Hi guys,

We were learning about metric functions. I don't understand why d-infinity(0,1) is a square.

Similarly, why is a unit ball in R3 using the infinity metric a cube?

Can someone please explain this to me? Thanks very much.

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- Mar 8th 2011, 02:27 AMsakodoQuestion on the infinity metric
Hi guys,

We were learning about metric functions. I don't understand why d-infinity(0,1) is a square.

Similarly, why is a unit ball in R3 using the infinity metric a cube?

Can someone please explain this to me? Thanks very much. - Mar 8th 2011, 03:31 AMHallsofIvy
I am afraid you will have to explain exactly what "d-infinity" you mean. The only "infinity metrics" I know are on function spaces ($\displaystyle L_\infty$) and sequence spaces ($\displaystyle l_\infty$). I don't know of any "d-infinity" on Euclidean spaces.

- Mar 8th 2011, 04:16 AMsakodo
I am not sure if I am posting in the right section, but here it goes.

d-infinity is a distance function(a metric) defined as d(x,y)=max(|x1-y1|,|x2-y2|...)

It is also called the supremum metric.

Our lecturer told us that d-infinity(0,1) in R3 represents a cube, I had no idea why.

Definition and examples of metric spaces

In the above link, if you scroll down to example 5, it says d-infinity(0,1) in R2 is a square.

Can someone please explain why that is so?

Thanks very much. - Mar 8th 2011, 04:40 AMPlato
First, the link does help. Now this completely standard material.

Second, What you first posted is misleading.

The set $\displaystyle \left\{ {x \in R^3 :d_\infty (0,x) \le 1} \right\}$ is a cube with center at the origin.

Here are some points in that set $\displaystyle (1,0,0),~(0.5,-0.6,0),(1,-1,1)$. - Mar 8th 2011, 04:49 AMAckbeet
Just a side note, sakodo: if you're unsure of where a thread belongs, report your own post (lower left corner, looks like a triangle with an exclamation mark inside), and a moderator will move it for you. I would recommend that this thread go to the Analysis subforum.

- Mar 8th 2011, 09:16 AMOpalg
It doesn't actually say "d-infinity(0,1) in R2 is a square". It says that the unit circle $\displaystyle \{x\in\mathbb{R}^2:d(x,0)=1\}$ is a square, for that metric d.

The unit circle is the set of points whose distance from the origin (as measured by the d-metric) is 1. If $\displaystyle x = (x_1,x_2)$ then $\displaystyle d(x,0) = \max\{|x_1|,|x_2|\}$. The equation $\displaystyle d(x,0)=1$ says that*either*$\displaystyle |x_1| = 1$ and $\displaystyle |x_2|\leqslant1$,*or*$\displaystyle |x_1| \leqslant 1$ and $\displaystyle |x_2| =1$. If you think about what that means, it says that the point x lies on the square with vertices at $\displaystyle (\pm1,\pm1)$. - Mar 8th 2011, 11:45 AMsakodo
Thanks for the reply guys. I think I kind of get it now.

So in R2, the unit circle of the d-metric is a set of points of which the maximum components are less than 1. This means the bounds for this metric are the lines x=1,-1 and y=1,-1. Thus the unit circle is a square. Is this correct? Same argument goes for R3.

But the unit ball B(0,1) , using the infinity metric, would be a square. I think I mixed the two up.

Thanks a lot guys.