I'm given the following function
$$S=\{x>0:(x,\sin(1/x))\in \mathbb R^2\}$$
and asked to identify its interior, cluster points, closure, and boundary.
Interior points: I know the set of interior points is the largest open set contained completely within $S$. Would I then consider the interior of $S$ to be the coordinates of the graph $y=\sin(1/x)$ for $x\in(0,\infty)$? That is, the interior of $S$ is simply $S$?
Cluster points: This is where I start getting tripped up. Since the limit $\lim_{x\to\infty}\sin (1/x)$ doesn't exist, would I not then say there does not exist a cluster point with x-component $0$? Additionally, I know $\lim_{x\to\infty}\sin(1/x)=0$. Does this imply $(\infty,0)$ is a cluster point? I'm not exactly sure if a coordinate with an infinite component is allowed.
The closure of $S$ is just going to be the union of $S$ and its cluster points so I won't need help there on top of help with identifying cluster points.
Lastly, would the boundary of $S$ simply be its interior since any epsilon neighbourhood in $\mathbb R^2$ of a point along the graph would contain points along the graph and outside the graph?
Note that this should be $$S=\{(x,\sin(1/x))\in \mathbb R^2 :x>0\}$$ which is the graph of the function $f(x) = \sin (1/x)$ on $(0,\infty)$
The interior of the curve is empty because there is no open ball which fits in the graph.
The cluster points are points on the graph union with the points on the vertical segment $\{0\} \times [-1,1]$ because every point on that segment is the limit of a sequence of points on the graph.
The boundary is the graph itself union with the vertical segment $\{0\}\times [-1,1]$
The closure is the same as the boundary.