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 $\forall x \in \partial S \; \forall \varepsilon > 0 \; \exists y_1,y_2 \in B(x,\varepsilon) : y_1 \in S, y_2 \in S^\complement$?
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Yes, that's the boundary.
If $x=(x_1,x_2)$ belongs to that set, there are two possibilities:
- $x_1=0$: 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 $S$.
- $x_1>0$: Then $x_2=\sin\left(\frac1{x_1}\right)$. Then $(x_1,x_2-\frac\varepsilon2)\in S^\complement\cap B(x,\varepsilon)$ and $\left(x_1,x_2+\frac\varepsilon2\right)\in S\cap B(x,\varepsilon)$.
answered May 12 '19 at 11:01
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$\begingroup$ Thank you very much! Is there any way to prove, that the boundary is'nt a superset of $\partial S$? $\endgroup$ – Tim May 12 '19 at 12:06
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$\begingroup$ 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. $\endgroup$ – José Carlos Santos May 12 '19 at 12:09
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$\begingroup$ Am I thinking unnecessary complicated or is my answer below the right way to do so? $\endgroup$ – Tim May 12 '19 at 16:28
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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 $x \in (S^C)^o$.
- $x_1 < 0, x_2 \in \mathbb{R}$. $\Rightarrow x \in S^C$. Let $y \in B(x, \varepsilon)$ with $\varepsilon=\frac{|x_1|}{2}$. $y_1 < x_1+\varepsilon < 0$. $\Rightarrow y \in S^C$.
- $x_1 = 0, x_2<-1$. $\Rightarrow x \in S^C$. Let $y \in B(x, \varepsilon)$ with $\varepsilon=\frac{|x_2+1|}{2}$. $y_2 < x_2+\varepsilon < 1$. $\Rightarrow y \in S^C$.
- $x_1 > 0, x_2 > \sin(\frac{1}{x_1})$. $\Rightarrow x \in S$. Let $y \in B(x, \varepsilon)$ with $\varepsilon=\min\{ \frac{|x_1|}{2}, \frac{1}{2}\min_{x>0} \{ \sqrt{(x-x_1)^2+(\sin(\frac{1}{x})-x_2)^2} \} \}$.$y_1 > x_1-\varepsilon > 0$. $y_2 > x_2-\varepsilon \geq \sin(\frac{1}{y_1})$. $\Rightarrow y \in S$.
- $x_1 > 0, x_2 < \sin(\frac{1}{x_1})$. $\Rightarrow x \in S^C$. 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?
