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Open Ball Clarification

  1. Apr 2, 2014 #1
    1. The problem statement, all variables and given/known data
    What is the set of the unit ball centered at 0 = (0,0) in the (x1, y2)-plane, where d1(x,y)=|x1-y1|+|y1-y2| is in the metric space (R2, d1)


    2. Relevant equations
    An open ball is Br(x)={y[itex]\in[/itex]X|d(x,y)<r}


    3. The attempt at a solution
    I know the solution is this: B1(0)={(x1,x2)[itex]\in[/itex]R2 | |x1| + |x2| < 1}
    but I'm not sure how to get that answer. Here is what my answer would have been (by the definition of the open ball):
    B1(0)={y[itex]\in[/itex]R2| |-y1|+|x2|<1}
    I know that's wrong. So basically, could someone just explain the correct answer? Thanks!
     
  2. jcsd
  3. Apr 2, 2014 #2
    Given a point ##\mathbf{x}=(x_1,x_2)\in\mathbb{R}^2##, can you write down the algebraic expression for ##d_1(\mathbf{0},\mathbf{x})##? For example the (simplified) algebraic expression for ##d_2(\mathbf{0},\mathbf{x})## would be ##\sqrt{x_1^2+x_2^2}##.

    Please take care to proofread your response. It looks as though there a multiple errors in your original post, and it's hard to tell if they are typographical errors or errors in understanding. It makes it difficult to help.
     
  4. Apr 2, 2014 #3
    I have attached a picture of the notes where I got this from. Sorry for the typos. The only one I can see is d1(x,y)=|x1-y1|+|y1-y2| should be d1(x,y)=|x1-y1|+|x2-y2|
     

    Attached Files:

  5. Apr 2, 2014 #4
    Using the definition precisely as quoted, there is only one typo in your answer: [tex]\begin{align*} B_1(0) &= \{y\in X: d_1(0, y) < 1\}\\
    &= \{(y_1, y_2)\in X: |0 - y_1| + |0 - y_2| < 1\}\end{align*}[/tex] The symbol "x" may replace the symbol "y" when you are done, since the symbol "x" has no special meaning other than notifying the reader as to the proper position of quantities in the definition. You then only need to simplify the terms |-x1| and |-x2|.
     
    Last edited: Apr 2, 2014
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