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Let $\displaystyle A_n=\left\{(x,y)\in\mathbb{R}^2:2+\frac1n\le|x|+|y|<3+\frac{1}{n^2}\right\}\cup\left\{\left(7,6+\frac1n\right)\right\}$ and $\displaystyle A=\bigcup_{n=1}^\infty A_n.$

Since $2+1/n>2+1/(n+1)$, $3+1/(n+1)^2<3+1/n^2$ and the $(7,6+1/n)$ are isolated points, am I right in stating $$A=\left\{(x,y)\in\mathbb{R}^2:2<|x|+|y|<4\right\}\cup\bigcup_{n=1}^\infty\left\{\left(7,6+\frac1n\right)\right\}?$$From this I would get $A^0=A\setminus\bigcup_{n=1}^\infty\left\{\left(7,6+1/n\right)\right\},$ $A'=\left\{(x,y)\in\mathbb{R}^2:2\le|x|+|y|\le4\right\}$, $∂A=\left\{(x,y)\in\mathbb{R}^2:|x|+|y|=2 \ \text{or} \ |x|+|y|=4 \right\}$, correct?

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    $A^0$: yes. $\partial A$: no. What does $A'$ stand for? – 2017-02-09
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    @zipirovich: Thanks for answering, I guess $A$ is right. $A'$ is the set of accumulation points. As for $∂A$, what is it then? I'm not getting it – 2017-02-09

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$A^0$: correct.

$\partial A$: no. The $\left\{(x,y)\in\mathbb{R}^2:|x|+|y|=2 \text{ or } |x|+|y|=4 \right\}$ part is correct, but there's more. By definition, a boundary point of $A$ is a point $p$ such that any open neighborhood of $p$ contains points both in $A$ and not in $A$. Note that each of the points $(7,6+1/n)$ satisfies this condition. Moreover, along the line $x=7$ they converge (or accumulate — hold this thought!) towards the point $(7,6)$, which also satisfies this condition.

Similarly for $A'$: your answer is almost correct, but it's missing one more point.

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    Damn $(7,6)$! :) Thank you. – 2017-02-09