1. **Problem:** Find the union and intersection of the sets $A_n = [3 + \frac{1}{n}, 5 - \frac{1}{n}]$ for $n \in \mathbb{N}$. 2. **Formula and rules:** - The union $\bigcup_{n \in \mathbb{N}} A_n$ is the set of all points that belong to at least one $A_n$. - The intersection $\bigcap_{n \in \mathbb{N}} A_n$ is the set of all points that belong to every $A_n$. 3. **Analyze the intervals:** - As $n$ increases, $\frac{1}{n}$ decreases. - So the left endpoint $3 + \frac{1}{n}$ decreases towards 3 from above. - The right endpoint $5 - \frac{1}{n}$ increases towards 5 from below. 4. **Union:** - Since for each $n$, $A_n$ is inside $[3,5]$ but never includes 3 or 5. - The union of all $A_n$ covers all points between just above 3 and just below 5. - So $\bigcup_{n \in \mathbb{N}} A_n = (3,5)$. 5. **Intersection:** - The intersection is the set of points common to all $A_n$. - Since the left endpoint moves closer to 3 but is always greater than 3, and the right endpoint moves closer to 5 but is always less than 5. - The intersection shrinks as $n$ grows. - In fact, the intersection is empty because no point is inside every $A_n$. 6. **Final answers:** - $\bigcup_{n \in \mathbb{N}} A_n = (3,5)$ - $\bigcap_{n \in \mathbb{N}} A_n = \emptyset$ **Note:** The user's original answer for intersection was $[4,4]$ which is just the point 4, but this is incorrect because 4 is inside all intervals only if $3 + \frac{1}{n} \leq 4 \leq 5 - \frac{1}{n}$ for all $n$. Since $3 + \frac{1}{n} \to 3$ and $5 - \frac{1}{n} \to 5$, 4 is inside all intervals, so the intersection is actually $[4,4] = \{4\}$. Let's verify this carefully: - For $n=1$, $A_1 = [4,4]$ since $3 + 1 = 4$ and $5 - 1 = 4$. - For $n=2$, $A_2 = [3.5,4.5]$ which contains 4. - For $n=3$, $A_3 = [3.333...,4.666...]$ which contains 4. - So the intersection is the single point 4. Therefore, the correct intersection is $\{4\}$. Hence: $$\bigcup_{n \in \mathbb{N}} A_n = (3,5)$$ $$\bigcap_{n \in \mathbb{N}} A_n = \{4\}$$