Interior, Limit and closure of Union of Sets

Question

Find the interior, limit and closure of the following union to state whether it is either/neither open and/or closed.

$$ A=\{0,1,\frac{1}{2},\frac{1}{3},...\}\cup \bigcup_{n=1}^{\infty} \left\{ \frac{1}{n}-\frac{1}{n(n+1)2^m} :m\in\mathbb{N} \right\} $$

I can tell that the first sequence has no interior points and {0} as limit point, second sequence as far as I think is interior of (0,1) and limit of [0,1]. However I am lost as to if that is correct and I am merely guessing. I would appreciate an explanation as I am merely trying to understand the interiors, limits, closure and open/closed sets.

$A^\circ = (0,1)$ $A'=[0,1]$ $\overline{A}=[0,1]$ A is open A is not closed

Answer 1

Find the (1) interior, (2) limit, and (3) closure of $A$ to state whether $A$ is either/neither open and/or closed, where

$$ A=\{0,1,\frac{1}{2},\frac{1}{3},...\}\cup \bigcup_{n=1}^{\infty} \left\{ \frac{1}{n}-\frac{1}{n(n+1)2^m} :m\in\mathbb{N} \right\} $$

We assume that $A$ is a subset of the set of real numbers $\mathbb{R}$ with the standard topology.

1. The interior of $A$ is empty because no open interval is included in $A$.

2. The set of all limit points of $A$ is $\{0,1,\frac{1}{2},\frac{1}{3},\dotsc\}$. (Note that $x$ is a limit point of $A$ if every neighborhood of $x$ intersects $A$ in some point other than $x$ itself.)

   It is easy to see that $0$ is a limit point of $A$. Moreover, for each $n\in\mathbb{N}$, the point $\frac{1}{n}$ is the limit of the sequence $\{\frac{1}{n}-\frac{1}{2^m}\frac{1}{n(n+1)}\mid m\in\mathbb{N}\}$.

3. The closure of $A$ is $A$ itself because $A$ contains all limit points of $A$.

Therefore $A$ is closed.