QUESTION 8 Complete the proof of the following statement using the drop-down menu: Let An 1+ for n 1, 2, 3, .. Then N A,= Ø. n=1 pf. (by )Assume 1 A, is not Ø. n=1 Then there exists an X € || An. So x EA, n21. n=1 Therefore X EA1, x EA2, x EA 3, ... Since x EA1=(1,2), we know Since X>1, we know X -1>[ By the AP, there exists an N EZ' such that 0< 1 Since
QUESTION 8 Complete the proof of the following statement using the drop-down menu: Let An 1+ for n 1, 2, 3, .. Then N A,= Ø. n=1 pf. (by )Assume 1 A, is not Ø. n=1 Then there exists an X € || An. So x EA, n21. n=1 Therefore X EA1, x EA2, x EA 3, ... Since x EA1=(1,2), we know Since X>1, we know X -1>[ By the AP, there exists an N EZ' such that 0< 1 Since
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**QUESTION 8**
Complete the proof of the following statement using the drop-down menu:
Let \( A_n = \left(1, 1 + \frac{1}{n}\right) \) for \( n = 1, 2, 3, \ldots \). Then \(\bigcap_{n=1}^{\infty} A_n = \emptyset \).
**Proof:**
Assume \(\bigcap_{n=1}^{\infty} A_n\) is not \(\emptyset\).
Then there exists an \( x \in \bigcap_{n=1}^{\infty} A_n \). So \( x \in A_n \) for \( n \geq 1 \).
Therefore \( x \in A_1, x \in A_2, x \in A_3, \ldots \)
Since \( x \in A_1 = (1, 2) \), we know [Drop-down menu].
Since \( x > 1 \), we know \( x - 1 > [Drop-down menu] \).
By the Archimedean Principle (AP), there exists an \( N \in \mathbb{Z}^+ \) such that \( 0 < \frac{1}{N} < [Drop-down menu] \).
Since \( \frac{1}{N} < x - 1 \), we get \( 1 + \frac{1}{N} < x \).
But then \( x \notin \left(1, 1 + \frac{1}{N}\right) = A_N \), contradicting \( x \in A_N \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F79599c56-a340-49a0-b0ff-829b3947a798%2F87679e29-c787-4a1d-9279-fb9c226c5375%2F19a463_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**QUESTION 8**
Complete the proof of the following statement using the drop-down menu:
Let \( A_n = \left(1, 1 + \frac{1}{n}\right) \) for \( n = 1, 2, 3, \ldots \). Then \(\bigcap_{n=1}^{\infty} A_n = \emptyset \).
**Proof:**
Assume \(\bigcap_{n=1}^{\infty} A_n\) is not \(\emptyset\).
Then there exists an \( x \in \bigcap_{n=1}^{\infty} A_n \). So \( x \in A_n \) for \( n \geq 1 \).
Therefore \( x \in A_1, x \in A_2, x \in A_3, \ldots \)
Since \( x \in A_1 = (1, 2) \), we know [Drop-down menu].
Since \( x > 1 \), we know \( x - 1 > [Drop-down menu] \).
By the Archimedean Principle (AP), there exists an \( N \in \mathbb{Z}^+ \) such that \( 0 < \frac{1}{N} < [Drop-down menu] \).
Since \( \frac{1}{N} < x - 1 \), we get \( 1 + \frac{1}{N} < x \).
But then \( x \notin \left(1, 1 + \frac{1}{N}\right) = A_N \), contradicting \( x \in A_N \).
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