A polygonal path with straight sections parallel to the axis Question:We say that $f: I \rightarrow \mathbb{R^n}$ is parallel to the i'th axis when it has the form $f(t) = a + te_{i}, t \in I$. If $U \subset \mathbb{R^n}$ is connected open set, prove that any two points $a, b \in U$ can be linkeds by polygonal path contained in $U$ whose straight sections are parallel to the axis.
My idea: I know that any two points $a, b \in U$ can be linked by polygonal path contained in $U$; take a straight segment of the path, we can cover it with open balls and then to take a finite number from them. For each ball (in finite number) take his closure (closed ball), we have subdivided the segment in finite number of segments contained in some closed ball. My doubt: I can say that for each segment contained in some ball the result is valid for his end points?,if yes then the general result follows (the balls have intersections, then result is valid for the original segment e then for the polygonal path), if no how I can to prove that the result is valid in closed ball?
 A: We just need to prove the result for a closed ball $B\subset \mathbb{R}^n$ centered at the origin and with radius $r$. The general case follows from translation.
Let $a=(a_1, ...,a_n)$ and $b=(b_1, ..., b_n)$ two point in $B$ that we want to connect by a polygonal path with straight sections parallel to the axis (call this "$\star$-way"). By definition, we have $B=\{x\in\mathbb{R}^n;|x|\leq r\}$. The result is now simple:


*

*The point $a=(a_1, ..., a_n)$ can be connected to $(0, a_2, ..., a_n)$ by the $\star$-way. In fact, we have $|(0, a_2, ..., a_n)|\leq |a|\leq r$, what implies $(0, a_2, ..., a_n)\in B$, and $B$ is convex, what implies this line segment is contained in $B$;

*Similarly, the point $(0, a_2, ..., a_n)$ can be connected to $(0, 0, a_3, ..., a_n)$ by the $\star$-way;

*$\vdots$

*Finally, the point $(0, ..., 0, a_n)$ can be connected to the origin $(0, ..., 0)$ by the $\star$-way.


The juxtaposed path connect $a$ to the origin $(0, ..., 0)$ by the $\star$-way. The same can be done with $b$, that is, $b$ is connected with the origin by the $\star$-way. So the juxtaposed path connect $a$ to $b$ by the $\star$-way and we're done.
A: Let's call the special kind of polygonal path we're interested in a TCP (taxi-cab path). Given the TCP result for a ball, I think the following may be simpler than using non TCP polygonal paths and overlapping balls and such: Fix any $a\in U.$ Define $E$ to be the set of points $ b\in U$ such that there is a TCP in $U$ from $a$ to $b.$ Since some $B(a,r)\subset U,$ $E$ is nonempty by the result for balls. Let $b\in E.$ Then for some $r>0,$ $B(b,r)\subset U.$ There is a TCP from $a$ to $b$ within $U.$ And there is a TCP from $b$ to any other point in $B(b,r).$ Therefore $B(b,r)\subset E.$ Hence $E$ is open. To show $E$ is closed in $U,$ suppose $b_1,b_2, \dots \in U$ and $b_k \to b,$ where $b \in U.$ Again some $B(b,r)\subset U.$ For large $k,b_k \in B(b,r).$ Fix such a $k.$ Then there there is a TCP from $a$ to $b_k$ within $U.$ And there is a TCP within $B(b,r)$ from $b_k$ to $b.$ Putting these TCP's together shows $b\in E.$ Hence $E$ is closed in $U.$ Thus $E$ is nonempty and clopen in $U.$ Since $U$ is connected, $E = U,$ and we're done.
A: Firstly, we recall that the topology of $\mathbb{R} ^n$ is generated by open cubes. The result is trivial for open cubes.
Now, let $U$ be an arbitrary open connected set. Define the following equivalence relation on $U$: $x \sim y $ iff there exists such a polygonal path between both. Every equivalence class is open, since the result is true for open cubes. Since $U$ is connected, there must be at most one, and the result follows.
