Determine limit points, isolated points and boundary points of $S=\{0\}\cup \{1,1/2,1/3,\dots \}$. In the set $S=\{0\}\cup \{1,1/2,1/3,\dots \}$, each of the points $1/k$ is an isolated point, but $0$ is not an isolated point because there are other points in $S$ as close to $0$ as desired. Then it is a limit point. What are the boundary points? I think they are all the real numbers between $0$ and $1$, i.e. the range $[0,1]$. Is it true? Is $S$ a close set? Is $S$ open?
Moreover, what are the boundary points of $\{0\}\cup [1,2]$?
 A: Yes, $S$ is even compact so closed. Being countable it has empty interior so all points of $S$ are boundary points ( closure minus interior).
A: I assume both $S$ and $T=\{0\}\cup [1,2]$ are subsets of $\Bbb R$ endowed with the usual topology.
By definition, a boundary $\partial A$ of a set $A$ equals $\overline{A}\setminus\operatorname{int} A$. It is easy to check that the set $S$ is closed, it has the only limit point $0$, which is already in $S$ (in fact, $S$ is closed being a compact subset of a Hausdorff space $\Bbb R$). That is $\overline{S}=S$. On the other hand,  the set $S$ contains no non-empty open intervals, so the interior $\operatorname{S}$  is empty. That is $\partial S= \overline{S}\setminus\operatorname{int} S=S$.
It is easy to see that the set $T$ has the open complement, so $T$ is closed ($T$ is also closed being a compact).  On the other hand,
$\operatorname{int} T$ is the union ol all open intervals containing in $S$, that is  $\operatorname{int} T=(1,2)$. That is $\partial T= \overline{T}\setminus\operatorname{int} T=(\{0\}\cup [1,2])\setminus (1,2)=\{1,2,3\}$.
