0

We know that with the euclidean metric the open balls in $\mathbb{R}^2$ are circles without the frontier, of course.

My question is if there exist a known metric in $\mathbb{R}^2$ such that the open balls have the appearance as triangles or hexagons?

Thank you for the support.

LH8
  • 513
  • Why don't you try $d \left( \left( x_1, y_1 \right), \left( x_2, y_2 \right) \right) = \left| x_1 - x_2 \right| + \left| y_1 - y_2 \right|$ or $d \left( \left( x_1, y_1 \right), \left( x_2, y_2 \right) \right) = \max \left\lbrace \left| x_1 - x_2 \right|, \left| y_1 - y_2 \right| \right\rbrace$? Try making open balls of radius $1$ centered at origin with these metrics. You will find your answer. – Aniruddha Deshmukh Oct 15 '18 at 18:48
  • Open balls in the taxicab metric are squares, see for example this link. – user505379 Oct 15 '18 at 18:49
  • You may refer https://people.uwec.edu/mbirika/M316_Notes/Metric_Spaces_Open_Balls_and_Limit_Points.pdf for more information – Aniruddha Deshmukh Oct 15 '18 at 18:49

1 Answers1

1

I guess that you want a metric which induces the standard topology on $\mathbb{R}^2$.

If you want that the metric is induced by a norm, then you can find an answer here: Norm induced by convex, open, symmetric, bounded set in $\Bbb R^n$. . Note that any open norm-ball centered at $0$ must be bounded, open, convex and centre symmetric (for bounded recall that all norms on $\mathbb{R}^n$ are equivalent). This excludes triangles, but allows hexagons.

Edited:

Let $H$ be an open hexagon with centre $0$ and vertices lying on the unit circle. It is bounded, open, convex and centre symmetric. As in the above link define $$\|x\|_{hex} = \inf \{k>0 : x/k \in H \} .$$ This is a norm such that $H = \{ x \in \mathbb{R}^2 \mid \|x\|_{hex} < 1 \}$. Note that $$\{ x \in \mathbb{R}^2 \mid \|x\|_{hex} < r \} = r H ,$$ the latter being defined as $r H = \{r h \mid h \in H \}$. This is again an open hexagon stretched by the factor $r$. Our norm induces the metric $$d_{hex}(x,y) = \|x - y\|_{hex} .$$ Then $$B_{hex}(x;r) = \{ y \in \mathbb{R}^2 \mid d_{hex}(x,y) < r \} = x + B_{hex}(0;r) = x +r H$$ which is an open hexagon with centre $x$.

The same construction can be performed for any bounded, open, convex and centre symmetric set $A$. Then all open balls with respect to the metric $d_A$ are copies of some $rA$ (which is obtained by stretching $A$).

Paul Frost
  • 87,968