Show that any connected open subset of $E^n$ is arcwise connected. 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?
 A: 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 $.
A: 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 $p_n=p'$. 
