For the set A, I have the following thoughts that I want to verify:
A is neither closed nor open on $R$. It isn't open as take for example $x = 1.5, r = 0.1$, then $x \in A$ but $B_r(x)$ (open ball centered at x) is not a subset of A. A is not closed as it doesn't contain all of it's limit points, for example, it doesn't contain $x=\{ 1\}$.
The limit points of A are $[0,1]$
int $A$ = $(0,1)$, (closure) $\overline{A} = [0,1]$, $\partial A = \{ 0, 1\} \cup \{ 1 + 1/n : n \in N\}$
If we then consider the set $A$ relative to $(0, \infty)$ instead, then the limit points are instead $(0,1]$, the interior points remain the same, the closure is $(0,1]$, and $\partial A = \{ 1\} \cup \{ 1 + 1/n : n \in N\}$
I'm wondering if this is correct, or if there are any issues with my reasoning?
typo which was pointed out in one of the answers
In the first case, it should be : $\overline{A} = [0,1]\cup \{ 1 + 1/n : n \in N\}$
and in the second: $\overline{A} = (0,1]\cup \{ 1 + 1/n : n \in N\}$
The closure of $A$ always contains all elements of $A$ (not only the limitpoints).
We have $\overline A=A\cup A'$ where $A'$ denotes the collection of limit points of $A$.
So also the elements $1+\frac1n$ in this case.
Further it is correct.