Question

For a given real number \( a \), let \( a^+ = \max\{a, 0\} \) and \( a^- = \max\{-a, 0\} \). If \( \{x_n\}_{n \geq 1} \) is a sequence of real numbers, then which of the following statements is/are true?

• A. If \( \{x_n\}_{n \geq 1} \) converges, then both \( \{x_n^+\}_{n \geq 1} \) and \( \{x_n^-\}_{n \geq 1} \) converge
• B. If \( \{x_n\}_{n \geq 1} \) converges to 0, then both \( \{x_n^+\}_{n \geq 1} \) and \( \{x_n^-\}_{n \geq 1} \) converge to 0
• C. If both \( \{x_n^+\}_{n \geq 1} \) and \( \{x_n^-\}_{n \geq 1} \) converge, then \( \{x_n\}_{n \geq 1} \) converges
• D. If \( \{x_n^2\}_{n \geq 1} \) converges, then both \( \{x_n^+\}_{n \geq 1} \) and \( \{x_n^-\}_{n \geq 1} \) converge

Hint:

- If a sequence converges, the positive and negative parts also converge.
- The converse is also true: if the positive and negative parts converge, the sequence itself converges.

Answer 1

The Correct Option is A

1) Understanding the sequence and the terms:
Given that \( x_n^+ \) and \( x_n^- \) are the positive and negative parts of \( x_n \), respectively, we know the following properties: if \( x_n \) converges to a limit, both \( x_n^+ \) and \( x_n^- \) also converge. Similarly, if both \( x_n^+ \) and \( x_n^- \) converge, then \( x_n \) converges.
2) Analysis of the options:
(A) If \( \{x_n\}_{n \geq 1} \) converges, then both \( \{x_n^+\}_{n \geq 1} \) and \( \{x_n^-\}_{n \geq 1} \) converge:
This is true. If \( x_n \) converges to some limit \( L \), then both \( x_n^+ \) and \( x_n^- \) must also converge to \( L^+ \) and \( L^- \), respectively.
(B) If \( \{x_n\}_{n \geq 1} \) converges to 0, then both \( \{x_n^+\}_{n \geq 1} \) and \( \{x_n^-\}_{n \geq 1} \) converge to 0:
This is true. If \( x_n \) converges to 0, then \( x_n^+ \) and \( x_n^- \) will both converge to 0.
(C) If both \( \{x_n^+\}_{n \geq 1} \) and \( \{x_n^-\}_{n \geq 1} \) converge, then \( \{x_n\}_{n \geq 1} \) converges:
This is true. If both the positive and negative parts of the sequence converge, the original sequence must also converge.
(D) If \( \{x_n^2\}_{n \geq 1} \) converges, then both \( \{x_n^+\}_{n \geq 1} \) and \( \{x_n^-\}_{n \geq 1} \) converge:
This is false. Convergence of \( x_n^2 \) does not necessarily imply that both \( x_n^+ \) and \( x_n^- \) converge.
The correct answers are (A), (B), and (C).