- #1
stauros
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Prove if the following set is open
[tex]\displaystyle{A=\begin{cases}
\vec{x}=(x_1,...,x_n)\in\mathbb{R}^n:x_n>0\\
\end{cases}}[/tex] .
I have written the following proof and please correct me if i am wrong
Let : [tex]\displaystyle{\vec{x}\in A}[/tex]
Then we have : [tex]\displaystyle{\vec{x}=(x_1,...,x_n)}[/tex] with [tex]\displaystyle{x_n>0}[/tex]
Choose [tex]\epsilon[/tex] such that [tex]\displaystyle{0<\epsilon<x_n}[/tex] and then [tex]\displaystyle{B(\vec{x},\epsilon)\subseteq A}[/tex]
This happens because if [tex]\displaystyle{\vec{y}=(y_1,...,y_n)\in B(\vec{x},\epsilon)}[/tex] then [tex]\displaystyle{||\vec{y}-\vec{x}||<\epsilon}[/tex]
and [tex]\displaystyle{y_i\in\left(x_i-\epsilon,x_i+\epsilon\right)}[/tex]
Then we have [tex]\displaystyle{y_n\in\left(x_n-\epsilon,x_n+\epsilon\right)}[/tex] and thus [tex]\displaystyle{y_n>0}[/tex][/quote]
[tex]\displaystyle{A=\begin{cases}
\vec{x}=(x_1,...,x_n)\in\mathbb{R}^n:x_n>0\\
\end{cases}}[/tex] .
I have written the following proof and please correct me if i am wrong
Let : [tex]\displaystyle{\vec{x}\in A}[/tex]
Then we have : [tex]\displaystyle{\vec{x}=(x_1,...,x_n)}[/tex] with [tex]\displaystyle{x_n>0}[/tex]
Choose [tex]\epsilon[/tex] such that [tex]\displaystyle{0<\epsilon<x_n}[/tex] and then [tex]\displaystyle{B(\vec{x},\epsilon)\subseteq A}[/tex]
This happens because if [tex]\displaystyle{\vec{y}=(y_1,...,y_n)\in B(\vec{x},\epsilon)}[/tex] then [tex]\displaystyle{||\vec{y}-\vec{x}||<\epsilon}[/tex]
and [tex]\displaystyle{y_i\in\left(x_i-\epsilon,x_i+\epsilon\right)}[/tex]
Then we have [tex]\displaystyle{y_n\in\left(x_n-\epsilon,x_n+\epsilon\right)}[/tex] and thus [tex]\displaystyle{y_n>0}[/tex][/quote]