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A closed unit disk around P is the set of points whose distance from P is less than or equal to 1. In other words, $D(P)=\{Q:|P-Q|\leq 1\}$. Mathematically, the point removing the point $(0, 0)$ does not satisfy this condition and thus the disk is not closed. But intuitively, I am unable to rationalize this.

Any guidance is greatly appreciated!

Edit: My confusion stems from the fact that this new disk is not closed, but is also not open as per my understanding. So I am unable to rationalize the nature of this set.

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    What do you mean by "Mathematically, the point $(0,0)$ does not satisfy this condition"? – azif00 Jul 05 '21 at 22:07
  • You may want to reformulate your question, at this moment it doesn't make any sense. – cos_dm_math21 Jul 05 '21 at 22:09
  • Going from the question in your title: ${(0,0)}$ is a closed set, if you consider the subspace topology of $D \subseteq \mathbb{R}^2$. Thus, the complement $D \setminus {(0,0)}$ is open. – jasnee Jul 05 '21 at 22:11
  • @azif00 Okay that's true of course. Should I still leave my comment as it is, you think? – jasnee Jul 05 '21 at 22:15
  • @aziffo I edited my question to make more sense now - sorry for the confusion. –  Jul 05 '21 at 22:28
  • @cos_dm_math21 just fixed it! –  Jul 05 '21 at 22:28
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    A set may be neither closed nor open. For example, $(1,2]$ as a subset of the real line. – David K Jul 05 '21 at 22:33

2 Answers2

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Have you seen the characterization of closed sets as sets containing all their limit points? If you remove the origin from your closed disc $D$, then any sequence converging to it previously will now converge to a point outside of $D$, thus it is not closed. To give a concrete example, if you take the space $\mathbb{R}^2$ and look at the unit closed disk, you can take the sequence $(x_n)_{n\geq 1}$ where $x_n=(\frac{1}{n},0)$. This sequence converges to $(0,0)$ but it's fully contained in your closed unit disk, showing that puncturing the origin from that disk results in a non-closed set.

Here's a link to the limit points characterization: Prove that a set is closed iff it contains all its accumulation points

Saegusa
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You can just work from the definition. A set $X$ is closed iff the complementary $X^C$ is open. But the complementary of your set is the exterior of the disk and the point $(0, 0)$. This set is not open because any open disk around $(0,0)$ is not contained in $X^C$ (this is, $(0,0)$ is not an interior point of $X^C$). Therefore, $X$ is not closed.

Zanzag
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