# Find ball B

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• Sep 27th 2009, 09:48 AM
thaopanda
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?
• Sep 27th 2009, 03:20 PM
redsoxfan325
Quote:

Originally Posted by thaopanda
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 $x$.
• Sep 27th 2009, 03:57 PM
thaopanda
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
• Sep 27th 2009, 04:00 PM
redsoxfan325
$B_2(x)$ is also all of $X$, because the distance between every point is equal to $1$ (i.e. $<2$).
• Sep 27th 2009, 04:13 PM
Plato
Quote:

Originally Posted by thaopanda
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 $x \in X$.

$B_r(x) = \left\{ {\begin{array}{rl}
{\{ x\} ,} & {0 \leqslant r < 1} \\
{X,} & {1 \leqslant r} \\ \end{array} } \right.\,$
• Sep 27th 2009, 05:07 PM
thaopanda
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 $B_{\frac{\pi}{12}}(1)$?
• Sep 27th 2009, 07:22 PM
redsoxfan325
Quote:

Originally Posted by thaopanda
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)

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