We are given the following problem:

Let $P = \{ p_1 = 2, p_2 = 3, p_3 = 5, p_4 = 7, \dots \}$ be the set of prime numbers. Define the sets $A_n = \{ m \in \mathbb{N} \mid m \text{ leaves a remainder 1 when divided by the } n\text{th prime } p_n \}$.

Now, we need to analyze the two sets $X$ and $Y$:

• $X = \bigcup_{j \geq 1} \bigcap_{i \geq j} A_i$
• $Y = \bigcap_{j \geq 1} \bigcup_{i \geq j} A_i$

Understanding the sets $A_n$

Each set $A_n$ contains natural numbers that leave a remainder of 1 when divided by the $n$-th prime $p_n$. For example:

• $A_1 = \{ 1, 3, 5, 7, 9, 11, 13, \dots \}$ (all numbers that leave a remainder of 1 when divided by 2).
• $A_2 = \{ 1, 4, 7, 10, 13, 16, 19, \dots \}$ (all numbers that leave a remainder of 1 when divided by 3), and so on.

Set $X = \bigcup_{j \geq 1} \bigcap_{i \geq j} A_i$

This set is formed by taking the intersection of all $A_i$ starting from some index $j$, and then taking the union of these intersections over all $j \geq 1$.

• For $j = 1$, we take the intersection of all the $A_i$ for $i \geq 1$. Since every number in $A_1$ leaves a remainder of 1 when divided by 2, and every number in $A_2$ leaves a remainder of 1 when divided by 3, and so on, the only number that leaves a remainder of 1 when divided by every prime is $1$. Therefore, $\bigcap_{i \geq 1} A_i = \{ 1 \}$.
• For $j = 2$, we take the intersection of $A_2, A_3, A_4, \dots$, which again will result in $\{ 1 \}$, because 1 is the only number that leaves a remainder of 1 when divided by every prime from the 2nd prime onward.
• The same holds for all $j \geq 1$, so $X = \{ 1 \}$.

Set $Y = \bigcap_{j \geq 1} \bigcup_{i \geq j} A_i$

This set is formed by taking the union of $A_i$ for $i \geq j$, and then taking the intersection of these unions over all $j \geq 1$.

• For $j = 1$, we take the union of all the $A_i$ for $i \geq 1$, which is $\mathbb{N}$, because every number belongs to at least one $A_i$ (since every number leaves a remainder of 1 when divided by some prime).
• For $j = 2$, we take the union of $A_2, A_3, A_4, \dots$, which is still $\mathbb{N}$, because every number will eventually leave a remainder of 1 when divided by some prime.
• The same holds for all $j \geq 1$, so $Y = \mathbb{N}$.

Conclusion

• $X = \{ 1 \}$
• $Y = \mathbb{N}$

Therefore, the correct answer is:

$\boxed{X = \{ 1 \} \text{ and } Y = \mathbb{N}.}$

Do you want further details or have any questions?

Related Questions:

1. What are the properties of sets defined using primes?
2. How do the sets $A_n$ change as $n$ increases?
3. How is the Chinese remainder theorem related to this problem?
4. What is the significance of taking unions and intersections in set theory?
5. Can the result $X = \{1\}$ be generalized to other modular conditions?

Tip:

When solving problems involving infinite intersections or unions, focus on the smallest elements and how they behave across all sets to simplify reasoning.