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 $1+p_2<q_2<p_2-1$ 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 $W \subset \mathbb{R} \times [1+p_2,p_2-1]$

  • $\begingroup$ 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. $\endgroup$ Dec 7 '20 at 22:11
  • $\begingroup$ These already was correct $\endgroup$
    – user795628
    Dec 7 '20 at 22:28
  • $\begingroup$ ? You just edited it $\endgroup$ 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

$$B(p,\epsilon)=\begin{cases} \{x\}\times(y-\epsilon,y+\epsilon),&\text{if }0<\epsilon\le 1\\ \Bbb R^2,&\text{if }\epsilon>1\,. \end{cases}$$


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