# Determine the interior, closure and boundary of a set under usual topology

Let $$(\mathbb{R},\mathcal{T})$$ be a topological space and $$\mathcal{T}$$ is an usual topology. Determine the interior, the closure and the boundary of the set $$A=\lbrace{ \frac{n}{n+1}, n \in \mathbb{N}\rbrace}$$ in $$\mathbb{R}$$ with $$\mathbb{N}=\lbrace{ 0,1,\cdots \rbrace}$$.

I know its interior is empty set because there is no open set in A. But how about its closure and boundary ? Thank you.

• $\overline{A} = A\cup \{1\}$ and $\partial A = \overline{A}\setminus \text{Int}(A) = \overline{A}$. Note that $1$ is limit of $a_n = \frac{n}{n+1}$. – Jakobian Jul 10 at 17:20
• Could you explain more clearly why its closure is $A \cup \lbrace{ 1 \rbrace}$ – Nguyen Thy Jul 10 at 17:25
• Closure of $A$ divides into two parts. There's isolated points of $A$ (which belong to $A$), so there is a neighbourhood of them which contains only that point, or limits points of $A$, so for every neighbourhood there is infinitely many points contained in it from $A$. Every point in $A$ is isolated, it's easy to see that. And it's easy to see that in this case, limit points of $A$ coincide with limit points of sequence $a_n = \frac{n}{n+1}$. That is, the only limit point which is the limit of $a_n$. – Jakobian Jul 10 at 17:56

$$A$$ is countable so does not contain an open interval/ball. So $$\operatorname{int}(A)=\emptyset$$.
$$\overline{A}=A \cup \{1\}$$, as the sequence $$\frac{n}{n+1}$$ converges to $$1$$, so the latter is a limit point of $$A$$.