Izabella Ponce

2022-06-24

Let $\left({a}_{1},{a}_{2},...,{a}_{n}\right)\in {\mathbb{R}}^{\mathbb{n}}$ and $b\in \mathbb{R}$. Prove that the set:
${F}_{1}:=\left\{\left({x}_{1},{x}_{2},...,{x}_{n}\right)\in {\mathbb{R}}^{\mathbb{n}}|\sum _{i=1}^{n}{a}_{i}{x}_{i}\ge b\right\}$
is closed in ${\mathbb{R}}^{\mathbb{n}}$

Ryan Newman

Expert

First prove that if $f\left(x\right):{\mathbb{R}}^{\mathbb{n}}\to \mathbb{R}$ is continuous then the set $\left\{x\in {\mathbb{R}}^{\mathbb{n}}|f\left(x\right)\ge b\right\}$ is closed in ${\mathbb{R}}^{\mathbb{n}}$. So let $f\left({x}_{1},{x}_{2},..{x}_{n}\right)=\sum _{i=1}^{n}{a}_{i}{x}_{i}$. If $h,k$ be to function which are continuous then $h+k$ is continuous. so it is enough to show that $f\left(x\right)=ax$ is continuous which is obvious ($\delta \le \frac{ϵ}{|a|}$)