Is $M=\left \{(x,y)\in (0,\infty )\times\mathbb{R} : y=\sin(\frac{1}{x}) \right \}$ a closed set in space $((0,\infty )\times\mathbb{R} ,\rho_{e})$, $\rho_{e}$ - Euclidean metric ? I think that open set could be built "around" function $y=\sin(\frac{1}{x})$ on domain $(a,\infty)$ where $a>0$ with sum of open balls, but I don't know if it's true, and I really don't have an idea what to do in the main case of - that is for $a=0$, mayby by induction ?
This is much less important : And what happens (in short) in the case of other metrics or when we change $(0,\infty )\times\mathbb{R}$ into $[0,\infty )\times\mathbb{R}$ in definition of $M$ and metric space ($M$ have to be expanded to $x=0$ case, and it seems that $\forall\ {a\in (0,1)}\Bigl(\ \exists\ {b_{n}\subset M}\text{ such that }\lim\limits_{n\to \infty }b_{n}=(0,a)\Bigr)$, so $M$ is going to be not closed because there exist seuences with limit points outside set $M$ for every value $y$ of $M$ for $x=0$ which we choosed).
$\rho_{e}$ - Euclidean metric