I have to prove that $U = \mathbb{R}^2\setminus \{(0,0)\}$ is open in $\mathbb{R}^2$. I know that $U$ is open if $U = \operatorname{inw}(U)$ (collection of all interior points of $U$) (this is the Dutch notation, not sure about the English one). I know the definition of a interior point too, but I don't know how I can solve this question. I think you can use the balls to cover $U$ including the point $(0,0)$?
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1$U$ is the complement of a closed set. Also you could construct for each $u\in U$ a ball $B_r(u)$ such that $B_r(u)\subset U$. Or try to solve this for $\mathbb R\setminus{0,0}$. – daw Dec 18 '20 at 09:29
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1For any $u \in U$ the open ball $B(u,|u|) $ is contained in $U$. – Kavi Rama Murthy Dec 18 '20 at 09:32
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3A subset of $\mathbf R ^{\color{red}2}$ open in $\mathbf R$??? – Bernard Dec 18 '20 at 09:52
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Hint: For each $(x,y)\neq(0,0)$ in $\mathbb R^2$ the open ball of radius $\sqrt{x^2+y^2}$ centered at $(x,y)$ does not contain the origin.
Christoph
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Points in $\mathbb{R}^2 $ with the euclidean topology are closed sets (every $T_1$ space satisfies this claim). The complement of a closed set is open by definition.
Crash Bandicoot
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