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.

  • $\begingroup$ $\overline{A} = A\cup \{1\}$ and $\partial A = \overline{A}\setminus \text{Int}(A) = \overline{A}$. Note that $1$ is limit of $a_n = \frac{n}{n+1}$. $\endgroup$ – Jakobian Jul 10 at 17:20
  • $\begingroup$ Could you explain more clearly why its closure is $A \cup \lbrace{ 1 \rbrace}$ $\endgroup$ – Nguyen Thy Jul 10 at 17:25
  • 1
    $\begingroup$ 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 $a_n$. $\endgroup$ – 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 $A$.

The boundary is closure minus interior, so just the same as the closure.


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