Open and connected in $R^n$ revised I am trying to understand the following: If we have an open and connected set in $R^n$ then it can be connected with line segments parallel to the axes.
I managed to prove this:
 If a set $U$ is open and connected in $\mathbb{R}^n$ then we can prove it is polygonally connected(there is a path formed from line segments completely contained in $U$).
My question now is how would I modify the path such that the line segments remain in $U$ and they are now parallel to the axes?
I would very much appreciate some help.
Thank you!
 A: First note that for any cube $C=[-r,r]^n\subseteq\mathbb R^n$ any point $c=(c_0, c_1,\ldots c_{n-1})\in C$ is polygonally connected to the center of $C$ along the axes.
$$(0,0,0,\ldots,0)\to(c_0,0,0, \ldots, 0)\to(c_0,c_1,0,\ldots0)\to\ldots\to(c_0,c_1,\ldots c_{n-1})$$
Let $G$ be any nonempty open connected set in $\mathbb R^n$ and let $a\in G$.
Now set $A=\{g\in G\mid\text{$g$ is polygonally connected to $a$ along the axes}\}$.
Note that for any any $b\in A$ there's a cube $C=b+[-r,r]^n\subseteq G$ because G is open, so $C\subseteq A$. This means $A$ is open.
And for any $b\in\overline A$ there is a cube $C=b+[-r,r]^n\subseteq G$, so there's also a point $c\in A\cap C$, so $b\in A$. This means $A$ is closed.
Therefore $A=G$, because it's a nonempty clopen subset of a connected set $G$.
A: For simplicity we can take the case of $R^2$. We want to go from (a,b) to (c,d) parallel to coordinate axes, but not get out of the given set. We move from (a,b) to (a,c) and test if the path (a,b)-(a,c) is in the set. If yes, we than move from (a,c) to (b,c) and test if (a,c)-(b-c) is in the set. If yes, we are done. But if (a,b)-(a,c) is not in the set, then we go from (a,b) to (a,c/2) and do the same test... We do the same thing for y-axis parallel lines.
