Problem 1. Let A be a set with cardinality n ∈ N. Prove that the set {B : B ⊆ A} hascardinality 2n.

Sure, here is a step-by-step proof for the problem:

Step 1: Define the set Let's define the set A with cardinality n. This means that there are n elements in set A.

Step 2: Define the power set The set {B : B ⊆ A} is the power set of A, often denoted as P(A). The power set of A is the set of all possible subsets of A.

Step 3: Count the subsets Each element in A has two possibilities: it can either be in a subset or not. Therefore, for n elements, there are 2^n possible combinations, or subsets.

Step 4: Conclude the proof Since there are 2^n possible subsets of A, the cardinality of the power set P(A) is 2^n. Therefore, the set {B : B ⊆ A} has cardinality 2^n.

This completes the proof.

Sure, here is a step-by-step proof:

1. Let's denote the set {B : B ⊆ A} as P(A), which is the power set of A. The power set of A is the set of all possible subsets of A.

2. We know that a set with n elements has 2^n subsets. This is because for each element in the set, we have two choices: either we include it in the subset, or we do not. Therefore, for n elements, we have 2^n possible combinations, hence 2^n subsets.

3. Therefore, the cardinality (or size) of the power set P(A) is 2^n.

4. Since the cardinality of A is given as n, we can replace n with |A| in the above expression.

5. Therefore, the cardinality of the power set P(A) is 2^|A|.

6. Hence, the set {B : B ⊆ A} has cardinality 2^n.

This completes the proof.