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Given metric space $M = (\mathbb{R}^2, d)$ where $d = \operatorname{max}\{|x_1 - y_1|, |x_2 - y_2|\}$, how can one measure distance from some arbitrary point $X$ to the line $y = 3$, let's say?

How it can be done, by use of the identity $$\operatorname{max}\{x, y\} = \frac{1}{2}(x + y+ |x - y|)\text{?}$$

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  • $\begingroup$ y=3 will look like y=3 no matter which metric as far as I can see. I think a geometric shape will change with the underlying metric only if its definition depends on the metric. A circle might be more interesting, where I define a circle as the set of points of equal distance to some center point. It will look like a square with this metric. $\endgroup$ – joedoe8585 Jan 20 '16 at 22:05
  • $\begingroup$ @joedoe8585 and if I would like to measure distance from point X to some point on line y = 3, how can I do so? $\endgroup$ – Accelerate to the Infinity Jan 20 '16 at 22:06
  • $\begingroup$ I am not sure how you plan to use your identity about $\max$, but do note that this problem is particularly easy for a vertical or horizontal line. $\endgroup$ – Rob Arthan Jan 21 '16 at 0:07
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Denote your point by $(x_0,y_0)$ and the line $y = 3$ by $l$. You have

$$ d((x_0,y_0),l) = \inf_{x \in \mathbb{R}} d((x_0,y_0), (x,3)) = \inf_{x \in \mathbb{R}} \max \{ |x_0 - x|, |y_0 - 3| \}. $$

It is clear that $\max \{ |x_0 - x|, |y_0 - 3| \} \geq |y_0 - 3|$ and so $d((x_0,y_0),l) \geq |y_0 - 3|$. By taking $x = x_0$, we see that the infimum is actually attained and $d((x_0,y_0),l) = |y_0 - 3|$. It is worth interpreting this calculation geometrically using the description of the circles in this norm as squares.

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  • $\begingroup$ Could you please give me a details about this solution? :-) Why did you set $x = x_0$? I still can't get it. $\endgroup$ – Accelerate to the Infinity Feb 1 '16 at 8:47
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    $\begingroup$ If you want to show that $\inf_{x \in \mathbb{R}} g(x) = C$, you can show that $g(x) \geq C$ and that there exists $x_0 \in \mathbb{R}$ so that $g(x_0) = C$. This shows that the infimum is actually a minimum (attained at $x = x_0$). This is what I did for $g(x) = \max \{ |x - x_0|, |y_0 - 3| \}$ and $C = |y_0 - 3|$. $\endgroup$ – levap Feb 1 '16 at 9:16
  • $\begingroup$ Got it, thanks! $\endgroup$ – Accelerate to the Infinity Feb 1 '16 at 9:58
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Hint: the "circles" of radius $r$ about $X$ in $M$ are squares with edges parallel to the axes that are centred on $X$ and have sides of length $r/2$. See https://en.wikipedia.org/wiki/Uniform_norm or https://en.wikipedia.org/wiki/Norm_(mathematics). To find the distance from $X$ to a line $l$ find the smallest of these squares that has a vertex on $l$.

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