t_2) \not\in B_{i_1}$, and there is some $i_2$ such that $\gamma(t_2) \in B_{i_2}$.

Now suppose we have $0=t_1 < \cdots < t_k < 1$ and $B_{i_1}, ..., B_{i_k}$ such that $\gamma(t)\in B_{i_1}\cup ...\cup B_{i_k}$ for $t \in [t_1,t_k]$. Let $t_{k+1} = \sup \{ t \in [0,1] | \gamma(\tau) \in B_{i_1}\cup ...\cup B_{i_k} \forall \tau \in [0,t] \}$. As above, if $t_{k+1} = 1$ we are finished, otherwise $\gamma(t_{k+1}) \not\in B_{i_1}\cup ...\cup B_{i_k}$, and there is some $i_{k+1}$ such that $\gamma(t_{k+1}) \in B_{i_{k+1}}$.

It is clear that this construction must terminate with some $t_{m} = 1$. Then form the polygonal path $p \to x_{k_1} \to \cdots \to x_{k_m} \to q $.

copper.hat
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  • This may be a silly question, but why is $C_p$ open when $S$ is open? Thanks so much! – user20391 Feb 16 '15 at 09:21
  • Suppose $x \in C_p$, then there is some $\epsilon>0$ such that $B(x,\epsilon) \subset S$ as well. Hence a straight line from $x$ to any point $y \in B(x,\epsilon)$ shows that the path $p \to x \to y$ connects $p$ to $y$. Hence we have $B(x,\epsilon) \subset C_p$, so it is open. – copper.hat Feb 16 '15 at 09:23
  • A similar argument shows that if $S \subset C_p$ is non empty, then it must be open. – copper.hat Feb 16 '15 at 09:25
  • I get the first reasoning but could you elaborate further on the second? Thanks! – user20391 Feb 16 '15 at 10:35
  • Suppose $q \in S \setminus C_p$. There is some $r>0$ such that $U=B(q,r) \subset S$. If $U \cap C_p \neq \emptyset$ then there is a path from $p$ to $q$, which is a contradiction. – copper.hat Feb 16 '15 at 15:33
  • Nice job! I always had thought about the proof explicitly invoking compactness, but only recently did I try to write it out and realise that the order of the $x_i$ (as well as which ones to use; I think but am not sure that using all $n$ is in general untenable) takes some care -- I had thought that simply taking them in the same order that $\gamma$ traverses them (i.e., arranging them in order of increasing $\inf\gamma^{-1}(x_k)=\min\gamma^{-1}(x_k)$) would do until I noticed a counter-example. I guess the proof really does have to be that technical (more than I would've liked) to be correct. – Vandermonde Jul 25 '15 at 03:25
  • My favorite/second-cleanest rendition is to consider the (minimum) distance $\varepsilon$ between (a point of) $\Gamma$ and (a point of) $\partial S$ (showing this is well-defined and strictly positive) and just use segments shorter than this. – Vandermonde Jul 25 '15 at 03:33
  • Also, I've finally put my finger on what makes the proof seem complicated: I've identified it to be the following up of the use of compactness with a hacky (I tend to regard these as hacky, at least) iterative argument, which makes one or the other feel somewhat wasted (even though the iterative part requires the finiteness which the other part yields), whereas considering $\sup {t \in [0,1] : \text{there exists a polygonal path joining } \gamma(0) = p \text{ and } \gamma(t) }$ uses compactness alone (at the price of being admittedly less explicit/constructive). – Vandermonde Jul 25 '15 at 18:54
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For every point $p$ consider $B_p$ an open ball centered at $p$ and contained in $U$ - our open set. I claim that for every $p$, $p'$, there exist a sequence of point $p_0 = p$, $p_1$, $\ldots$, $p_n=p'$ so that $B_{p_i} \cap B_{p_{i+1}} \ne \emptyset$, for all $0 \le i \le n-1$. For otherwise we could separate the balls $B_q$ into two groups so that balls from different groups do not intersect, and that would provide a partition of our set into two nonempty open sets. Take for every $i$ a point $q_i$ in the intersection $B_{p_i} \cap B_{p_{i+1}}$. Consider now the polygonal line $p_0 q_0 p_1 q_1 \ldots p_{n-1} q_{n-1} p_n$ contained in $U$ and joining $p_0=p$ and $p_n=p'$.

orangeskid
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