assume $(d,\mathbb R)$ be a mertic space such that $$d:\mathbb R\times \mathbb R \to [0,\infty)$$$$d(x,y)= \begin{cases} 0, & \text{if x=y} \\ max\{|x|,|y|\}, & \text{if x$\neq$y} \\ \end{cases}$$ How prove $\mathbb Q$ is close in this metric space and $\mathbb Q$ is not open in this metric space?
Thanks for any hint?
Let $A=\Bbb Q$.
If $x\in\overline{A}\setminus A$, there's a sequence $(a_n)\subseteq A$ with $$a_n\to x$$ $$\Rightarrow 0<d(a_n,x)\to0$$ $$\Rightarrow|x|\le d(a_n,x)\to0$$ $$\Rightarrow x=0$$ a contradiction.
$A$ can be any set of real numbers not containing $0$.
HINT: For any $x\in\Bbb R$ and $\epsilon>0$, the open ball of radius $\epsilon$ centred at $x$ is
$$\begin{align*} B(x,\epsilon)&=\{y\in\Bbb R:d(x,y)<\epsilon\}\\ &=\big\{y\in\Bbb R:y=x\text{ or }\max\{|x|,|y|\}<\epsilon\big\}\;. \end{align*}$$
Now $\max\{|x|,|y|\}\ge|x|$, so if $x\ne 0$, and $\epsilon\le|x|$, then $\max\{|x|,|y|\}\ge|x|\ge\epsilon$ for any $y\ne x$, and therefor $B(x,\epsilon)=\{x\}$. In other words, every non-zero real number is an isolated point in this space. To show that $\Bbb Q$ is closed, just use the fact that $0$ is rational.
To show that $\Bbb Q$ is not open, determine exactly what $B(0,\epsilon)$ looks like for $\epsilon>0$.
