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.