Are these variations on the topologist's sine curve closed? Let $S=\{(x,\sin(1/x)):0 < x \leq 1\}$ and $T=S \cup {(0,0)}$. Under the usual metric on $\Bbb R^2$, is $S$ closed or $T$ or both or none?
My attempt: $x$ is varying between 0 and 1, and $\sin(1/x)$ is varying between -1 and 1. Since $(0,0)$ is not there in $S$, so, it is not closed. But I wonder why $T$ is not closed.

The closure of $S$ in $R^2$ is $S\cup (\{0\}\times [-1,1]).$
For $b\in [-1,1]$ and  $n\in \mathbb N$ there exists $x_n\in [1/2\pi (n+1),1/2\pi n)$ with $\sin (x_n^{-1})=b,$ so the Euclidean distance from $(0,b)$ to $(x_n, \sin (x_n^{-1}))=(x_n,b)$ is $x_n,$ which is less than $1/2\pi n.$
