# Sequence with distinct points: derived set in the discrete metric

Let $A = \{1/n \mid \text{$n$is a positive integer}\}$ be a subset of $(\mathbb{R}, d)$. Find the derived set, the interior, the closure and the boundary of $A$ in the usual metric and the discrete metric.

I think they are $\{0\}$, empty set, $A$ and $A$ in the usual metric; in the discrete metric they should be empty set, $A$, $A$ and empty set.

Am I right?

If it's correct, then how can I show it? If not what is incorrect?

Usual metric: $0$ is in the derived set, so it also belongs to the closure (hence the closure of $A$ is not $A$). Does the set $A$ contain any open interval? So, what's the interior? Once you know the closure and the interior, can you tell what's the boundary?

Discrete metric: every set is open and closed. So, for instance, every set has empty boundary and no point is an accumulation point of any set.

Always $\overline{A} = A\cup A'$ so we get that the closure of $A$ in the standard topology includes $0$ as well.

Always $\operatorname{Bd}(A) = \overline{A} \setminus \operatorname{int}(A)$, so the boundary equals the closure in the standard topology.

For the discrete topology, all sets are open and closed, and $A' = \emptyset$ always. So $\overline{A} = \operatorname{int}(A) = A$,$\operatorname{Bd}(A) =\emptyset$.

I think you are close, except for needing to add $\{0\}$ to the closure and boundary in the usual topology... Indeed, it ($\{0\}$) is the derived set of $A$ and part of that of $\mathbb R\setminus A$ (in the usual topology)...

As far as how to show it, $a_n=\frac1n$is clearly a sequence in $A$ converging to $0$.

In the discrete topology,on the other hand, points are open, so int$A=A$. The sequence $(a_n)$ doesn't converge... So $\bar A=A.$

In the usual topology $\partial A=\bar A\cap\overline{\mathbb R\setminus A}=A\cup\{0\}$ Whereas in the discrete topolgy, we get $\partial A=A\cap\mathbb R\setminus A=\emptyset$ Etc...