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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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I want to give you some advice about the site: To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people are much more willing to help you if you show that you've tried the problem yourself. If this is homework, please add the [homework] tag; people will still help, so don't worry. Also, many would consider your post rude because it is a command ("Prove..."), not a request for help, so please consider rewriting it. –  Zev Chonoles Jul 7 '12 at 17:35
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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
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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
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1 Answer 1

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$[a,b]/n < \delta$.)

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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
    
Otherwise this proof might be seen as incomplete. –  Did Jul 14 '12 at 14:24
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