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I have a problem with solving the following problem: We are given a set that looks like this:

$\{x \in \mathbb{R}^n | b \leq a^Tx \leq c\}$ where $a \in \mathbb{R}^n$, $b, c \in \mathbb{R}$

My task is to find set of Border points and set of Inner points x in X. I was able to compute examples like this for simplier one and two dimensional functions where I can imagine, how the picture looks like but Im quite lost in this case. Can you help me out and suggest some simple method how to solve these examples for n-dimensional vectors (I have a lot of these to solve :D). Thank you!

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The word "set" does not imply any actual relation to set theory. –  Asaf Karagila Nov 6 '12 at 9:08

2 Answers 2

up vote 2 down vote accepted

Call $S$ this set. As the map $x\mapsto a^Tx$ is continuous, $S$ is closed. So we just have to compute its interior. We assume $a=\neq 0$. If $x$ is such that $b<a^Tx<c$, then we have $b<a^Ty<c$ if $\lVert x-y\rvert<\frac{\min\{b,c\}}{\lVert a\rVert}$, so $x\in\operatorname{int}(S)$. Conversely, if $x\in\operatorname{int}(S)$, we can find $\delta$ such that if $\lVert y\rVert<\delta $ then $b\leqslant a^T(x+y)\leqslant c$. Taking $y=\alpha x$ for $|\alpha|<\delta$ gives that $b<a^Tx<c$.

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Hint: Say $x$ lies in your set, and $b < a^T x < c$. Let $\epsilon = \text{max }\{a^T x - b, c - a^T x\} > 0$. Then, since the inner product is continouos, you can find some $\epsilon'$ such that $x + d$ also lies in your set if $||d|| < \epsilon'$. All such points $x$ are thus inner points of your set.

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