# Determine whether the set is closed or open ,both, or neither

Determine whether the set is closed or open ,both, or neither $$\left \{ (x,y)\in \mathbb{R}^2:y=x\sin \frac{1}{x},x\in\mathbb{R} \right \}$$

my idea:

since set of derived points is equal to given set hence the set is closed

is I am right what bout open..is this set is open?

• – Robert Z Dec 11 '17 at 13:15
• I would argue that the set is ill-defined and that the definition should say: $\left \{ (x,y)\in \mathbb{R}^2:y=x\sin \frac{1}{x},x\in\mathbb{R}\setminus\{0\} \right \}$ – user491874 Dec 11 '17 at 13:30
• @user8734617..you are right – Inverse Problem Dec 11 '17 at 13:35

Note that $x\sin\frac1x$ is a continuous function for $x\in(-\infty,0)\cup(0,\infty)$. If $x\ne0$, then for any sequence $x_n$ converging to $x$, $(x_n,x_n\sin\frac1{x_n})$ converges to $(x,x\sin\frac1x)$. We just need to compute the limits of this function for $x\rightarrow0$. Show that $\lim_{x\rightarrow0}x\sin\frac1x=0$. This $(0,0)$ is a limit point of this set. But $(0,0)$ doesn't lie in this set. Thus the set contains all its limit points except $(0,0)$.
• @AbishankaSaha What is the problem of sitting $(0,0)$ inside the set? $y=x\sin(\frac{1}{x}), x=0$ and y must be zero right? $x\sin(\frac{1}{x})=0\times$ a finite number$=0$ – user464147 Dec 11 '17 at 13:54
• See that $y=x\sin\frac1x$ is undefined at $x=0$. So no point of the form $(0,a)$ can be in the set – Abishanka Saha Dec 11 '17 at 14:14