# proof of any line joining two points lying in opposite half-spaces determined by a hyperplane in $\Bbb{R}^n$ intersects the hyperplane.

I am a student specializing in mathematics for economists. I have been struggling with proof question regarding hyperplane and was wondering if you could please give a helpful hand.

The question is: Prove that any line joining two points lying in opposite half-spaces determined by a hyperplane in $\Bbb{R}^n$ intersects the hyperplane.

Thank you!

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If you actually mean line, it is pretty simple. Set the hyperplane as the (n-1) axes of your coordinate system (with the nth being perpendicular - call it the "z" direction), and define the line between the points. The definition of the line obviously has a solution for z=0.

If you mean something a little more curvy than an actual line, the result still holds, but you need what is effectively a limit case of a generalisation of the Jordan Curve theorem.

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Let's let $H=\{x\in\Bbb{R}^n: \langle x,y\rangle=b\}$ be a hyperplane in $\Bbb{R}^n$, and let $u,v\in\Bbb{R}^n$ be such that $\langle u,y\rangle<b$ and $\langle v,y\rangle>b$ - in other words, $u$ and $v$ lie in opposite half-spaces. Then, let $z(t)=(1-t)u+tv,t\in[0,1]$ be the line segment joining $u$ to $v$. Now, define the function

$$f(t)=(1-t)(\langle u, y\rangle-b)+t(\langle v, y\rangle-b)$$

Notice that since $\langle u,y\rangle<b$ and $\langle v,y\rangle>b$, we have a continuous function of $t$ such that $f(0)<0$ and $f(1)>0$. Hence by the intermediate value theorem, there exists $t_0\in [0,1]$ such that $f(t)=0$; rearranging we would have $$(1-t_0)(\langle u, y\rangle-b)+t_0(\langle v, y\rangle-b)=0\\ \Rightarrow (1-t_0)\langle u,y\rangle+t_0\langle v,y\rangle=b\\ \Rightarrow \langle z(t_0),y\rangle=b$$

and hence $z(t_0)$ is on the hyperplane.

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