Izabella Ponce

2022-06-24

Let $({a}_{1},{a}_{2},...,{a}_{n})\in {\mathbb{R}}^{\mathbb{n}}$ and $b\in \mathbb{R}$. Prove that the set:

${F}_{1}:=\{({x}_{1},{x}_{2},...,{x}_{n})\in {\mathbb{R}}^{\mathbb{n}}|\sum _{i=1}^{n}{a}_{i}{x}_{i}\ge b\}$

is closed in ${\mathbb{R}}^{\mathbb{n}}$

${F}_{1}:=\{({x}_{1},{x}_{2},...,{x}_{n})\in {\mathbb{R}}^{\mathbb{n}}|\sum _{i=1}^{n}{a}_{i}{x}_{i}\ge b\}$

is closed in ${\mathbb{R}}^{\mathbb{n}}$

Ryan Newman

Beginner2022-06-25Added 26 answers

First prove that if $f(x):{\mathbb{R}}^{\mathbb{n}}\to \mathbb{R}$ is continuous then the set $\{x\in {\mathbb{R}}^{\mathbb{n}}|f(x)\ge b\}$ is closed in ${\mathbb{R}}^{\mathbb{n}}$. So let $f({x}_{1},{x}_{2},..{x}_{n})=\sum _{i=1}^{n}{a}_{i}{x}_{i}$. If $h,k$ be to function which are continuous then $h+k$ is continuous. so it is enough to show that $f(x)=ax$ is continuous which is obvious ($\delta \le \frac{\u03f5}{|a|}$)

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