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Let the universal set be the set \(\mathbf{R}\) of all real numbers and let \(A=\\{x \in \mathbf{R} \mid 0

Short Answer

Expert verified
a. \(A \cup B=\\{x \in \mathbf{R} \mid 0 2\\}\) d. \(A \cup C=\\{x \in \mathbf{R} \mid 0 2\\}\) i. \((A \cap B)^{c}=\\{x \in \mathbf{R} \mid x < 1 \mathrm{~or~} x > 2\\}\) j. \((A \cup B)^{c}=\\{x \in \mathbf{R} \mid x \leq 0 \mathrm{~or~} x \geq 4\\}\)

Step by step solution

01

Define the Sets A, B, and C

Set A contains all real numbers \(x\) that are greater than 0 and less than or equal to 2, denoted by \(0<x \leq 2\). Set B contains all real numbers \(x\) that are greater than or equal to 1 and less than 4, denoted by \(1 \leq x < 4\). Set C contains all real numbers \(x\) that are greater than or equal to 3 and less than 9, denoted by \(3 \leq x < 9\).
02

a. Find \(A \cup B\)

To find the union, \(A \cup B\), we need to find all the real numbers \(x\) that belong to either A or B or both. We see that \(A \cup B=\\{x \in \mathbf{R} \mid 0<x<4\\}\), since set A covers real numbers from 0 to 2 and set B covers real numbers from 1 to 4.
03

b. Find \(A \cap B\)

To find the intersection, \(A \cap B\), we need to find all the real numbers \(x\) that belong to both A and B. We see that \(A \cap B=\\{x \in \mathbf{R} \mid 1 \leq x \leq 2\\}\), since set A covers real numbers from 0 to 2 and set B covers real numbers from 1 to 4.
04

c. Find \(A^{c}\)

To find the complement of A, \(A^{c}\), we need to find all the real numbers \(x\) that do NOT belong to set A. We see that \(A^{c}=\\{x \in \mathbf{R} \mid x \leq 0 \mathrm{~or~} x > 2\\}\), which includes all real numbers that are less than or equal to 0 and greater than 2.
05

d. Find \(A \cup C\)

To find the union, \(A \cup C\), we need to find all the real numbers \(x\) that belong to either A or C or both. We see that \(A \cup C=\\{x \in \mathbf{R} \mid 0<x \leq 2 \mathrm{~or~} 3 \leq x < 9\\}\), which includes all real numbers greater than 0 and less than or equal to 2 and all real numbers greater than or equal to 3 and less than 9.
06

e. Find \(A \cap C\)

To find the intersection, \(A \cap C\), we need to find all the real numbers \(x\) that belong to both A and C. We see that \(A \cap C\) is an empty set, denoted by \(\\{\\}\) or \(\emptyset\), since there are no real numbers that belong to both A and C.
07

f. Find \(B^{c}\)

To find the complement of B, \(B^{c}\), we need to find all the real numbers \(x\) that do NOT belong to set B. We see that \(B^{c}=\\{x \in \mathbf{R} \mid x < 1 \mathrm{~or~} x \geq 4\\}\), which includes all real numbers that are less than 1 and greater than or equal to 4.
08

g. Find \(A^{c} \cap B^{c}\)

To find the intersection of the complements, \(A^{c} \cap B^{c}\), we need to find all the real numbers \(x\) that belong to both \(A^{c}\) and \(B^{c}\). We see that \(A^{c} \cap B^{c}\)=\\{x \in \mathbf{R} \mid x \leq 0 \mathrm{~or~} x \geq 4\\}
09

h. Find \(A^{c} \cup B^{c}\)

To find the union of the complements, \(A^{c} \cup B^{c}\), we need to find all the real numbers \(x\) that belong to either \(A^{c}\) or \(B^{c}\) or both. We see that \(A^{c} \cup B^{c}\)=\\{x \in \mathbf{R} \mid x \leq 0 \mathrm{~or~} x > 2\\}, which includes all real numbers less than or equal to 0 and greater than 2, essentially excluding the real numbers in A and B.
10

i. Find \((A \cap B)^{c}\)

To find the complement of the intersection, \((A \cap B)^{c}\), we need to find all the real numbers \(x\) that do NOT belong to \(A \cap B\). We see that \((A \cap B)^{c}\)=\\{x \in \mathbf{R} \mid x < 1 \mathrm{~or~} x > 2\\}
11

j. Find \((A \cup B)^{c}\)

To find the complement of the union, \((A \cup B)^{c}\), we need to find all the real numbers \(x\) that do NOT belong to \(A \cup B\). We see that \((A \cup B)^{c}\)=\\{x \in \mathbf{R} \mid x \leq 0 \mathrm{~or~} x \geq 4\\}

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Key Concepts

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

Universal Set and Subsets
In the realm of set theory within discrete mathematics, the concept of a \universal set\ plays a foundational role. The universal set, often denoted as \(U\), is the set that contains all the elements under consideration for a particular discussion or problem. Essentially, it's the 'biggest' set that includes every other set as its part.

Imagine the universal set as an all-encompassing ocean in which every other set floats as an island. Each of these 'islands', or subsets, contains elements that are also found in the universal set, thereby making them a part of the larger whole. A subset is formally defined as a set where every element within it is also found in another set (which could be the universal set itself). For instance, if our universal set, \(\mathbf{R}\), is the set of all real numbers, then set \(A = \{x \ \mathbf{R} \mid 0< x \leq 2\}\) is a subset of \(\mathbf{R}\) because all elements in \(A\) are contained within \(\mathbf{R}\).

To visualize this concept, imagine labeling a giant container with 'Real Numbers'. Inside you'd place smaller containers labeled 'A', 'B', and 'C', each with numbers that satisfy their respective conditions. Every number inside these smaller containers will also be somewhere in the giant 'Real Numbers' container, showing us that \(A\), \(B\), and \(C\) are all subsets of the universal set \(\mathbf{R}\).
Set Operations
Moving on to the operations we can perform with sets, these are much like arithmetic operations but applied to collections of elements. The most common set operations are union (represented by \(\cup\)) and intersection (represented by \(\cap\)).

The union of two sets, say \(A\) and \(B\), denoted by \(A \cup B\), is a new set that contains all the elements that are in either \(A\), \(B\), or both. In other words, it gathers everything together. For example, in the given exercise, \(A \cup B\) combines the numbers greater than 0 and less than 4, giving us a set that includes the entire range of both sets without repetition.

The intersection of two sets, on the other hand, is where things get more specific. Noted as \(A \cap B\), it gives us a set of elements that are common to both \(A\) and \(B\). It's like finding the overlap or shared territory between two regions. In our exercise case, \(A \cap B\) represents the real numbers between 1 and 2, inclusive, which are the numbers both sets share.

Understanding these operations is crucial because they form the basis for more complex expressions in set theory. For learners working through homework problems or simply trying to grasp the foundational concepts, picturing these operations visually, like combining or overlapping circles in a Venn diagram, can significantly enhance comprehension.
Complement of a Set
Finally, let us explore the concept of the complement of a set. It’s quite straightforward – if you have a universal set and a subset within it, the complement of that subset consists of everything in the universal set that is not in the subset.

Using notation, if we consider the complement of set \(A\), written as \(A^c\), this includes all elements in the universal set \(\mathbf{R}\) that don't satisfy the conditions to be in \(A\). It's as if we’ve scooped out the elements of \(A\) from \(\mathbf{R}\) and are left with the rest of the ocean. In the provided exercise, we have \(A^c = \{x \ \mathbf{R} \mid x \leq 0 \mathrm{~or~} x > 2\}\), indicating all real numbers except those between 0 (exclusive) and 2 (inclusive).

The complement can also apply to more intricate set expressions, such as the complement of the intersection or union of sets, often leading to results that might initially seem counterintuitive. For example, the complement of \(A \cap B\) is not simply the union of the complements; it's more extensive as it includes all elements not found in both \(A\) and \(B\). These concepts are beautifully linked and, when well-understood, can provide deep insights into the nature of mathematical sets and the relationships between them.

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Most popular questions from this chapter

Refer to the definition of symmetric difference given above. Prove each of \(40-45\), assuming that \(A, B\), and \(C\) are all subsets of a universal set \(U\). $$A \Delta B=B \Delta A$$

The following is a proof that for any sets \(A, B\), and \(C\), \(A \cap(B \cup C)=(A \cap B) \cup(A \cap C)\). Fill in the blanks. Proof: Suppose \(A, B\), and \(C\) are any sets. (1) \(A \cap(B \cup C) \subseteq(A \cap B) \cup(A \cap C):\) Let \(x \in A \cap(B \cup C)\). [We must show that \(x \in \underline{\text { (a) }}\).] By definition of intersection, \(x \in \underline{\text { (b) }}\) and \(x \in \frac{\text { (c) }}{\text { Thus }}\) \(x \in A\) and, by definition of union, \(x \in B\) or \((\mathrm{d})\) Case \(1(x \in A\) and \(x \in B):\) In this case, by definition of intersection, \(x \in \underline{(e)}\), and so, by definition of union, \(x \in(A \cap B) \cup(A \cap C) .\) Case \(2(x \in A\) and \(x \in C):\) In this case, (f) Hence in either case, \(x \in(A \cap B) \cup(A \cap C)\) [as was to be shown]. \([\) So \(A \cap(B \cup C) \subseteq(A \cap B) \cup(A \cap C)\) by definition of subset. (2) \((A \cap B) \cup(A \cap C) \subseteq A \cap(B \cup C):\) Let \(x \in(A \cap B) \cup(A \cap C)\). [We must show that \(\underline{\text { (a) }}\).] By definition of union, \(x \in \stackrel{\text { (b) }}{\text { or } x \in \stackrel{\text { (c) }}{\text { Let } x \in} \text {. }}\) Case \(1(x \in A \cap B):\) In this case, by definition of intersection, \(\frac{(\mathrm{d})}{}\) and \((\mathrm{e})\). Since \(x \in B\), then by definition of union, \(x \in B \cup C\). Hence \(x \in A\) and \(x \in B \cup C\), and so, by definition of intersection, \(x \in \underline{\text { (f) }}\). Case \(2(x \in A \cap C):\) In this case, \((g) .\) In either case, \(x \in A \cap(B \cup C)\) [as was to be shown]. [Thus \((A \cap B) \cup(A \cap C) \subseteq A \cap(B \cup C)\) by definition of subset. [Since both subset relations have been proved, it follows, by definition of set equality, that \(\stackrel{(\mathrm{h})}{-}\).]

Let \(S=\\{a, b, c\\}\) and let \(S_{a}\) be the set of all subsets of \(S\) that contain \(a\), let \(S_{b}\) be the set of all subsets of \(S\) that contain \(b\), let \(S_{c}\) be the set of all subsets of \(S\) that contain \(c\), and let \(S_{n}\) be the set whose only element is \(\emptyset\), Is \(\left\\{S_{a}, S_{b}, S_{c}, S_{\emptyset}\right\\}\) a partition of \(\mathscr{P}(S)\) ?

Prove each statement that is true and find a counterexample for each statement that is false. Assume all sets are subsets of a universal set \(U\). For all sets \(A\) and \(B\), if \(A \subseteq B\) then \(\mathscr{P}(A) \subseteq \mathscr{P}(B)\).

For all sets \(A\) and \(B,\left(\left(A^{c} \cup B^{c}\right)-A\right)^{c}=A\).

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