Prove that $A=\{x \in \mathbb{R} \ |\ \exists n\in \mathbb{N},\ x=2+\frac{1}{n}\}$ has 2 as its infimum.

To prove that $ 2 $ is the infimum of the set $ A = \left\{ x \in \mathbb{R} \mid \exists n \in \mathbb{N}, x = 2 + \frac{1}{n} \right\} $, we need to show two things:

1. **$ 2 $ is a lower bound of $ A $.**
2. **For any $ \epsilon > 0 $, there exists an element $ a \in A $ such that $ 2 \leq a < 2 + \epsilon $.**

### Step 1: $ 2 $ is a lower bound of $ A $.

Each element of the set $ A $ is of the form $ x_n = 2 + \frac{1}{n} $ for some $ n \in \mathbb{N} $. Since $ n $ is a natural number, $ n \geq 1 $, which implies that:

\[
x_n = 2 + \frac{1}{n} > 2.
\]

Thus, for every $ x_n \in A $, we have $ x_n \geq 2 $. Therefore, $ 2 $ is a lower bound of the set $ A $.

### Step 2: $ 2 $ is the greatest lower bound (infimum).

To show that $ 2 $ is the greatest lower bound, we need to show that for any $ \epsilon > 0 $, there exists some $ x_n \in A $ such that $ 2 \leq x_n < 2 + \epsilon $.

Given any $ \epsilon > 0 $, consider the inequality:

\[
x_n = 2 + \frac{1}{n} < 2 + \epsilon.
\]

This simplifies to:

\[
\frac{1}{n} < \epsilon,
\]

which is equivalent to:

\[
n > \frac{1}{\epsilon}.
\]

Since $ n \in \mathbb{N} $, we can always find some $ n $ that satisfies this condition. For instance, choose $ n > \frac{1}{\epsilon} $, and then $ x_n = 2 + \frac{1}{n} $ will satisfy $ 2 \leq x_n < 2 + \epsilon $.

Thus, for any $ \epsilon > 0 $, there exists an element $ x_n \in A $ such that $ x_n $ is arbitrarily close to $ 2 $ but greater than or equal to $ 2 $. This shows that $ 2 $ is the greatest lower bound of the set $ A $.

### Conclusion:

Since $ 2 $ is a lower bound of $ A $ and for any $ \epsilon > 0 $, there is an element in $ A $ that is within $ \epsilon $ of $ 2 $, it follows that $ 2 $ is the infimum of $ A $.

### Do you want more details or have any questions?

Here are 5 related questions:
1. Can we find the supremum of the set $ A $?
2. What happens if $ A $ is defined as $ \{ x \in \mathbb{R} \mid \exists n \in \mathbb{Z}^+, x = 2 - \frac{1}{n} \} $?
3. How does the infimum differ from the minimum of a set?
4. Can you give an example where the infimum is not a member of the set?
5. What is the infimum of $ \{ 1 + \frac{1}{n} \mid n \in \mathbb{N} \} $?

**Tip:** The infimum of a set does not necessarily have to belong to the set, unlike the minimum which must be an element of the set.

Learn how to prove that the set A = {x ∈ R | ∃n ∈ N, x = 2 + 1/n} has an infimum of 2. This problem involves key real analysis concepts such as bounded sets and infimum properties. Ideal for undergraduate students studying real analysis.

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