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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?

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  • $\begingroup$ See math.stackexchange.com/questions/108709/… $\endgroup$ – Robert Z Dec 11 '17 at 13:15
  • $\begingroup$ I would argue that the set is ill-defined and that the definition should say: $\left \{ (x,y)\in \mathbb{R}^2:y=x\sin \frac{1}{x},x\in\mathbb{R}\setminus\{0\} \right \}$ $\endgroup$ – user491874 Dec 11 '17 at 13:30
  • $\begingroup$ @user8734617..you are right $\endgroup$ – Inverse Problem Dec 11 '17 at 13:35
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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 $(0,0)$.

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  • $\begingroup$ yea i think then the set is not closed what about open? $\endgroup$ – Inverse Problem Dec 11 '17 at 13:33
  • $\begingroup$ @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$=0$ $\endgroup$ – user464147 Dec 11 '17 at 13:54
  • $\begingroup$ @AbhishankaSaha I might miss something. can you please help me. $\endgroup$ – user464147 Dec 11 '17 at 14:03
  • $\begingroup$ See that $y=x\sin\frac1x$ is undefined at $x=0$. So no point of the form $(0,a)$ can be in the set $\endgroup$ – Abishanka Saha Dec 11 '17 at 14:14
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Set is not open. Please see the figure. Apply the definition of open set enter image description here

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