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I am trying to understand closed sets.

Is the set $$ A =\{x \in \mathbb{R}^2: \lVert x \rVert_2 \leq 1 \} -\{0\} $$ closed?

I think it is not, because if we take the sequence $(\frac{1}{n})_{n=1}^{\infty}$ we can see it lies in A, but its limit is zero which is not in the set A. And because of that, A is not a closed set.

Is this right? Thank you.

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    $\begingroup$ Yes, correct, it's not closed for the reason you give. Nor is it open: there is no open ball of positive radius about e.g. $(1,0)$ contained in $A$. $\endgroup$ Commented Feb 5, 2021 at 15:08
  • $\begingroup$ Yes, assuming $||x||_2$ means the euclidean norm (so $A$ is the unit disc minus the origin point) then $0$ is a limit point of $A$ (as are all points in the unit disk) but as $0$ is not in $A$ it is not true that all the limit points of $A$ are in $A$ ($0$ is not but all the others are) so $A$ is not closed. $\endgroup$
    – fleablood
    Commented Feb 5, 2021 at 22:29

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It is almost correct, but you missed the fact that $(\forall n\in\Bbb N):\frac1n\notin\Bbb R^2$. Take the sequence $\left(\left(\frac1n,0\right)\right)_{n\in\Bbb N}$ instead.

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Another way to see it is via the fact that closed sets are (perhaps defined as, depends on your definition of open set) the complements of open sets. The complement of $A$ is not open because there is no open ball of positive radius contained in $A^\complement$ around $(0,0)$

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