Sketch the open ball at the origin $(0,0)$, and radius $1$.

Question

Given the norm $||(x,y)|| = 2|x| +\frac{1}{3}|y|$. Sketch the open ball at the on the origin $(0,0)$, and radius $1$.

I understand that the sketch of an open ball withina set looks like the image attached, in a general case, but have no idea how to sketch one applying the above norm to the situation.

i understand that in the case of $B_{r}(a)=\{x \in X | d(x,a) < r\}$ in this case, $a = (0,0)$, and $r = 1$. Could someone please help as to how to sketch it?

Thanks

Answer 1

The example you drew is not a general case at all. That is what an open ball looks like in $\Bbb R^2$ under the euclidian metric.

We want to cover any point in the plane $(x,y)$ such that $2|x|+\frac{1}{3}|y|<1$.

In the first quadrant, $x$ and $y$ are positive, so we have $2x+y/3<1$, or the area under $y<3-6x$. In the second quadrant, $x$ is negative and $y$ is positive, so use $-2x+y/3<1$, or the line $y<3+2x$.

Et cetera. It should be fairly easy to cover all four cases and then make sure the boundary points are correct.

Answer 2

The open ball centered at the origin, with radius $1$, is the region containing the center $(0,0)$ and bounded by the line-segments \begin{eqnarray} 2x+\dfrac13y&=&1, 0 \le x\le 0.5\\ -2x+\dfrac13y&=&1, -0.5\le x\le 0\\ -2x-\dfrac13y&=&1, -0.5\le x\le 0\\ 2x-\dfrac13y&=&1, 0\le x\le 0.5 \end{eqnarray} In other words, your open ball is the region bounded by the polygon with vertices at $$ (0.5,0), (0,3), (-0.5,0), (0,-3). $$