Let's solve the two parts of the problem shown in the image:

1. Finding the intersection of sets $I_n$:

We are given that: $I_n = \left( -\frac{1}{n}, \frac{1}{n} \right)$ for $n \in \mathbb{N}$, and we need to find the intersection of these intervals for all $n \in \mathbb{N}$, i.e.: $\bigcap_{n=1}^{\infty} I_n$

Solution:

• $I_n$ is the open interval $\left( -\frac{1}{n}, \frac{1}{n} \right)$.
• As $n$ increases, the interval $I_n$ becomes smaller. Specifically, $\frac{1}{n} \to 0$ as $n \to \infty$.
• Therefore, as we take the intersection of all these intervals, the limiting interval will be $(0, 0)$.

Since the interval $(0, 0)$ is an empty set (there are no points between 0 and 0), we have: $\bigcap_{n=1}^{\infty} I_n = \{0\}$

2. Showing that $\inf\left\{ \frac{1}{2^n} : n \in \mathbb{N} \right\} = 0$:

We are asked to show that: $\inf\left\{ \frac{1}{2^n} : n \in \mathbb{N} \right\} = 0$

Solution:

• The set $\left\{ \frac{1}{2^n} : n \in \mathbb{N} \right\}$ consists of values like $\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \dots$, which are positive numbers that keep getting smaller as $n$ increases.
• As $n \to \infty$, $\frac{1}{2^n} \to 0$.
• Since the values in the set are strictly positive but approach zero, the greatest lower bound (infimum) of the set is 0.

Thus, we have: $\inf\left\{ \frac{1}{2^n} : n \in \mathbb{N} \right\} = 0$

Would you like further details or explanations?

Here are 5 follow-up questions for practice:

1. How would the result of the intersection change if $I_n$ were closed intervals instead of open intervals?
2. What is the significance of the infimum in the context of a set of real numbers?
3. Can the intersection of an infinite sequence of closed intervals ever be empty?
4. How would the solution change if the intervals $I_n$ were shifted, say $I_n = \left( -\frac{1}{n} + 1, \frac{1}{n} + 1 \right)$?
5. How do we compute the supremum of a bounded set of real numbers?

Tip: The infimum of a set is the greatest number that is less than or equal to every element in the set. It may or may not belong to the set itself.