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Define the two sets $$ U:=\left\{(x,y)\in \mathbb{R}^2:\frac{1}{\sqrt{2}}<x^2+y^2\right\}\\[20pt] M:=\{(x,y)\in \mathbb{R}^2:x^2+y^2+\sin(x)=2\} $$

Then $ U $ is an open surrounding of $ M $.

I plotted both sets in Desmos to get an impression of these sets. My first attempt was to choose for each point $ a\in M $ an open ball

$$ U_{\frac{1}{10}}(a):=\left\{v\in \mathbb{R}^2:\|v-a\|_2<\frac{1}{10}\right\} $$

with radius $ \frac{1}{10} $.

But I already fail to show for all $ a:=(a_1,a_2)\in M $ the relation $ U_{\frac{1}{10}}(a)\subseteq U $. So if I take an arbitrary $ v:=(v_1,v_2)\in U_{\frac{1}{10}}(a) $ then I get at first $$ a_1^2+a_2^2+\sin(a_1)=2\\v_1^2+v_2^2+a_1^2+a_2^2-2(v_1a_1+a_2v_2)=(v_1-a_1)^2+(v_2-a_2)^2<\frac{1}{100}. $$

And I want to get the relation $ v_1^2+v_2^2>\frac{1}{\sqrt{2}} $. But how?

J. W. Tanner
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hallo007
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  • It might help to solve two sub-problems: (1) $M$ is a subset of $U$. (2) $U$ is open. Try working with each sub-problem separately, instead of trying to handle both at the same time. – leslie townes Feb 04 '24 at 21:44
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    By surrounding, do you mean neighborhood? – J. W. Tanner Feb 04 '24 at 23:37
  • Ok if I'm not wrong I could find a way: I consider an arbirtrary point $ a\in M $ and fix it. Take annother arbitrary point $ v $ of my open ball and I consider the estimation $ |a|_2=|a-v+v|_2\leq |v-a|_2+|v|_2<\frac{1}{10}+|v|_2 $. With $ a\in M $ I get $ a_1^2+a_2^2=2-\sin(a_1)\geq 2-1=1>\frac{1}{\sqrt{2}} $ which shows also $ M\subseteq U $. So I get $ |v|_2>|a|_2-\frac{1}{10}\geq 1-\frac{1}{10}=\frac{9}{10} $. So I have $ |v|_2^2>\frac{81}{100}>\frac{71+\frac{3}{7}}{100}=\frac{\frac{500}{7}}{100}=\frac{10}{14}=\frac{1}{1.4}=\frac{1}{\sqrt{1.96}}>\frac{1}{\sqrt{2}} $ – hallo007 Feb 06 '24 at 01:05

1 Answers1

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I think it's easy to look in the other way. Let $B = \mathbb{R}^2 - U$. Notice that $U$ is the set of points $(x,y)\in\mathbb{R^2}$ such that $|(x, y)| > r$, where $r = \sqrt{\frac{1}{\sqrt{2}}}$, so that $B$ will be the set of points which $|(x,y)| \leq r$, i.e. a closed ball of radius $r$ centered at the origin. Hence $B$ is a closed set and $U$, as it's complement, is an open set.

To show that $U$ is surrounding $M$, one just need to notice that any point $(x, y)\in M$ satisfies $|(x, y)|^2 = 2 - \sin x > 2 - 1 = 1 > r^2$. So, any point in $M$ has it's norm greater than $r$ and, therefore, belongs to $U$. This proves that $M\subset U$.

Paulo Estêvão
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