Consider the set S={n+(1/m) : m,n in N}.
show that N subset of S'. Use the definition of a compact set to show that S is not compact.

### Part 1: Show that \(\mathbb{N} \subseteq S'\)

We need to show that \(\mathbb{N} \subseteq S'\), where \(S' = \overline{S} \setminus S\) is the set of all limit points (or accumulation points) of \(S\).

#### Step 1: Understanding \(S\)
The set \(S = \{ n + \frac{1}{m} : m, n \in \mathbb{N} \}\) is the set of all numbers of the form \(n + \frac{1}{m}\), where \(n\) and \(m\) are natural numbers. So, for each \(n \in \mathbb{N}\), there are points like:
\[
n + \frac{1}{1}, n + \frac{1}{2}, n + \frac{1}{3}, \dots
\]
The points \(n + \frac{1}{m}\) approach \(n\) as \(m \to \infty\). Therefore, for every \(n \in \mathbb{N}\), there is a sequence in \(S\) that converges to \(n\).

#### Step 2: Every \(n \in \mathbb{N}\) is a limit point of \(S\)
Let \(n \in \mathbb{N}\). Consider the sequence \( \{ n + \frac{1}{m} \} \subseteq S \), which consists of points in \(S\) and converges to \(n\). Since there is a sequence in \(S\) converging to \(n\), it means that \(n\) is a limit point of \(S\). Hence, \(n \in S'\), the set of limit points of \(S\).

Therefore, we have shown that \( \mathbb{N} \subseteq S' \).

### Part 2: Show that \(S\) is not compact

To show that \(S\) is not compact, we will use the **definition of a compact set**. In metric spaces, a set is compact if it is both:
1. **Closed**: Contains all its limit points.
2. **Bounded**: Can be contained within some large finite interval.

#### Step 1: \(S\) is not bounded
The set \(S = \{ n + \frac{1}{m} : m, n \in \mathbb{N} \}\) contains numbers that get arbitrarily large as \(n\) increases. For example, for any natural number \(n\), the point \(n + \frac{1}{1} = n + 1\) belongs to \(S\). As \(n\) increases, these points become arbitrarily large.

Hence, \(S\) is **not bounded** because it contains numbers that grow without bound as \(n\) increases.

#### Step 2: \(S\) is not closed
We showed earlier that every natural number \(n \in \mathbb{N}\) is a limit point of \(S\). However, no natural number \(n\) is an element of \(S\) because \(S\) contains only numbers of the form \(n + \frac{1}{m}\), and for any \(m \in \mathbb{N}\), \(n + \frac{1}{m} \neq n\).

Therefore, \(S\) does **not** contain all its limit points, meaning \(S\) is **not closed**.

#### Conclusion
Since \(S\) is neither bounded nor closed, it is not compact.

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Would you like further details or have any questions on the steps above?  
Here are 5 related questions to deepen your understanding:

1. What is the formal definition of a limit point in metric spaces?
2. How would you prove that a set is closed in \(\mathbb{R}\)?
3. Can you provide an example of a set that is bounded but not closed?
4. What is the Heine-Borel theorem, and how does it relate to compactness?
5. How does compactness in \(\mathbb{R}\) differ from compactness in general metric spaces?

**Tip:** In \(\mathbb{R}\), a set is compact if and only if it is closed and bounded. This is a special property of Euclidean spaces known as the **Heine-Borel theorem**.

Consider the set S={n+(1/m) : m,n in N}. Show that N is a subset of S'. Use the definition of a compact set to show that S is not compact.

Explore how to prove that the natural numbers (N) are a subset of the limit points of a set S = {n + 1/m : m,n ∈ N}, and demonstrate using the definition of compactness that S is not compact. This solution covers key concepts from real analysis, such as limit points, boundedness, and the Heine-Borel theorem. Suitable for undergraduate students in real analysis.

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