Is the set defined by y=sin(1/x) open or closed?

"Is this set open, closed, or neither, and what is the boundary, interior, and closure of $X$?"

Firstly, openness and closedness: This is a one dimensional subset of the plane. Every non-empty open set in the plane is two dimensional, so $X$ is not open. (More explicitly: choose a point $a$ in $X$. Then no matter how small you make the radius of an open ball centred at $a$, it will not be contained in $X$ and hence $X$ is not open.)

As for whether or not it is closed, what you did was fine. Given any ball around $(0, 0)$, this will have a non-empty intersection with $X$, so $X^c$ is not open, so $X$ is not closed.

Hence $X$ is not open nor closed.

Boundary $(\partial X)$, interior $(\mathrm{int}(X))$, and closure $(\bar{X})$: The easiest one is the interior. Clearly this is empty because no open balls around a point in $X$ are in $X$. This means the boundary and closure will be equal, because $\bar{X}$ = $\partial X$ $\cup$ int($X$).

consider $a>0$. Then define $X_a =\{(x,y)\in\mathbb{R}^2:x≥a,y=\sin(1/x)\}$. It is easy to see that this is closed, so it is equal to its boundary and closure. Hence we only need to consider the case where $x=0$.

To find a closure, it is easier to know the answer beforehand and then verify that it's correct (a bit like showing the limit of a function). By drawing a pciture, you'll see rapidly increasing oscillations at $x$ approached $0$, so the closure is equal to $Y =\{(x,y)\in\mathbb{R}^2:x>0,y=\sin(1/x)\}\cup\{(0,y):y\in[-1,1]\}$. (i.e. you're adding a vertical segment at 0 which is the "limit of the oscillation".)

The answer's getting a bit long and I'm running out of energy, so you can do the rest. To show that this is indeed $\bar{X} = \partial X$, first you must show that every point in $Y$ is a point of closure. (i.e. any ball centered on a point in $Y$ has nonempty intersection with $X$.) This tells you that $Y \subset \bar{X}$.

Now you must show that $Y$ is closed. An equivalent definition of the closure of a set $A$ is "the smallest closed set containing $A$". Hence this shows that $\bar{X} \subset Y$, because it is obvious that $X\subset Y$. Combining these results gives $Y = \bar{X} = \partial X$. Good luck!

this set is not closed. A closed set has all its limit points, but this as x aproches 0 you can chose a sequence such that sin(1/x) will approch 0(or any other point in [0,1]) so (0,0) is a limit point of this set but its not in X

Remark: A function is equal to it graph.

(1). A non-empty function whose domain and range are subsets of $\mathbb R$ can never be an open subset of $\mathbb R^2.$ For suppose $(x,y)\in f$ and $B$ is an open ball centered at $(x,y)$ with $B\subset f.$ Then for some $r>0 $ we have $$\{x\}\times [-r+y,r+y]\subset B\subset f,$$ which implies that $f$ is not a function.

(2). The same argument shows that $f$ has empty interior.

(3)....$\{0\}\times [-1,1] \subset Cl(\{(x,\sin 1/x):x>0\}).$ For $y=\arcsin t\in [-1,1]$ with $t\in [-\pi /2,\pi /2] $, consider $\{(x_n,\sin 1/x_n):n\in \mathbb N\}$ where $x_n=(2\pi n+t)^{-1}.$ So $X$ is not closed.