# Let $[a,b]\subset U$(open set in $\mathbb R^n$).Prove that there is a polygonal path(parallel to the coordinate axes) from $a$ to $b$.

Let $a,b\in\mathbb{R}^n$ be two points, and assume that the line segment $[a,b]=\{ta+(1-t)b\mid t\in[0,1]\}$ is completely contained in an open subset $U\subseteq \mathbb{R}^n$ then there is a polygonal path $\lambda:[0,1]\to U$ such that $\lambda(0)=a \, , \lambda(1)=b$ whose linear segments are parallel to the coordinate axes.

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What do you mean by saying that the interval $[a,b]$ is contained in a subset of $\mathbb{R}^n$? –  Arturo Magidin Jul 7 '12 at 17:39
Given two points $a,b\in U$ such that $[a,b]\subset U$ prove that $a$ and $b$ can be connected otherwise...$[a,b]=\{ (1-t)a+tb:0 \le t\le 1 \}$ –  felipeuni Jul 7 '12 at 17:48
@Arturo I think, I think he means the straight line segment joining the points $\,a,b\in U\subset \Bbb R^n\,$ is completely contained in $\,U\,$...I don't know though why wouldn't he say this so. –  DonAntonio Jul 7 '12 at 17:53
Okay, I've added the definitions. Now, @felipeuni, how about asking a question instead of issuing a command? –  Arturo Magidin Jul 7 '12 at 17:58
Since $[a,b]$ is compact and totally contained in $U$, there is a $\delta>0$ such that $[a,b]$ is contained in the "tube" $\cup_{x\in [a,b]} B(x,\delta)$, where $B(x,\delta)$ is the closed disc centered at $x$ of radius $\delta$. Now start at $a$ horizontally until you reach the border of $B(a,\delta)$. Go vertically until you reach $[a,b]$. Continue like this until reach you $b$. Note that to reach $b$ you may need to adapt how far you go horizontally in the last step. (This may be made more precise by finding a $n$ such that length$[a,b]/n < \delta$.)
An argument to show that the process actually reaches $b$ and does not accumulate on some $c$ in $(a,b)$ halfway, might be added. –  Did Jul 8 '12 at 7:35