### Problem Statement:**9.** Suppose $ B \subseteq A $, $ B \neq A $, and $ B \sim A $. Prove that $ A $ is infinite.---### Explanation and Solution:**Given:**- $ B \subseteq A $: Set $ B $ is a subset of set $ A $.- $ B \neq A $: Set $ B $ is not equal to set $ A $; hence, $ B $ is a proper subset of $ A $.- $ B \sim A $: Set $ B $ is equivalent to set $ A $, meaning there exists a bijection (one-to-one and onto function) between $ B $ and $ A $.**Objective:**- Prove that set $ A $ is infinite.**Proof:**To prove that $ A $ is infinite, assume that $ A $ is finite and derive a contradiction.1. **Assumption (for contradiction)**: Suppose $ A $ is finite. - This implies that $ A $ has a finite number of elements, say $ n $.2. **Proper subset implication**: - Since $ B \subseteq A $ and $ B \neq A $, $ B $ must have fewer elements than $ A $, i.e., the number of elements in $ B $ is less than $ n $.3. **Cardinality argument**: - If $ B $ is a proper subset of $ A $ and $ A $ is finite, the number of elements in $ B $ must be $ m $, where $ 0 \leq m < n $.4. **Contradiction with equivalence**: - Given $ B \sim A $, there exists a bijection between $ B $ and $ A $. - For finite sets, a bijection requires both sets to have the same number of elements for one-to-one correspondence. - Hence, $ B $ should have $ n $ elements if it is equivalent to $ A $, but $ B $ has only $ m $ elements ($ m < n $), which is a contradiction.5. **Conclusion**: - Therefore, the assumption that set $ A $ is finite leads to a contradiction. - Hence, $ A $ must

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Description: ### Problem Statement:**9.** Suppose \( B \subseteq A $, $ B \neq A $, and $ B \sim A $. Prove that $ A $ is infinite.---### Explanation and Solution:**Given:**- $ B \subseteq A $: Set $ B $ is a subset of set $ A $.- $ B \neq A $: Se

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Answered: Suppose B C A, B + A, and B ~ A. Prove…

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Suppose B C A, B + A, and B ~ A. Prove that A is infinite.

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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Statements and Logic

Logic is the analysis of general patterns of thoughts without regard to any specific sense of context.

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Question

### Problem Statement:

**9.** Suppose $ B \subseteq A $, $ B \neq A $, and $ B \sim A $. Prove that $ A $ is infinite.

---

### Explanation and Solution:

**Given:**
- $ B \subseteq A $: Set $ B $ is a subset of set $ A $.
- $ B \neq A $: Set $ B $ is not equal to set $ A $; hence, $ B $ is a proper subset of $ A $.
- $ B \sim A $: Set $ B $ is equivalent to set $ A $, meaning there exists a bijection (one-to-one and onto function) between $ B $ and $ A $.

**Objective:**
- Prove that set $ A $ is infinite.

**Proof:**

To prove that $ A $ is infinite, assume that $ A $ is finite and derive a contradiction.

1. **Assumption (for contradiction)**: Suppose $ A $ is finite.
- This implies that $ A $ has a finite number of elements, say $ n $.

2. **Proper subset implication**:
- Since $ B \subseteq A $ and $ B \neq A $, $ B $ must have fewer elements than $ A $, i.e., the number of elements in $ B $ is less than $ n $.

3. **Cardinality argument**:
- If $ B $ is a proper subset of $ A $ and $ A $ is finite, the number of elements in $ B $ must be $ m $, where $ 0 \leq m < n $.

4. **Contradiction with equivalence**:
- Given $ B \sim A $, there exists a bijection between $ B $ and $ A $.
- For finite sets, a bijection requires both sets to have the same number of elements for one-to-one correspondence.
- Hence, $ B $ should have $ n $ elements if it is equivalent to $ A $, but $ B $ has only $ m $ elements ($ m < n $), which is a contradiction.

5. **Conclusion**:
- Therefore, the assumption that set $ A $ is finite leads to a contradiction.
- Hence, $ A $ must

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