这是一份ed.ac.uk爱丁堡大学ECNM11073作业代写的成功案例
Given a metric space $\mathbf{X}$ and a distance $d(\mathbf{x}, \mathbf{y})$ defined on $\mathbf{X}^{n}$, an $\epsilon$-neighbourhood (or $\epsilon$-ball) of the point $\mathbf{x} \in \mathbf{X}$ is given by:
$$
N_{\epsilon}(\mathbf{x})={\mathbf{y} \in \mathbf{X} / d(\mathbf{x}, \mathbf{y})<\epsilon}
$$
and $\epsilon$ is a real finite positive number (usually small).
shows an $\epsilon$-neighbourhood in $\mathbf{E}^{1}$, and Fig. $3.5(b)$ in $\mathbf{E}^{2}$. We also define a deleted neighbourhood of $x$ as $N_{-}^{\prime} \epsilon(x)=N_{\epsilon}(x)-{x}$, i.e. the $\epsilon$-neighbourhood minus the point $x$ itself.
ECNM11073 COURSE NOTES ：
The solution of a system of linear inequalities:
$$
a_{i 1} x_{1}+a_{i 2} x_{2}+\ldots+a_{i n} x_{n} \leqslant b_{i} \quad i=1, \ldots, m \quad \forall x_{i} \geqslant 0
$$
form a convex set in $\mathbf{R}^{n}$. This can be shown as follows. Let $C=$ ${x / A x0}$ and let $x, y \in C$ then
$\mathbf{A}[a x+(1-a) y] \leqslant a b+(1-a) b=b$, i.e. $\mathbf{A x}0$. Therefore $\mathbf{z} \in \mathbf{C}$, hence the solution set is convex.