# Boundary of Set defined with $\sin(\frac{1}{x})$

Let $$S=\{(x,y) \in (0,\infty) \times (-1,\infty) \mid y \geq \sin(\frac{1}{x})\}$$ in $$(\mathbb{R}^2,\lVert \cdot \rVert_\infty)$$. I think that the boundary $$\partial S= \{ (0,y) \mid y \in (-1, \infty) \} \cup \{ (x,y) \mid x \in (0, \infty), y =\sin(\frac{1}{x}) \}$$, but I'm not sure about this as $$\lim_{x \to 0} \sin(1/x)$$ is undefined. And how can I prove, that $$\forall x \in \partial S \; \forall \varepsilon > 0 \; \exists y_1,y_2 \in B(x,\varepsilon) : y_1 \in S, y_2 \in S^\complement$$?

Yes, that's the boundary.

If $$x=(x_1,x_2)$$ belongs to that set, there are two possibilities:

1. $$x_1=0$$: then $$x=(0,x_2)$$ with $$x_2\geqslant-1$$. If you take $$\varepsilon>0$$, then $$B(x,\varepsilon)$$ will contain points whse first coordinate is negative, and those points do not belong to $$S$$. If $$x_2>1$$, $$\left(\frac\varepsilon2,x_2\right)\in S\cap B(x,\varepsilon)$$. And if $$x_2\in[-1,1]$$, then since any number from $$[-1,1]$$ is the limit of some sequence $$\left(\sin\left(\frac1{x_n}\right)\right)_{n\in\mathbb N}$$, the ball $$B(x,\varepsilon)$$ will contain elements of $$S$$.
2. $$x_1>0$$: Then $$x_2=\sin\left(\frac1{x_1}\right)$$. Then $$(x_1,x_2-\frac\varepsilon2)\in S^\complement\cap B(x,\varepsilon)$$ and $$\left(x_1,x_2+\frac\varepsilon2\right)\in S\cap B(x,\varepsilon)$$.
• Thank you very much! Is there any way to prove, that the boundary is'nt a superset of $\partial S$? – Tim May 12 '19 at 12:06
• Sure. For each $x$ outside the set that you described, prove that $x\in\mathring S$ or that $x\in\mathring{S^\complement}$. It's not particulary hard. – José Carlos Santos May 12 '19 at 12:09
• Am I thinking unnecessary complicated or is my answer below the right way to do so? – Tim May 12 '19 at 16:28
• It looks correct to me. – José Carlos Santos May 12 '19 at 16:34

For proving, that $$\partial S$$ is a subset of the boundary I have to show: Let $$x \in (\partial S)^C$$, it is $$x \in S^o$$ or $$x \in (S^C)^o$$.

1. $$x_1 < 0, x_2 \in \mathbb{R}$$. $$\Rightarrow x \in S^C$$. Let $$y \in B(x, \varepsilon)$$ with $$\varepsilon=\frac{|x_1|}{2}$$. $$y_1 < x_1+\varepsilon < 0$$. $$\Rightarrow y \in S^C$$.
2. $$x_1 = 0, x_2<-1$$. $$\Rightarrow x \in S^C$$. Let $$y \in B(x, \varepsilon)$$ with $$\varepsilon=\frac{|x_2+1|}{2}$$. $$y_2 < x_2+\varepsilon < 1$$. $$\Rightarrow y \in S^C$$.
3. $$x_1 > 0, x_2 > \sin(\frac{1}{x_1})$$. $$\Rightarrow x \in S$$. Let $$y \in B(x, \varepsilon)$$ with $$\varepsilon=\min\{ \frac{|x_1|}{2}, \frac{1}{2}\min_{x>0} \{ \sqrt{(x-x_1)^2+(\sin(\frac{1}{x})-x_2)^2} \} \}$$.$$y_1 > x_1-\varepsilon > 0$$. $$y_2 > x_2-\varepsilon \geq \sin(\frac{1}{y_1})$$. $$\Rightarrow y \in S$$.
4. $$x_1 > 0, x_2 < \sin(\frac{1}{x_1})$$. $$\Rightarrow x \in S^C$$. Let $$y \in B(x, \varepsilon)$$ with $$\varepsilon=\min_{x>0} \{ \sqrt{(x-x_1)^2+(\sin(\frac{1}{x})-x_2)^2} \}$$. $$y_2 < x_2+\varepsilon \leq \sin(\frac{1}{y_1})$$. $$\Rightarrow y \in S^C$$.

Can I simplify the cases in any way?