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Math Help - Find ball B

  1. #1
    Member thaopanda's Avatar
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    Find ball B

    Let X be an arbitrary set and consider d: X \times X \rightarrow R defined by d(x,y) := \left \{\begin{array}{cc}0,&\mbox{ if } x = y \\1, & \mbox{ if } x \neq y \end{array}\right.

    I need to find B_{1}(x) where x \in X. Don't know how... help?
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by thaopanda View Post
    Let X be an arbitrary set and consider d: X \times X \rightarrow R defined by d(x,y) := \left \{\begin{array}{cc}0,&\mbox{ if } x = y \\1, & \mbox{ if } x \neq y \end{array}\right.

    I need to find B_{1}(x) where x \in X. Don't know how... help?
    That depends...are you defining B_1(x)=\{y\in X: d(x,y)<1\} or B_1(x)=\{y\in X: d(x,y)\leq 1\}

    In the first case, B_1(x) is simply the point x, because the only point with distance strictly less than 1 from x is x itself.

    In the second case, B_1(x) is all of X, because every point has distance less than or equal to 1 from x.
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  3. #3
    Member thaopanda's Avatar
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    Well, I'm not given anymore than what I wrote to the problem, so I'm assuming it would be the second case.

    So is it that B_{1}(x) = X? And how would B_{2}(x) look? I think I'm just lost as to how you would define it or something...

    awesome username btw not only do the red sox rock, but 325 is my birthday, hehe
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  4. #4
    Super Member redsoxfan325's Avatar
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    B_2(x) is also all of X, because the distance between every point is equal to 1 (i.e. <2).
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  5. #5
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    Quote Originally Posted by thaopanda View Post
    Let X be an arbitrary set and consider d: X \times X \rightarrow R defined by d(x,y) := \left \{\begin{array}{cc}0,&\mbox{ if } x = y \\1, & \mbox{ if } x \neq y \end{array}\right.

    I need to find B_{1}(x) where x \in X.
    B_r(x) = \left\{ {\begin{array}{rl}<br />
   {\{ x\} ,} & {0 \leqslant r < 1}  \\<br />
   {X,} & {1 \leqslant r}  \\ \end{array} } \right.\,
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  6. #6
    Member thaopanda's Avatar
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    I had another question, same concept:

    let d: R \times R \rightarrow R.
    the function this time is |arctan x - arctan y|, what is the B_{\frac{\pi}{12}}(1)?
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  7. #7
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by thaopanda View Post
    I had another question, same concept:

    let d: R \times R \rightarrow R.
    the function this time is |arctan x - arctan y|, what is the B_{\frac{\pi}{12}}(1)?
    Since \arctan(1)=\frac{\pi}{4}, B_{\pi/12}(1)=\left\{y\in\mathbb{R}: \left|\frac{\pi}{4}-\arctan(y)\right|<\frac{\pi}{12}\right\}

    Solving for y, we know that:

    \frac{\pi}{6}<\arctan(y)<\frac{\pi}{3}

    Taking the tangent of all three parts, we get:

    \tan(\pi/6)<y<\tan(\pi/3) \implies \frac{\sqrt{3}}{3}<y<\sqrt{3}

    So B_{\pi/12}(1)=\left\{y\in\mathbb{R}: \frac{\sqrt{3}}{3}<y<\sqrt{3}\right\}
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