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Chapter 5: Problem 9

Is the following proposition true or false? Justify your conclusion with a proof or a counterexample. For all sets \(A\) and \(B\) that are subsets of some universal set \(U,\) the sets \(A \cap B\) and \(A-B\) are disjoint.

Short Answer

Expert verified
The proposition is true. For any sets A and B, the intersection of A 鈭� B and A - B is empty, making them disjoint sets. This is because if x 鈭� A 鈭� B, then x 鈭� A and x 鈭� B, while for any element y 鈭� A - B, y 鈭� A and y 鈭� B. Since x 鈭� B and y 鈭� B, there cannot be any common element between A 鈭� B and A - B, and their intersection is empty.

Step by step solution

01

Recall the Definitions

Recall the definitions of intersection and difference of sets: 1. Intersection (A 鈭� B): It is the set containing all the elements that are common to both A and B. 2. Difference (A - B): It is the set containing all the elements in A that are not in B. Now we will move on to the next step of finding if this proposition is true or false.
02

Consider an Example

Let's consider an example to see if this proposition holds. Let A = {1, 2, 3} and B = {2, 3, 4}. This means that U should at least be {1, 2, 3, 4}. Now let's find A 鈭� B and A - B: 1. A 鈭� B = {2, 3} 2. A - B = {1} Now, let's find the intersection of these two sets: (A 鈭� B) 鈭� (A - B) = {} (empty set) In this example, the proposition holds true since the intersection of A 鈭� B and A - B is an empty set, which means they are disjoint.
03

Generalize the Proposition

Let's try to generalize this with any arbitrary sets A and B. For any sets A and B, consider any element x 鈭� A 鈭� B. By definition of intersection, x 鈭� A and x 鈭� B. Now, let's consider A - B. For any element y 鈭� A - B, by definition of set difference, y 鈭� A and y 鈭� B. Since x 鈭� B and y 鈭� B, there cannot be any common element between A 鈭� B and A - B, hence their intersection must be empty.
04

Conclusion

With our analysis, we have shown that the intersection of A 鈭� B and A - B is empty for any arbitrary sets A and B, making them disjoint sets. Hence, the given proposition is true.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Intersection of Sets
Imagine two circles overlapping, with the area where they both overlap representing the intersection of those circles or, in set theory, the intersection of two sets. Formally, the intersection of sets A and B, denoted as \( A \cap B \), includes all elements that are shared between both sets. To put it simply, it's like finding common friends in two different friendship groups.

When we work through an example, say set A = {1, 2, 3} and set B = {2, 3, 4}, the intersection \( A \cap B \) is {2, 3} - these are the members that show up to parties hosted by both groups A and B.
Difference of Sets
Now consider a setting where we want to highlight what's unique to one group and not shared with another. In set theory, the difference between sets A and B, written as \( A - B \), consists of elements that are only in set A but not in set B. It's like planning a solo trip without the friends you often travel with.

Using our previous sets as an example, set A = {1, 2, 3} and set B = {2, 3, 4}, the difference \( A - B \) is {1}, since only the number 1 is exclusive to set A and doesn't appear in set B's travel plans.
Disjoint Sets
Disjoint sets are like non-overlapping circles; they have no common elements, or in friend group terms, they do not share any common friends. When two sets have no overlap, or in other words, their intersection is empty, they are described as disjoint.

Following this, if you take our intersected set {2, 3} and the difference set {1}, these two sets have no common elements - like two groups with entirely different tastes, they never meet up. Their intersection is the empty set (\( \{\} \)), reaffirming their status as disjoint sets.
Proof by Example
To clarify complex theories, sometimes an example serves as the best explanation, much like a practical demonstration in a science class. Proof by example involves showing that a statement holds true in a particular case, though it's important to note that this alone doesn't always prove a statement to be true in all cases, but it can occasionally provide sufficient reasoning for particular propositions.

In our scenario, by showing that the sets \( A \cap B \) and \( A - B \) are disjoint for specific sets A and B, it effectively supports the proposition that this is true for any subsets of the universal set \( U \).
Mathematical Reasoning
Mathematical reasoning is the glue that holds our understanding of mathematical concepts together. It involves the process of thinking logically and systematically to arrive at a conclusion, based on given premises or known facts.

In set theory, reasoning allows us to generalize from specific instances. As seen in the proposition that sets \( A \cap B \) and \( A - B \) are always disjoint, it is the careful reasoning from the definitions and properties of set operations that enables us to deduce that this statement holds true universally for all sets A and B within a universal set \( U \), demonstrating the power of mathematical reasoning.

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Most popular questions from this chapter

Let \(A\) be a subset of some universal set \(U\). Prove each of the following (from Theorem 5.20): (a) \( \left(A^{c}\right)^{c}=A\) \((\mathbf{c}) \emptyset^{c}=U\) (b) \(A-\emptyset=A\) (d) \(U^{c}=\emptyset\)

Let \(A, B,\) and \(C\) be subsets of a universal set \(U\). (a) Draw a Venn diagram with \(A \subseteq B\) and \(B \subseteq C .\) Does it appear that \(A \subseteq C ?\) (b) Prove the following proposition: If \(A \subseteq B\) and \(B \subseteq C,\) then \(A \subseteq C\) This may seem like an obvious result. However, one of the reasons for this exercise is to provide practice at properly writing a proof that one set is a subset of another set. So we should start the proof by assuming that \(A \subseteq B\) and \(B \subseteq C .\) Then we should choose an arbitrary element of \(A\).

Let \(C=\\{x \in \mathbb{Z} \mid x \equiv 7(\bmod 9)\\}\) and \(D=\\{x \in \mathbb{Z} \mid x \equiv 1(\bmod 3)\\}\). (a) List at least five different elements of the set \(C\) and at least five elements of the set \(D\). (b) Is \(C \subseteq D\) ? Justify your conclusion with a proof or a counterexample. (c) Is \(D \subseteq C\) ? Justify your conclusion with a proof or a counterexample.

For each positive real number \(r,\) define \(T_{r}\) to be the closed interval \(\left[-r^{2}, r^{2}\right]\) That is, \(T_{r}=\left\\{x \in \mathbb{R} \mid-r^{2} \leq x \leq r^{2}\right\\}\) Let \(\Lambda=\\{m \in \mathbb{N} \mid 1 \leq m \leq 10\\}\). Use either interval notation or set builder notation to specify each of the following sets: * (a) \(\bigcup_{k \in \Lambda} T_{k}\) (c) \(\bigcup_{r \in \mathbb{R}^{+}} T_{r}\) (e) \(\bigcup_{k \in \mathbb{N}} T_{k}\) *(b) \(\bigcap_{k \in \Lambda} T_{k}\) (d) \(\bigcap_{r \in \mathbb{R}^{+}} T_{r}\) (f) \(\bigcap_{k \in \mathbb{N}} T_{k}\)

Prove Theorem 5.25, Part (4): \((A \cup B) \times C=(A \times C) \cup(B \times C)\).
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