When working with metric spaces we usually have to sketch absolute value inequalities. I can determine the open balls and everything but the sketching part is difficult, for example for a metric defined as $$d((x_1,y_1),(x_2,y_2))=|x_1+ y_1- x_2- y_2|+|-x_1+ y_1+ x_2- y_2|$$ if I want the ball $$B_2(0,0) = \{(x,y) \in \mathbb{R}^2: d((x,y),(0,0))<2\}=\{(x,y) \in \mathbb{R}^2: |x+y|+|-x+y| <2\}$$
I have no idea how to draw these things; although I know what an absolute value function looks like. Any help is appreciated.
Let us use the following linear change of coordinates
$$\begin{bmatrix}u \\ v\end{bmatrix} = \begin{bmatrix}1 & 1 \\ -1 & 1\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}.$$
Then
$$B_2(0,0) = \{(x,y) \mid |u| + |v| < 2 \}$$
which should be easy to draw in terms of $u$ and $v$. Then you just have to undo this change of coordinates, which in this case is nothing but a rotation by $-\pi/4$ with scaling, since we can decompose it into these two parts:
$$\begin{bmatrix}1 & 1 \\ -1 & 1\end{bmatrix}= \sqrt{2}\begin{bmatrix}\cos \alpha & -\sin \alpha \\ \sin \alpha& \cos \alpha\end{bmatrix}$$
for $\alpha = -\pi/4$. (In other cases we might have to consider another transformation might not have such an easy description.)
In the $u,v$ coordinate system this is the $l^1$ ball of radius $2$ which is the case $p=1$ in the image below, just scaled by a factor of $2$.
By rotating it back you get an upright square in the $x,y$ coordinate system. I leave it to you to calculate its size.