https://file.aiquickdraw.com/mathbot/file/1728063867051gwzk9gt6.jpg

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.

If I_n = (-1/n, 1/n) for n ∈ N, find the set ∩∞n=1 I_n. Show that inf{1/2^n : n ∈ N} = 0.

Learn how to find the intersection of an infinite set of open intervals I_n = (-1/n, 1/n) and show that the infimum of the set {1/2^n : n ∈ N} is 0. This problem focuses on core concepts in real analysis, including set theory and infimum properties.

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