Fix ε > 0. Show that if (an)n∈N,(bn)n∈N are eventually ε-close, then (an)n∈N is
bounded if and only if (bn)n∈N is also bounded.

Let \( \epsilon > 0 \), and suppose that the sequences \( (a_n)_{n \in \mathbb{N}} \) and \( (b_n)_{n \in \mathbb{N}} \) are eventually \( \epsilon \)-close. This means that there exists some \( N \in \mathbb{N} \) such that for all \( n \geq N \), we have:

\[
|a_n - b_n| < \epsilon.
\]

We want to prove that \( (a_n)_{n \in \mathbb{N}} \) is bounded if and only if \( (b_n)_{n \in \mathbb{N}} \) is also bounded.

### Step 1: Proving the "if" direction

#### Assumption:
Assume \( (a_n)_{n \in \mathbb{N}} \) is bounded. By definition, this means there exists a constant \( M_a > 0 \) such that for all \( n \in \mathbb{N} \),

\[
|a_n| \leq M_a.
\]

Since \( (a_n) \) and \( (b_n) \) are eventually \( \epsilon \)-close, for all \( n \geq N \), we know:

\[
|a_n - b_n| < \epsilon.
\]

Thus, for \( n \geq N \),

\[
|b_n| = |b_n - a_n + a_n| \leq |b_n - a_n| + |a_n| < \epsilon + M_a.
\]

This shows that for all \( n \geq N \), \( |b_n| \) is bounded by \( \epsilon + M_a \).

For the remaining terms \( n < N \), since there are finitely many terms, the sequence \( (b_n) \) for \( n < N \) is also bounded. Let \( M_b' = \max \{|b_n| : n < N\} \). Define:

\[
M_b = \max(M_b', \epsilon + M_a).
\]

Thus, \( |b_n| \leq M_b \) for all \( n \in \mathbb{N} \), meaning \( (b_n) \) is bounded.

### Step 2: Proving the "only if" direction

#### Assumption:
Assume \( (b_n)_{n \in \mathbb{N}} \) is bounded. By definition, this means there exists a constant \( M_b > 0 \) such that for all \( n \in \mathbb{N} \),

\[
|b_n| \leq M_b.
\]

Again, since \( (a_n) \) and \( (b_n) \) are eventually \( \epsilon \)-close, for all \( n \geq N \), we know:

\[
|a_n| = |a_n - b_n + b_n| \leq |a_n - b_n| + |b_n| < \epsilon + M_b.
\]

This shows that for all \( n \geq N \), \( |a_n| \) is bounded by \( \epsilon + M_b \).

For the remaining terms \( n < N \), since there are finitely many terms, the sequence \( (a_n) \) for \( n < N \) is also bounded. Let \( M_a' = \max \{|a_n| : n < N\} \). Define:

\[
M_a = \max(M_a', \epsilon + M_b).
\]

Thus, \( |a_n| \leq M_a \) for all \( n \in \mathbb{N} \), meaning \( (a_n) \) is bounded.

### Conclusion:

We have shown that if the sequences \( (a_n)_{n \in \mathbb{N}} \) and \( (b_n)_{n \in \mathbb{N}} \) are eventually \( \epsilon \)-close, then \( (a_n) \) is bounded if and only if \( (b_n) \) is bounded.

---

Would you like further clarification on any part of this proof? Here are five related questions to consider:

1. What does it mean for two sequences to be eventually close?
2. How does eventual closeness influence the properties of two sequences?
3. How would the proof change if the sequences were not eventually close?
4. Can you construct an example of two eventually close sequences?
5. How does this result relate to the convergence of sequences?

**Tip**: Always make use of the fact that finite sets of terms in a sequence are trivially bounded, since bounding finite sets is easy.

Fix ε > 0. Show that if (an)n∈N,(bn)n∈N are eventually ε-close, then (an)n∈N is bounded if and only if (bn)n∈N is also bounded.

Learn how to prove that two eventually ε-close sequences are bounded under specific conditions. This detailed step-by-step solution explores key concepts in real analysis, including bounded sequences, eventual closeness, and core theorems relevant to the topic. Suitable for undergraduate-level students studying real analysis.

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