Is $\{0\}$ is open in $(\mathbb{R}^2 , p)$?

Question

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

Answer 1

No. because $B(0,r) = \{ x : d(0,x) < r \}$
is not a subset of $\{0\}$ because $(0,\frac{r}{2})$ in $B(0,r).$

Answer 2

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 $\{x\in \mathbb{R}^2:d(x,0)<r\}$)