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Open set

  1. Oct 7, 2012 #1
    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]
     
  2. jcsd
  3. Oct 7, 2012 #2

    HallsofIvy

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    That makes no sense because there is no one number labeled "[itex]x_n[/itex]". What you mean to say is that [itex]\epsilon< min(x_n)[/itex].


    Can you prove this? That is, after all the whole point of the exercise! In particular, what is the definition of [itex]||\vec{y}-\vec{x}||[/itex]?

     
  4. Oct 7, 2012 #3

    Yes you right,how about the inequality: [tex]\displaystyle{0<\epsilon<x_k, \forall 1\le k \le n}[/tex].

    But i think the center point of the problem is that:

    [itex]|x_{i}-y_{i}|\leq ||x_{i}-y_{i}||<\epsilon[/itex] using the Euclidian norm
     
    Last edited: Oct 7, 2012
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