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Every hyperplane divides $\mathbb{R}^n$ into two "half space": the set of points "on and above" the hyperplace, $H^+ = \{ \mathbf{x} \mid \mathbf{a} \cdot \mathbf{x} \geq \alpha \}$, and the set of points "on and below" the hyperplace, $H^- = \{ \mathbf{x} \mid \mathbf{a} \cdot \mathbf{x} \leq \alpha \}$. Prove that each of these two half spaces is a closed, convex set.

Open Sets in $\mathbb{R}^n$
$S \subset \mathbb{R}^n$ is an open set if, for all $\mathbf{x} \in S$, there exists some $\varepsilon > 0$ such that $B_\varepsilon ( \mathbf{x} ) \subset S$.

Closed Sets in $\mathbb{R}^n$
$S$ if a closed set if its complement, $S^c$, is an open set.

I'm trying to prove the question above using the definitions provided in the textbook (Advanced Microeconomic Theory by Jehle and Reny) that posed this question but I'm having trouble finding the required $\varepsilon$ such that you can draw a ball around every element in the complement of the closed half-space and that ball is still entirely contained in the complement of the closed half-space thereby proving that the complement of closed half-space is open, and the closed half space is indeed closed. Any proof that uses the above definition with regard to open and closed set is greatly appreciated.

user642796
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Chun-Kai
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2 Answers2

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A half space is nothing but the preimage of $[\alpha, \infty)$ (or $(-\infty, \alpha]$), for some $\alpha$, under the map

$$ x \stackrel{\phi}{\mapsto} \langle x, a \rangle. $$

This map is a continuous/bounded linear functional, so its pre-image of a closed set is closed.

To show this map is continuous with respect to the usual topologies, use Cauchy-Schwarz:

$$ |\langle x, a \rangle| \leq \|x\| \cdot \|a\|. $$

This tells you $\phi$ is actually more than continuous:

$$ |\phi(x) - \phi(y)| \leq \|a\| \|x-y\|. $$

Actually, linear maps between finite dimensional locally convex topological vector spaces are automatically continuous.

Michael
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Let us consider the complement of $H^+$ which is $C=\{x:a'x < \alpha\}$. Then for every $x \in C$ let $\Delta_x = \alpha - a'x >0$. You can choose $\epsilon$ as $\frac{\Delta_x}{2}$. Similarly you can do it for $H^-$.

Stefano
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  • Hi, so by choosing epsilon as (α−a′x)/2 , then how do you show that every element in this ball is in C, that is a′y <α for every y in Y = {y : d (x , y) < (α−a′x)/2 } where d(.,.) is the standard Euclidean Metric. Thanks @Stefano – Chun-Kai Mar 04 '14 at 22:42
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    @Chun-Kai actually it is easier if you choose $\epsilon$ as $\frac{\Delta_x}{2||a||}$. In this way, $a'y=a'(x+h)=a'x+a'h=-\Delta_x + \alpha + ||a|| ||h|| \cos(a,h) <= -\Delta_x + \alpha + ||a||\frac{\Delta_x}{2||a||}1 <\alpha$. $h$ is $y-x$ and is enclosed in the ball. – Stefano Mar 04 '14 at 23:19
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    Sorry, I ment to say that $h$ is $y-x$, the distance between $y$ and $x$, therefore its norm is $<=\epsilon$. Also, since I used the definition of scalar product I should have written $a \cdot x$. – Stefano Mar 04 '14 at 23:26
  • Hi, awesome it works. Is there any algorithm that you use to come up with such workable epsilon ? Or is it just through experience? I have been stuck on this for several hours until you came, thanks. @Stefano – Chun-Kai Mar 05 '14 at 00:04
  • @Chun-Kai I suggest always to draw a picture. This is why the first time I told you to use $\frac{\Delta_x}{2}$. Then you start 'doing the math' and you may realize that using another quantity can simplify the computation. – Stefano Mar 05 '14 at 15:17