Let n be any counting number. Using the two questions above as a guide, how

many subsets does the set {1,2,3,...,n−1,n} have? Prove it as well.
First question:Including itself, how many subsets does the set {1, 2, 3} have? List them.
Second Question: Including itself, how many subsets does the set {1, 2, 3, 4} have? List them.

This is called the power set.Any set of n elements has 2^n subsets.

Just sum all the powers of 2 from 0 to n-1

To find the number of subsets of a set with n elements, including itself, we can use the following approach:

1. First, let's consider the simpler case of {1, 2, 3}.

- To find the number of subsets, we need to determine the number of choices we have for each element of the set: either we include it in a particular subset or exclude it.
- For each element, there are two choices, either include it or exclude it.
- For a set with 3 elements, we have 2 choices for the first element, 2 choices for the second element, and 2 choices for the third element.
- Therefore, the total number of subsets is obtained by multiplying the number of choices for each element: 2 x 2 x 2 = 8.
- The subsets of {1, 2, 3} are: {}, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, and {1, 2, 3}.

2. Now, let's consider {1, 2, 3, 4}.

- Similar to the previous case, each element has two choices: include or exclude it.
- For a set with 4 elements, we have 2 choices for the first element, 2 choices for the second element, 2 choices for the third element, and 2 choices for the fourth element.
- Therefore, the total number of subsets is obtained by multiplying the number of choices for each element: 2 x 2 x 2 x 2 = 16.
- The subsets of {1, 2, 3, 4} are: {}, {1}, {2}, {3}, {4}, {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}, {1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}, and {1, 2, 3, 4}.

3. Now, let's generalize this pattern to any counting number n.

- For a set with n elements, each element has two choices: include or exclude it.
- Therefore, the total number of subsets is obtained by multiplying the number of choices for each of the n elements: 2^n.
- In other words, there are 2^n subsets of a set with n elements, including itself.

To prove this mathematically, we can use mathematical induction:

- Base Case: For n=1, the set {1} has 2^1 = 2 subsets, which are {} and {1}.
- Inductive Hypothesis: Assume that for some positive integer k, the set {1, 2, ..., k} has 2^k subsets.
- Inductive Step: We need to show that the set {1, 2, ..., k+1} also has 2^(k+1) subsets.
- To do this, we can observe that the subsets of {1, 2, ..., k+1} can be obtained by either excluding the (k+1)-th element or including it in each subset of {1, 2, ..., k}.
- Since the subsets of {1, 2, ..., k} are 2^k in number (by the inductive hypothesis), we have twice as many subsets when including the (k+1)-th element.
- Therefore, the number of subsets of {1, 2, ..., k+1} is 2^k * 2 = 2^(k+1), which completes the induction step.

Hence, we have proved that for any counting number n, the set {1, 2, ..., n} has 2^n subsets, including itself.