TITLE: Is $\{0\}$ is open in $(\mathbb{R}^2 , p)$?
QUESTION [0 upvotes]: Associate with $\mathbb{R}^2 $ the euclidean metric and denote  its  by $d$ . Define   on $\mathbb{R}^2 \times \mathbb{R}^2 $ a  function  $p$  by 
$$p(x,y) = \begin{cases} 0, \text{x=y} \\ d(x,0) + d(y,0) ,  x \neq y \end{cases}$$ where  $0 = (0,0)$
Now my question is that 
Is $\{0\}$  is open in $(\mathbb{R}^2 , p)$?
My attempt : I thinks  yes because we have  $B(a,r) = \{ x \in \mathbb{R}^2 : d(x,0) + d(a,0)  < r\}$ and   $B(0,r)  \subset \{0\}$
Is its true ?
Any hints/solution will be appreciated
thanks u

REPLY [2 votes]: If $\{0\}$ is open, then there exists $r>0$ such that $\{x\in \mathbb{R}^2:p(x,0)<r\}\subseteq \{0\}$, i.e., $\{x\in \mathbb{R}^2:d(x,0)<r\}\subseteq\{0\}$, which is not possible as $d$ is the Euclidean norm.( For example, $(\frac{r}{2\sqrt{2}},\frac{r}{2\sqrt{2}})$ is in $\{x\in \mathbb{R}^2:d(x,0)<r\}$)
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