# Finding the interior, boundary, closure and set of limit points

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?

• 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 $\overline{A}\supseteq A\supseteq \text{int}(A)$. – Keen-ameteur Mar 27 at 14:52
• The discrete case is simpler I think, since singletons are open in that case and convergence is trivial there. – Keen-ameteur Mar 27 at 14:54
• Nikita. In your first line you have $n\in \mathbb{N}$. First bullet: You say every $z\in \mathbb{Z}$, is this ok? – Peter Szilas Mar 27 at 15:17
• 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. – Никита Васильев Mar 27 at 15:26
• Nikita.No worries. A little practice will make it easier:) – Peter Szilas Mar 27 at 15:40

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 $$A'=\emptyset = \partial A$$.

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.

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