Chapter 1 Solutions

Here are the solutions to the exercises in Chapter 1 of Mathematical Physics: A Modern Introduction to Its Foundations by Sadri Hassani.

• 1.1
• 1.2

1.1

Show that the number of subsets of a set containing elements is .

There are a few ways to show this. One way is to use a combinatorial argument. In a set with elements, each element can either be included in a subset or not. Thus, for each element, there are two choices: to include it or not. Since there are elements, the total number of subsets is given by multiplying the number of choices for each element, which is .

Another way to show this is by using induction.

Base Case: For , the set is empty, and there is only one subset, which is the empty set itself. Thus, the number of subsets is . Inductive Step: Assume that for a set with elements, the number of subsets is . Now, consider a set with elements. We can think of this set as the union of a set with elements and an additional element. The subsets of this new set can be divided into two categories: those that do not include the additional element and those that do. The number of subsets that do not include the additional element is equal to the number of subsets of the original set with elements, which is by the inductive hypothesis. The number of subsets that include the additional element is also equal to the number of subsets of the original set with elements, which is again . Thus, the total number of subsets for the set with elements is .

By the principle of mathematical induction, we conclude that the number of subsets of a set containing elements is for all non-negative integers .

1.2

Let , , and be sets in a universal set . Show that:

1. and implies .
2. iff iff .
3. and implies .
4. .

Hint: To show the equality of two sets, show that each set is a subset of the other.

Let's prove each part step by step.

1. Assume and .

   The former means that for every element , we have . The latter means that for every element , we have . Combining these two statements, we see that for every element , we have and thus . Therefore, .

2. We will prove the equivalence in two parts.

   • () Assume . This means that for every element , we have . So is a subset of . Moreover, there are no elements in that are not in , so .

     Also, since every element of is in , the union is just itself, so .

   • () Assume . This means that every element in is also in , so .

     Similarly, assume . This means that every element in is also in , so .

   Therefore, we have shown in both directions that iff iff .

   It is insightful to draw a Venn diagram to visualize this.

3. Assume and .

   We want to show that . From the previous part, we know that implies . Since , we have . Therefore, .

4. We will prove the equality by showing that each side is a subset of the other.