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Well, if we have a set $A = \{n + \frac{1}{k}, k,n \in \mathbb{N}\}$ and the task is to find boundary, closure, interior and set of limit points of it in given space:

  1. $\mathbb{R}$ with euclidean metric
  2. $\mathbb{R}$ with discrete metric

Starting off with the first one, is the following correct:

  • the set of limit points $A'$ is basically every $n \in \mathbb{N}$, since each of them has a sequence $\frac{1}{k}$ converging to it

  • every point in the interior has to be there with some open ball - but for arbitrary $\epsilon$ only $z \in \mathbb{N}$ have open balls with them, so the int $A$ is the same as the set of limit points

  • the closure $\overline{A}$ = int $A$

  • the boundary is $\emptyset$

Is that a proper solution? How does that change when the metric is discrete?

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  • $\begingroup$ How do you conclude that integers have open balls around them? I agree with what $A'$ is in the first case bu disagree with everething else. Notice also that $\overline{A}\supseteq A\supseteq \text{int}(A)$. $\endgroup$ – Keen-ameteur Mar 27 at 14:52
  • $\begingroup$ The discrete case is simpler I think, since singletons are open in that case and convergence is trivial there. $\endgroup$ – Keen-ameteur Mar 27 at 14:54
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    $\begingroup$ Nikita. In your first line you have $n\in \mathbb{N}$. First bullet: You say every $z\in \mathbb{Z}$, is this ok? $\endgroup$ – Peter Szilas Mar 27 at 15:17
  • $\begingroup$ Thanks for pointing out - I have edited my (wrong) attempt to be more correct. I am ashamed though as the task was so trivial after the answers from two guys below. $\endgroup$ – Никита Васильев Mar 27 at 15:26
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    $\begingroup$ Nikita.No worries. A little practice will make it easier:) $\endgroup$ – Peter Szilas Mar 27 at 15:40
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The limit points of $A$, denoted $A'$, you have identified (almost) correctly as $\Bbb N$ (not $\Bbb Z$, unless you meant $n \in \Bbb Z$ in your original problem); the closure thus equals, as always, $A \cup A' = A \cup \Bbb N$. The interior is empty as no open interval can sit inside the countable set $A$.

Always $\partial A = \overline{A}\setminus A^\circ = \overline{A}=A \cup \Bbb N$.

In the discrete topology always $\overline{A} = A^\circ$ for any $A$ and thus $A'=\emptyset = \partial A$.

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Every convergent sequence of elements of $A$ converges to an element of $A$ or to a natural number. Therefore, $\overline A=A\cup\mathbb N$. And $\mathring A=\emptyset$, since $A$ contains no interval. So, $\partial A=\overline A\setminus\mathring A=A\cup\mathbb N$. Finally, yes, the set of limit points of $A$ is $\mathbb N$.

Things are simpler with respect to the discrete metric, since then every set is both closed an open. So, $\overline A=\mathring A=A$. In particular, $\delta A=\emptyset$. And in a discrete metric space, the set of limit points of any subset is empty.

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  • $\begingroup$ $\overline A = A \cup \Bbb N$, not $A$. $A' \subseteq \overline{A}$. $\endgroup$ – Henno Brandsma Mar 27 at 15:11
  • $\begingroup$ I've edited my answer. Thank you. $\endgroup$ – José Carlos Santos Mar 27 at 15:20
  • $\begingroup$ Carlos. $\partial A =\overline{A}$ , not $A$? $\endgroup$ – Peter Szilas Mar 27 at 15:57
  • $\begingroup$ @PeterSzilas Sure! I've edited my answer (again). Thank you. $\endgroup$ – José Carlos Santos Mar 27 at 15:58

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