Let

P={p1=2,p2=3,p3=5,p4=7,⋯}

be the set of primes. Let

An={m∈N∣m leaves a remainder 1 when it is divided by the nth prime pn}.

Consider the following sets:

(a) X=∪j≥1∩i≥jAi

(b) Y=∩j≥1∪i≥jAi

Which one of the following statements is true?

Group of answer choices

X={1} and Y=N.

X=N and Y=N.

X={1} and Y={1}.

X=N and Y={1}.

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.

Let P = {p1 = 2, p2 = 3, p3 = 5, p4 = 7, ⋯} be the set of primes. Let An = {m ∈ N | m leaves a remainder 1 when it is divided by the nth prime pn}. Consider the following sets: (a) X = ∪j≥1∩i≥jAi (b) Y = ∩j≥1∪i≥jAi. Which one of the following statements is true? Group of answer choices X = {1} and Y = N. X = N and Y = N. X = {1} and Y = {1}. X = N and Y = {1}.

Explore a complex set theory problem involving primes and modular arithmetic. Learn how to solve for sets X and Y defined by unions and intersections of remainders when divided by primes. This problem illustrates key mathematical concepts such as the Chinese Remainder Theorem and is suitable for advanced math students.

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