Prove that (i) if a set has only one element , then it has 2 subsets .
(ii) If `BsubA` and if B has one elements less than that of A, prove that A has twice as many subsets as B .
(iii) Deduce from these two results that a set with 2 elements has `2^(2)` subsets , a set with 3 elements has `2^(3)` subsets , and so on .
How many subsets does a set with n elements have ?

Answer

To solve the given problem, we will break it down into three parts as stated in the question.

### Part (i): Prove that if a set has only one element, then it has 2 subsets.

1. **Define the Set**: Let \( S = \{ a \} \), where \( a \) is the only element in the set.
2. **Identify Subsets**: The subsets of \( S \) are:
- The empty set \( \emptyset \)
- The set itself \( S = \{ a \} \)
3. **Count the Subsets**: Therefore, the total number of subsets of the set \( S \) is:
\[ \text{Number of subsets} = 2 \]

### Part (ii): Prove that if \( B \subset A \) and \( B \) has one element less than \( A \), then \( A \) has twice as many subsets as \( B \).

1. **Define Sets**: Let \( A \) have \( n \) elements and \( B \) have \( n-1 \) elements.
- For example, let \( B = \{ a_1, a_2, \ldots, a_{n-1} \} \) and \( A = \{ a_1, a_2, \ldots, a_{n-1}, a_n \} \).
2. **Count Subsets of \( B \)**: The number of subsets of \( B \) is:
\[ \text{Number of subsets of } B = 2^{n-1} \]
3. **Count Subsets of \( A \)**: The number of subsets of \( A \) is:
\[ \text{Number of subsets of } A = 2^n \]
4. **Relationship Between Subsets**: Since each subset of \( B \) can either include or exclude the additional element \( a_n \) from \( A \), we can form two subsets for each subset of \( B \):
- One that includes \( a_n \)
- One that does not include \( a_n \)
5. **Conclusion**: Therefore, the number of subsets of \( A \) is:
\[ \text{Number of subsets of } A = 2 \times \text{Number of subsets of } B \]
\[ 2^n = 2 \times 2^{n-1} \]

### Part (iii): Deduce that a set with \( n \) elements has \( 2^n \) subsets.

1. **Base Cases**: We have already established:
- For \( n = 1 \): \( 2^1 = 2 \) subsets
- For \( n = 2 \): \( 2^2 = 4 \) subsets
- For \( n = 3 \): \( 2^3 = 8 \) subsets
2. **Inductive Step**: Assume that for a set with \( k \) elements, the number of subsets is \( 2^k \).
- For a set with \( k+1 \) elements, by the previous part, it will have:
\[ 2^{k+1} = 2 \times 2^k \]
3. **Conclusion**: Thus, by induction, a set with \( n \) elements has \( 2^n \) subsets.

### Final Answer:
A set with \( n \) elements has \( 2^n \) subsets.

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