# Describe the open balls of the metric $d_v$ given by $d_v(p,q)= 1$ or $d_v=|p_2-q_2|$

For points $$p=(p_1,p_2)$$ and $$q=(q_1,q_2)$$ in $$\mathbb{R}^2$$ define $$d_v(p,q)=\begin{cases} 1\, \text{if p_1 \neq q_1 or |p_2-q_2| \geq 1} \\ |p_2-q_2| \, \text{if p_1=q_1 and |p_2-q_2|<1}\end{cases}$$

Describe the open balls in the metric $$d_v$$.

By definition of open ball we have that $$B(p,\varepsilon)= \lbrace q\in \mathbb{R}^2 \mid d(p,q)< \varepsilon \rbrace$$, from here we had two cases

Case I

$$B(p,\varepsilon)= \lbrace q\in \mathbb{R}^2 \mid 1< \varepsilon \rbrace$$ then

$$p_1 \neq q_2$$ or $$|p_2-q_2| \geq 2$$. If $$p_1 \neq q_2$$ then $$q=(q_1,q_2)$$ be in the open ball for any $$U\subset \mathbb{R}^2 \setminus \lbrace x=p_1 \rbrace$$.

Case II

$$|p_2-q_2|\geq 1$$ then $$1+p_2 therefore $$q$$ are in the open ball for $$\mathbb{R} \times [1-p_2,p_2-1]$$.

Finally the open balls for $$d_v$$ are $$\lbrace p_1 \rbrace \times (p_2-1,p_2+1)$$

$$U\subset \mathbb{R}^2 \setminus \lbrace x=p_1 \rbrace$$ or $$W \subset \mathbb{R} \times [1+p_2,p_2-1]$$

• Definition of $d_v$ is either incomplete or has a typo at $p_1\ne q_2$. Two cases of $d_v$ are not complements. Dec 7 '20 at 22:11
• These already was correct
– user795628
Dec 7 '20 at 22:28
• ? You just edited it Dec 7 '20 at 22:30

I’m afraid that I can make little sense of your reasoning. The nature of an $$\epsilon$$-ball in this metric depends entirely on whether $$\epsilon\le 1$$.
Let $$p=\langle x,y\rangle$$. Then
$$B(p,\epsilon)=\begin{cases} \{x\}\times(y-\epsilon,y+\epsilon),&\text{if }0<\epsilon\le 1\\ \Bbb R^2,&\text{if }\epsilon>1\,. \end{cases}$$