Is the set defined by y=sin(1/x) open or closed? Let $X = \{(x,y) \in \mathbb{R^2}: x > 0, y = sin(1/x)\}$. Is this set open, closed, or neither, and what is the boundary, interior, and closure of $X$?
I just want to make sure I am thinking about these problems in a reasonable manner. I am also trying to apply this to other concepts I have learned.
Is $X$ open?
X is open if every $x \in X$ is an interior point. Take an arbitrary $p \in X$ so that $p = (p_x, p_y)$ then $p_x > 0$ and $p_y = sin(1/p_x)$. Define an open ball $B_r(p)$, and take $q \in B_r(p)$ be an arbitrary point. Is $q \in X$? Is it possible for $q$ to not be in X? Since $y = sin(1/x)$ is defined and is continuous for all $x \neq 0$,then there will always exist a $q_y = sin(1/q_x)$, provided $q_x \neq 0$. How do I know $q_x \neq 0$? Otherwise, I propose that X is open. 
Alternatively, X is open if whenever $f: \mathbb{R^2 \to R}$ is continuous and $U \subseteq R$ so that if $U$ is open, then $f^{-1}(U)$ is also open. Could I construct such a $U$ to prove that it's open? My only concern is that if $U = (-1,1)$, then $f^{-1}(U) = (-\infty, \infty) = \mathbb{R}$ which is not open nor closed. 
Is $X$ closed? 
I am not sure about this one. I want to say that is not closed, since $X^c$ is not open. It is not open because $(0,0)$ defines a boundary point which lies in the set $X^c$.
Thanks!
 A: "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!
A: 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  
A: 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.
