Given that $S$ is defined as a subset of $\mathbb{R}$, let's break down the set $S$ and calculate the required properties.

Problem Setup

For $n \geq 1$, let:

$A_n = \left[\frac{1}{2n+1}, \frac{1}{2n}\right]$

The set $S$ is defined as:

$S = \bigcup_{n=1}^{\infty} A_n$

This means that $S$ is the union of the intervals $A_n$.

1. The Interior of $S$, $\text{int}(S)$

The interior of a set is the largest open set contained within $S$. Each $A_n$ is a closed interval, and since these intervals are disjoint, there is no open set contained entirely within $S$. Therefore, the interior of $S$ is empty:

$\text{int}(S) = \emptyset$

2. The Exterior of $S$, $\text{ext}(S)$

The exterior of $S$ consists of all points that are not in the closure of $S$. Since $S$ is made up of closed intervals that are isolated from each other, the exterior of $S$ is everything in $\mathbb{R}$ except the points in these intervals:

$\text{ext}(S) = \mathbb{R} \setminus S$

3. The Complement of $S$, $\mathbb{R} \setminus S$

The complement of $S$ is the set of all points in $\mathbb{R}$ that are not in $S$. Since $S$ consists of a union of closed intervals, the complement will consist of all the points outside these intervals. This can be written as:

$\mathbb{R} \setminus S = \bigcup_{n=1}^{\infty} \left( \left( \frac{1}{2n}, \frac{1}{2n-1} \right) \cup (-\infty, 0) \cup (1, \infty) \right)$

4. The Boundary of $S$, $\text{bd}(S)$

The boundary of a set is the set of points that can be approached both from within the set and from outside the set. For each interval $A_n = \left[\frac{1}{2n+1}, \frac{1}{2n}\right]$, the boundary consists of the two endpoints $\frac{1}{2n+1}$ and $\frac{1}{2n}$. Therefore, the boundary of $S$ is:

$\text{bd}(S) = \left\{ \frac{1}{2n+1}, \frac{1}{2n} \mid n \geq 1 \right\}$

5. The Derived Set $S'$

The derived set $S'$ consists of the limit points of $S$. A limit point of $S$ is a point where every neighborhood around that point contains at least one point of $S$ distinct from the point itself. In this case, each interval $A_n$ consists of isolated points, meaning there are no accumulation points. Therefore, the derived set $S'$ is empty:

$S' = \emptyset$

6. The Isolated Points of $S$

The isolated points of $S$ are those points in $S$ that do not have any other points of $S$ arbitrarily close to them. Since $S$ consists of disjoint intervals, every point in $S$ is isolated. Therefore, the set of isolated points of $S$ is:

$\text{Isolated Points of } S = S$

7. The Closure of $S$, $\text{cl}(S)$

The closure of $S$ is the smallest closed set containing $S$, which consists of all the points in $S$ together with the limit points of $S$. Since there are no limit points, the closure of $S$ is simply $S$ itself:

$\text{cl}(S) = S$

Summary of Results:

1. Interior of $S$: $\text{int}(S) = \emptyset$
2. Exterior of $S$: $\text{ext}(S) = \mathbb{R} \setminus S$
3. Complement of $S$: $\mathbb{R} \setminus S = \text{All points outside the intervals } A_n$
4. Boundary of $S$: [ \text{bd}(S) = \