The problem in the image provides the following setup:

We are given the set of prime numbers $P = \{2, 3, 5, 7, 9, 11, 13, \dots\}$. For integers $i \geq 2$, sets $A_i$ are defined as:

$A_i = \{ n \in \mathbb{N} \mid \gcd(n, p_i) = 1 \}$ where $p_i$ is the $i$-th prime number.

Example sets:

• $A_1 = \{1, 3, 5, 7, \dots\}$ is the set of odd positive integers (numbers not divisible by 2).
• $A_2 = \{1, 2, 4, 5, 7, 8, \dots\}$ is the set of positive integers not divisible by 3.
• $A_3 = \{1, 2, 3, 4, 6, 7, 8, 9, 11, \dots\}$ is the set of integers not divisible by 5.

Now, we are asked to evaluate the following two set operations:

1. $(a) \ \bigcup_{j=1}^\infty \ \bigcap_{i=j}^\infty A_i$
2. $(b) \ \bigcap_{j=1}^\infty \ \bigcup_{i=j}^\infty A_i$

Let's break these down.

(a) $\bigcup_{j=1}^\infty \bigcap_{i=j}^\infty A_i$

This expression represents the union of intersections of sets $A_i$ from index $i=j$ onward. Essentially, for each $j$, we look at the elements common to all sets $A_i$ starting from $A_j$ and onwards, and then take the union of those results for all $j$.

The set $A_i$ consists of numbers that are coprime with the $i$-th prime number. For increasing values of $j$, the intersections $\bigcap_{i=j}^\infty A_i$ will become increasingly restrictive because the numbers must avoid being divisible by a growing number of primes. Therefore, as $j$ increases, the intersection will eventually become empty because no natural number is coprime with all primes (besides 1).

Thus, the union of these intersections is just $\{1\}$, since 1 is the only number that is coprime with all primes.

Conclusion for (a): The result of (a) is $\{1\}$.

(b) $\bigcap_{j=1}^\infty \bigcup_{i=j}^\infty A_i$

This represents the intersection of unions of sets $A_i$, starting from index $i=j$ onward. For each $j$, the union $\bigcup_{i=j}^\infty A_i$ includes all numbers not divisible by $p_j, p_{j+1}, p_{j+2}, \dots$, meaning the union covers most natural numbers, excluding only those divisible by increasingly large primes.

As $j$ increases, the prime numbers being excluded become larger, but since any natural number is divisible by only finitely many primes, every number except for multiples of infinitely many primes (essentially 1) will eventually be included in one of these unions.

Thus, the intersection of these unions will also be $\{1\}$.

Conclusion for (b): The result of (b) is also $\{1\}$.

Final Answer:

Both (a) and (b) result in the set $\{1\}$, which means the correct answer is that both sets (a) and (b) are the set of natural numbers that are coprime with all primes, which is $\{1\}$.

Let me know if you want further details or have any questions.

Here are 5 related questions to expand on the topic:

1. How do the sets $A_i$ change as $i$ increases?
2. What would happen if the operation in (a) and (b) was swapped (union first, then intersection)?
3. What are the properties of numbers that belong to the set $\{1\}$?
4. How does the greatest common divisor (gcd) help in defining the set $A_i$?
5. How would this problem change if we considered only a finite number of primes?

Tip: Always try visualizing set operations using Venn diagrams to understand their unions and intersections better.