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sets.xml
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sets.xml
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<?xml version="1.0" encoding="UTF-8" ?>
<!-- This file is part of the book -->
<!-- -->
<!-- Logic and Proof for Teachers -->
<!-- -->
<!-- Copyright (C) 2019 Lesa L. Beverly, Kimberly M. Childs, Deborah A. Pace, Thomas W. Judson -->
<!-- -->
<!-- See the file COPYING for copying conditions. -->
<chapter xml:id="sets" xmlns:xi="http://www.w3.org/2001/XInclude">
<title>Sets</title>
<introduction>
<p>An underlying idea of mathematics is the concept of a collection of objects or a set.</p>
</introduction>
<assemblage>
<title>TEXAS STATE BOARD FOR EDUCATOR CERTIFICATION (SBEC): MATHEMATICS STANDARDS COVERED</title>
<p><ul>
<li>STANDARD V: MATHEMATICAL PROCESSES: The mathematics teacher understands and uses mathematical processes to reason mathematically, to solve mathematical problems, to make mathematical connections within and outside of mathematics, and to communicate mathematically.</li>
<li>STANDARD VI: MATHEMATICAL PERSPECTIVES: The mathematics teacher understands the historical development of mathematical ideas, the interrelationship between society and mathematics, the structure of mathematics, and the evolving nature of mathematics and mathematical knowledge.</li>
</ul></p>
</assemblage>
<definition xml:id="sets-definition-set">
<p>A <term>set</term> is a collection of distinct objects, which are called the <term>elements</term> of the set. We will use capital letters to denote sets, for example <m>a</m>, <m>B</m>, <m>S</m>, <m>T</m>, etc. If <m>x</m> is an element of a set <m>S</m>, we use the notation <m>x \in S</m>. As is frequently done in mathematics, if <m>x</m> is not an element of a set <m>S</m>, we denote this by <m>x \notin S</m>. Two sets, <m>A</m> and <m>B</m>, are <term>equal</term> (denoted <m>A = B</m>) if and only if they have precisely the same elements.</p>
<idx><h>set</h></idx>
<idx><h>elements</h></idx>
</definition>
<definition xml:id="sets-definition-subset">
<p>A set which is comprised of some (or all) of the elements of <m>\mathbb U</m>, some <term>universal set</term>, is called a subset of <m>\mathbb U</m> and is denoted <m>S \subseteq \mathbb U</m>. If <m>A \subseteq \mathbb U</m> and <m>B \subseteq \mathbb U</m>, we further state that <m>A</m> is a <term>subset</term> of <m>B</m>, denoted <m>A \subseteq B</m> or <m>A \subseteq B</m>, if each element of <m>A</m> is also an element of <m>B</m>. Moreover, if <m>A \subseteq B</m> but <m>A \neq B</m>, we say that <m>A</m> is a <term>proper subset</term> of <m>B</m> and write <m>A \subsetneq B</m>.</p>
</definition>
<idx><h>Subset</h></idx>
<idx><h>Proper subset</h></idx>
<idx><h>Universal set</h></idx>
<example xml:id="sets-example-subset">
<p>Let <m>\mathbb U = \{a, b, c, \ldots, z\}</m>, <m>S = \{ a, b, s, u \}</m>, <m>T = \{a, u\}</m>, and <m>W = \{ b,d \}</m>. The <m>S \subseteq \mathbb U</m>, <m>T \subseteq \mathbb U</m>, <m>W \subseteq \mathbb U</m>, and <m>T \subseteq S</m>. However, <m>W \not\subseteq S</m>, <m>W \not\subseteq T</m>, <m>T \not\subseteq W</m>, and <m>S \not\subseteq W</m>. We read <m>W \not\subseteq S</m> as <q><m>W</m> is not a subset of <m>S</m>.</q> In addition, <m>S \subset \mathbb U</m>, <m>T \subset \mathbb U</m>, <m>W \subset \mathbb U</m>, and <m>T \subset S</m>.</p>
</example>
<p>When referring to the definition of subset, you should note that to prove <m>A \subseteq B</m>, one would naturally consider an arbitrary element <m>x</m> of <m>A</m> and prove that <m>x \in B</m>. In other words, we must prove the conditional statement <q>If <m>x \in A</m>, then <m>x \in B</m>.</q></p>
<fact>
<p>Let <m>A</m> and <m>B</m> be sets. Then <m>A = B</m> if and only if <m>A \subseteq B</m> and <m>B \subseteq A</m>. We sometimes refer to this method of proof as <q>proofs by double inclusion.</q></p>
</fact>
<idx><h>Double inclusion</h></idx>
<p>Without entering into a philosophical argument, we define <m>\emptyset</m> to be the set with no elements, called the <term>empty set</term>, the <term>void set</term>, or the <term>null set</term>. This set is extremely useful and should be understood carefully.</p>
<idx><h>Empty set</h></idx>
<idx><h>Null set</h></idx>
<idx><h>Void set</h></idx>
<fact>
<p>Let <m>\mathbb U</m> be any universal set and <m>A \subseteq \mathbb U</m>. Then <m>\emptyset \subseteq A</m> and <m>\emptyset \subseteq \mathbb U</m>.</p>
</fact>
<proof>
<p>By definition of subset, we must show each element in <m>\emptyset</m> is also an element of <m>A</m> (or <m>\mathbb U</m>, if we are proving <m>\emptyset \subseteq \mathbb U</m>); however, since there are no elements in to check, the definition is satisfied.</p>
<p>Let us consider the proof above logically. Recall that <m>A \subseteq B</m> is a conditional statement (if <m>x \in A</m>, then <m>x \in B</m>). So we need to examine <q>If <m>x \in \emptyset</m>, then <m>x \in A</m>.</q> But <m>\emptyset</m> has no elements, so <m>x \in \emptyset</m> is false, meaning the conditional is true.</p>
</proof>
<p>The next definition introduces several basic ways of combining sets which are used throughout mathematics.</p>
<definition xml:id="sets-definition-set-operations">
<p>Let <m>\mathbb U</m> be the universal set with <m>A</m> and <m>B</m> subsets of <m>\mathbb U</m>. Then
<ul>
<li><m>A \cup B = \{ x \in \mathbb U \mid x \in A \text{ or } x \in B \}</m>, which is called <m>A</m> <term>union</term> <m>B</m>.</li>
<li><m>A \cap B = \{ x \in \mathbb U \mid x \in A \text{ and } x \in B \}</m>, which is called <m>A</m> <term>intersect</term> <m>B</m>.</li>
<li><m>\overline{A} = \{ x \in \mathbb U \mid x \notin A \}</m>, which is called the <term>complement</term> of <m>A</m>.</li>
<li><m>A \setminus B = \{ x \in \mathbb U \mid x \in A \text{ and } x \notin B \}</m>, which is read <m>A</m> <term>minus</term> <m>B</m>.</li>
<li><m>A</m> and <m>B</m> are set to be <term>disjoint</term> if <m>A \cap B = \emptyset</m>.</li>
</ul></p>
</definition>
<idx><h>Union</h></idx>
<idx><h>Intersection</h></idx>
<idx><h>Complement</h></idx>
<idx><h>Set minus</h></idx>
<idx><h>Disjoint</h></idx>
<p>While not sufficient for proof, Venn diagrams can be useful in visualizing these concepts (<xref first="figure-sets-venn-union" last="figure-sets-venn-complement" />).</p>
<figure xml:id="figure-sets-venn-union">
<caption>The shaded area represents <m>A \cup B</m></caption>
<image width="40%" xml:id="sets-venn-union">
<description>two shaded overlapping circles</description>
<latex-image>
<!-- CDATA prevents certain LaTeX code from being interpreted as xml -->
<![CDATA[
\begin{tikzpicture}[scale=1]
\draw [very thick] (0,0) rectangle (6,4);
\draw [fill=blue!20] (2,2) circle(1.5);
\draw [fill=blue!20] (4,2) circle(1.5);
\draw [very thick] (2,2) circle(1.5);
\draw [very thick] (4,2) circle(1.5);
\node at (2,3) {$A$};
\node at (4,3) {$B$};
\node at (0.75,3.5) {$\mathbb U$};
\end{tikzpicture}]]>
</latex-image>
</image>
</figure>
<figure xml:id="figure-sets-venn-intersect">
<caption>The shaded area represents <m>A \cap B</m></caption>
<image width="40%" xml:id="sets-venn-intersect">
<description>two overlapping circles with the common part shaded</description>
<latex-image>
<!-- CDATA prevents certain LaTeX code from being interpreted as xml -->
<![CDATA[
\begin{tikzpicture}[scale=1]
\draw [very thick] (0,0) rectangle (6,4);
\begin{scope}
\clip (2,2) circle (1.5);
\clip (4,2) circle (1.5);
\fill[color=blue!20] (0,0) rectangle (6,4);
\end{scope}
\draw [very thick] (2,2) circle(1.5);
\draw [very thick] (4,2) circle(1.5);
\node at (2,3) {$A$};
\node at (4,3) {$B$};
\node at (0.75,3.5) {$\mathbb U$};
\end{tikzpicture}]]>
</latex-image>
</image>
</figure>
<figure xml:id="figure-sets-venn-minus">
<caption>The shaded area represents <m>A \setminus B</m></caption>
<image width="40%" xml:id="sets-venn-minus">
<description>two overlapping circles with the non-overlapping part of one shaded</description>
<latex-image>
<!-- CDATA prevents certain LaTeX code from being interpreted as xml -->
<![CDATA[
\begin{tikzpicture}[scale=1]
\draw [very thick] (0,0) rectangle (6,4);
\draw [fill=blue!20] (2,2) circle(1.5);
\draw [fill=white] (4,2) circle(1.5);
\draw [very thick] (2,2) circle(1.5);
\draw [very thick] (4,2) circle(1.5);
\node at (2,3) {$A$};
\node at (4,3) {$B$};
\node at (0.75,3.5) {$\mathbb U$};
\end{tikzpicture}]]>
</latex-image>
</image>
</figure>
<figure xml:id="figure-sets-venn-complement">
<caption>The shaded area represents <m>\overline{A}</m></caption>
<image width="40%" xml:id="sets-venn-complement">
<description>two overlapping circles with the area outside one circle shaded</description>
<latex-image>
<!-- CDATA prevents certain LaTeX code from being interpreted as xml -->
<![CDATA[
\begin{tikzpicture}[scale=1]
\fill[color=blue!20] (0,0) rectangle (6,4);
\draw [very thick] (0,0) rectangle (6,4);
\draw [fill=white] (2,2) circle(1.5);
\draw [very thick] (2,2) circle(1.5);
\draw [very thick] (4,2) circle(1.5);
\node at (2,3) {$A$};
\node at (4,3) {$B$};
\node at (0.75,3.5) {$\mathbb U$};
\end{tikzpicture}]]>
</latex-image>
</image>
</figure>
<assemblage>
<title>TAKS CONNECTION</title>
<p>How might a student apply the concept of set intersections to answer the following question taken from the 2006 Texas Assessment of Knowledge and Skills (TAKS) Grade 5 Mathematics test?</p>
<p>The Venn diagram below is used to classify counting numbers according to a set of rules.</p>
<sidebyside width="60%">
<image xml:id="sets-venn-taks-2">
<description>three overlapping circles</description>
<latex-image>
<!-- CDATA prevents certain LaTeX code from being interpreted as xml -->
<![CDATA[
\begin{tikzpicture}[scale=1]
\draw [very thick] (0,0) rectangle (6,5.5);
\draw [very thick] (2,3.5) circle(1.5);
\draw [very thick] (4,3.5) circle(1.5);
\draw [very thick] (3,2) circle(1.5);
\node at (3,3) {?};
\node at (5,4) {$3$};
\node at (4,4.5) {$6$};
\node at (5,3) {$9$};
\node at (3,4) {$12$};
\node at (3.6,2.7) {$15$};
\node at (4.2,2.3) {$30$};
\node at (2,1.5) {$5$};
\node at (4,1.5) {$10$};
\node at (2.3,2.7) {$20$};
\node at (3.5,1) {$25$};
\node at (2.5,1) {$35$};
\node at (2,4) {$4$};
\node at (2.3,4.5) {$8$};
\node at (1.5,3.5) {$16$};
\node at (1.5,4.5) {$28$};
\node at (1,4) {$32$};
\end{tikzpicture}]]>
</latex-image>
</image>
</sidebyside>
<p>Which one of the following numbers belongs in the region of the diagram marked by the question mark?
<ul>
<li>A. 45</li>
<li>B. 50</li>
<li>C. 60</li>
<li>D. 65</li>
</ul></p>
</assemblage>
<definition xml:id="sets-definition-power-set">
<p>If <m>A</m> is a set, then the <term>power set</term> of <m>A</m> is the set <m>\mathcal P(A) = \{ B \mid B \subseteq A \}</m>.</p>
</definition>
<idx><h>Power set</h></idx>
<example xml:id="sets-example-power-set">
<p>If <m>A = \{ 1,2\}</m>, then <m>\mathcal P(A) = \{ \emptyset, \{1 \}, \{2\}, A\}</m>. If <m>B = \{ a, b, c \}</m>, then <m>\mathcal P(B) = \{ \emptyset, \{ a \}, \{ b \}, \{ c \}, \{ a, b \}, \{ a, c \}, \{ b, c \}, B \}</m>.</p>
</example>
<definition xml:id="sets-definition-cartesian-product">
<p>If <m>A</m> and <m>B</m> are sets, then <m>A \times B = \{ (a, b) \mid a \in A \text{ and } b \in B \}</m> is called the <term>Cartesian product</term> or the <term>cross product</term> of <m>A</m> and <m >B</m>.</p>
</definition>
<idx><h>Cartesian product</h></idx>
<idx><h>Cross product</h></idx>
<example xml:id="sets-example-cartesian-product">
<p>If <m>A = \{ 1,2\}</m> and <m>B = \{ a, b, c \}</m>, then <m>A \times B = \{ (1, a), (1, b), (1, c), (2, a), (2, b), (2, c)\}</m>.</p>
</example>
<fact xml:id="sets-fact-sets">
<p>Let <m>A</m>, <m>B</m>, and <m>C</m> be subsets of a universal set <m>\mathbb U</m>. Then
<ol>
<li><m>A \subseteq A \cup B</m>.</li>
<li><m>A \cap B \subseteq A</m>.</li>
<li><m>A \setminus B \subseteq A</m>.</li>
<li><m>A \setminus B</m> and <m>B \setminus A</m> are disjoint.</li>
<li><m>A</m> and <m>\overline{A}</m> are disjoint.</li>
<li>If <m>A \subseteq B</m> and <m>B \subseteq C</m>, then <m>A \subseteq C</m>.</li>
<li><m>A \cup B = B \cup A</m>.</li>
<li><m>A \cap B = B \cap A</m>.</li>
<li><m>A \cap (B \cap C) = (A \cap B) \cap C</m>.</li>
<li><m>A \cup (B \cup C) = (A \cup B) \cup C</m>.</li>
<li><m>A \cap (B \cup C) = (A \cap B) \cup (A \cap C)</m>.</li>
<li><m>A \cup (B \cap C) = (A \cup B) \cap (A \cup C)</m>.</li>
<li><m>A \subseteq B</m> if and only if <m>\overline{B} \subseteq \overline{A}</m>.</li>
<li><m>A \setminus B = A \cap \overline{B}</m>.</li>
<li><m>\overline{A \cup B} = \overline{A} \cap \overline{B}</m>.</li>
<li><m>\overline{A \cap B} = \overline{A} \cup \overline{B}</m>.</li>
<li><m>\overline{\overline{A}} = A </m>.</li>
<li><m>A \cup \emptyset = \emptyset \cup A = A</m>.</li>
<li><m>A \cap \emptyset = \emptyset \cap A = \emptyset</m>.</li>
<li><m>A \cup \mathbb U = \mathbb U \cup A = \mathbb U</m>.</li>
<li><m>A \cap \mathbb U = \mathbb U \cap A = A</m>.</li>
</ol></p>
</fact>
<example xml:id="sets-example-fact-3">
<p>We will prove (3) in <xref ref="sets-fact-sets" />. That is, we will show <m>A \setminus B \subseteq A</m>.</p>
<p><em>Proof:</em> Let <m>x \in A \setminus B</m>. Then <m>x \in A</m> but <m>x \notin B</m>. By the definition of a subset <m>A \setminus B \subseteq A</m>. That is, <q>if <m>x \in A \setminus B</m>, then <m>x \in A
</m>.</q></p>
</example>
<example xml:id="sets-example-fact-7">
<p>We will prove (7) in <xref ref="sets-fact-sets" />. That is, we will show <m>A \cup B = B \cup A</m>.</p>
<p><em>Proof:</em> Let <m>x \in A \cup B</m>. Then <m>x \in A</m> or <m>x \in B</m> by the definition of <m>A \cup B</m>. Hence, <m>x \in B</m> or <m>x \in A</m>. By the definition of union <m>x \in B \cup A</m>. We have argued by direct proof that <q><m>x \in A \cup B</m>, then <m>x \in B \cup A</m>.</q> Thus, <m>A \cup B \subseteq B \cup A</m>. By reversing the proof, we can show that <m>B \cup A \subseteq A \cup B</m>. Therefore, <m>A \cup B = B \cup A</m></p>
</example>
<example xml:id="sets-example-fact-16">
<p>We will prove (16) in <xref ref="sets-fact-sets" />. That is, we will show <m>\overline{A \cap B} = \overline{A} \cup \overline{B}</m>.</p>
<p><em>Proof:</em> To show <m>\overline{A \cap B} = \overline{A} \cup \overline{B}</m>, we again use the method of double inclusion. Let <m>x \in \overline{A \cap B}</m>. Then <m>x \notin A \cap B</m>. Recalling the definition of intersection, we notice that the statement <m>x \notin A \cap B</m> parallels the logic statement <m>\negate( p \wedge q)</m>. DeMorgan's Laws for logic tell us that <m>[\negate( p \wedge q)] \leftrightarrow [\negate p \; \vee \negate q]</m>. So we now have <m>x \notin A</m> or <m>x \notin B</m>; hence, <m>x \in \overline{A}</m> or <m>x \in \overline{B}</m>. Thus, by definition of union, <m>x \in A \cup B</m>. Consequently, <m>\overline{A \cap B} \subseteq \overline{A} \cup \overline{B}</m>.</p>
<p>To see that <m>\overline{A} \cup \overline{B} \subseteq \overline{A \cap B}</m> let <m>y \in \overline{A} \cup \overline{B}</m>. Then <m>y \in \overline{A}</m> or <m>y \in \overline{B}</m>, which means that <m>y \notin A</m> or <m>y \notin B</m>. Arguing as above, we have <m>y \notin A \cap B</m>. Thus, <m>y \in \overline{A \cap B}</m>, and <m>\overline{A} \cup \overline{B} \subseteq \overline{A \cap B}</m>.</p>
</example>
<p>As indicated, the proofs of the other parts of this fact are left as exercises.</p>
<paragraphs xml:id="sets-paragraphs-historical-note">
<title>Historical Note</title>
<p>Unlike most areas of mathematics, which arise as a result of the cumulative efforts of many mathematicians, sometimes over several generations, set theory is the creation of a single individual. Georg Ferdinand Ludwig Phillip Cantor (27 June 1806<ndash />18 March 1871) was a German mathematician, born in Russia. Use internet and/or library resources to research his major contributions to the area of set theory, specifically the concept of cardinality and Cantor's Theorem.</p>
<idx><h>Cantor, Georg Ferdinand Ludwig Phillip</h></idx>
</paragraphs>
<exercises>
<exercise xml:id="exercise-sets-setops">
<statement>
<p>Let <m>\mathbb U = \{1, 2, \ldots 10 \}</m>, <m>A = \{ 1, 2, 3, 4, 5 \}</m>, <m>B = \{ 2, 3, 5 \}</m>, <m>D = \{ 1, 3, 5, 7, 9 \}</m>, and <m>E = \{ 2, 4, 6, 8, 10 \}</m>. Label each of the statements below as True or False and explain.
<ol cols="2">
<li><m>D</m> and <m>E</m> are disjoint.</li>
<li><m>A \subseteq E</m>.</li>
<li><m>B \subset A</m>.</li>
<li><m>A \subset \mathbb U</m>.</li>
<li><m>(A \cap D) \subset A</m>.</li>
<li><m>(A \cup E) \subset \mathbb U</m>.</li>
<li><m>(D \cup E) \subset \mathbb U</m>.</li>
<li><m>(D \cup E) = \mathbb U</m>.</li>
<li><m>6 \in D</m>.</li>
<li><m>\{ 8 \} \in E</m>.</li>
<li><m>\emptyset \subseteq B</m>.</li>
<li><m>\{ 2, 3 \} \subseteq B</m>.</li>
<li><m>5 \subseteq B</m>.</li>
<li><m>\emptyset \subseteq \emptyset</m>.</li>
<li><m>\overline{D} = E</m>.</li>
<li><m>\overline{B \cup D} \subseteq E</m>.</li>
</ol></p>
</statement>
</exercise>
<exercise>
<statement>
<p>Using the sets in <xref ref="exercise-sets-setops" />, find:
<ol cols="2">
<li><m>\overline{B}</m></li>
<li><m>D \setminus A</m></li>
<li><m>\overline{D} \cap \overline{A}</m></li>
<li><m>\overline{D \cup A}</m></li>
<li><m>E \cap (B \cup D)</m></li>
<li><m>\overline{A} \cap (B \setminus D)</m></li>
<li><m>\overline{D \cap E}</m></li>
<li><m>\overline{D \setminus E}</m></li>
</ol></p>
</statement>
</exercise>
<exercise>
<statement>
<p>Prove (1) in <xref ref="sets-fact-sets" /></p>
</statement>
</exercise>
<exercise>
<statement>
<p>Prove (2) in <xref ref="sets-fact-sets" /></p>
</statement>
</exercise>
<exercise>
<statement>
<p>Prove (4) in <xref ref="sets-fact-sets" /></p>
</statement>
</exercise>
<exercise>
<statement>
<p>Prove (5) in <xref ref="sets-fact-sets" /></p>
</statement>
</exercise>
<exercise>
<statement>
<p>Prove (6) in <xref ref="sets-fact-sets" /></p>
</statement>
</exercise>
<exercise>
<statement>
<p>Prove (8) in <xref ref="sets-fact-sets" /></p>
</statement>
</exercise>
<exercise>
<statement>
<p>Prove (9) in <xref ref="sets-fact-sets" /></p>
</statement>
</exercise>
<exercise>
<statement>
<p>Prove (10) in <xref ref="sets-fact-sets" /></p>
</statement>
</exercise>
<exercise>
<statement>
<p>Prove (11) in <xref ref="sets-fact-sets" /></p>
</statement>
</exercise>
<exercise>
<statement>
<p>Prove (12) in <xref ref="sets-fact-sets" /></p>
</statement>
</exercise>
<exercise>
<statement>
<p>Prove the second half of Part (13) in <xref ref="sets-fact-sets" /></p>
</statement>
</exercise>
<exercise>
<statement>
<p>Prove (14) in <xref ref="sets-fact-sets" /></p>
</statement>
</exercise>
<exercise>
<statement>
<p>Prove (15) in <xref ref="sets-fact-sets" /></p>
</statement>
</exercise>
<exercise>
<statement>
<p>Prove (17) in <xref ref="sets-fact-sets" /></p>
</statement>
</exercise>
<exercise>
<statement>
<p>Prove (18) in <xref ref="sets-fact-sets" /></p>
</statement>
</exercise>
<exercise>
<statement>
<p>Prove (19) in <xref ref="sets-fact-sets" /></p>
</statement>
</exercise>
<exercise>
<statement>
<p>Prove (20) in <xref ref="sets-fact-sets" /></p>
</statement>
</exercise>
<exercise>
<statement>
<p>Prove (21) in <xref ref="sets-fact-sets" /></p>
</statement>
</exercise>
<p>For each of the following problems, assume <m>A</m>, <m>B</m>, <m>C</m>, and <m>D</m> are subsets of some universal set <m>\mathbb U</m>. In problems <xref first="sets-exercise-21" last="sets-exercise-33" />, prove that the statements are true.</p>
<exercise xml:id="sets-exercise-21">
<statement>
<p>If <m>C \subseteq A</m> or <m>C \subseteq B</m>, then <m>C \subseteq A \cup B</m>.</p>
</statement>
</exercise>
<exercise xml:id="sets-exercise-22">
<statement>
<p>If <m>A \subseteq C</m> and <m>B \subseteq C</m>, then <m>A \cap B \subseteq C</m>.</p>
</statement>
</exercise>
<exercise xml:id="sets-exercise-23">
<statement>
<p><m>A \setminus A = \emptyset</m>.</p>
</statement>
</exercise>
<exercise xml:id="sets-exercise-24">
<statement>
<p><m>A \setminus (A \setminus B) \subseteq B</m>.</p>
</statement>
</exercise>
<exercise xml:id="sets-exercise-25">
<statement>
<p>If <m>A</m> or <m>B</m> are disjoint, then <m>B \subseteq \overline{A}</m>.</p>
</statement>
</exercise>
<exercise xml:id="sets-exercise-26">
<statement>
<p>If <m>A \subseteq C</m> and <m>B \subseteq D</m>, then <m>A \setminus D \subseteq C \setminus B</m>.</p>
</statement>
</exercise>
<exercise xml:id="sets-exercise-27">
<statement>
<p><m>A \setminus (B \setminus C) = A \cap (\overline{B} \cup C)</m>.</p>
</statement>
</exercise>
<exercise xml:id="sets-exercise-28">
<statement>
<p><m>(A \setminus B) \setminus C = A \setminus (B \cup C )</m>.</p>
</statement>
</exercise>
<exercise xml:id="sets-exercise-29">
<statement>
<p><m>A \setminus (B \cup C) = (A \setminus B) \cap (A \setminus C)</m>.</p>
</statement>
</exercise>
<exercise xml:id="sets-exercise-30">
<statement>
<p><m>A \subseteq B</m> if and only if <m>A \cap B = A</m>.</p>
</statement>
</exercise>
<exercise xml:id="sets-exercise-31">
<statement>
<p><m>A \subseteq B</m> if and only if <m>A \cup B = B</m>.</p>
</statement>
</exercise>
<exercise xml:id="sets-exercise-32">
<statement>
<p><m>B \subseteq A</m> if and only if <m>A \cup \overline{B} = \mathbb U</m>.</p>
</statement>
</exercise>
<exercise xml:id="sets-exercise-33">
<statement>
<p>If <m>A \subseteq B</m> and <m>C \subseteq D</m>, then
<ol>
<li><m>A \cup C \subseteq B \cup D</m> and</li>
<li><m>A \cap C \subseteq B \cap D</m>.</li>
</ol></p>
</statement>
</exercise>
<exercise>
<statement>
<p>Prove or disprove:
<ol>
<li>If <m>A \cup B = A \cup C</m>, then <m>B = C</m>.</li>
<li>If <m>A \cap B = A \cap C</m>, then <m>B = C</m>.</li>
</ol></p>
</statement>
</exercise>
<exercise>
<statement>
<p>Define <m>A \bigtriangleup B = (A \setminus B) \cup (B \setminus A)</m>. Prove or disprove:
<ol>
<li><m>A \bigtriangleup B = B \bigtriangleup A</m>.</li>
<li><m>A \bigtriangleup (B \bigtriangleup C) = (A \bigtriangleup B) \bigtriangleup C</m>.</li>
<li><m>A \bigtriangleup \emptyset = A</m>.</li>
<li><m>A \bigtriangleup A = \emptyset</m>.</li>
<li><m>A \bigtriangleup B = (A \cup B) \setminus (A \cap B)</m>.</li>
<li><m>A \cap (B \bigtriangleup C) = (A \cap B) \bigtriangleup (A \cap C)</m>.</li>
<li><m>A \cup (B \bigtriangleup C) = (A \cup B) \bigtriangleup (A \cup C)</m>.</li>
<li><m>A \cap B = \emptyset</m> if and only if <m>A \bigtriangleup B = A \cup B</m>.</li>
</ol></p>
</statement>
</exercise>
<exercise>
<statement>
<p>Draw Venn diagrams to illustrate various parts of <xref ref="sets-fact-sets" />.</p>
</statement>
</exercise>
<exercise>
<statement>
<p>For sets <m>A = \{1,2, 3, 4, 5\}</m>, <m>B = \{ 3, 5, 7, 9\}</m>, and <m>C = \{a, b \}</m>, find:
<ol>
<li><m>\mathcal P(C)</m>.</li>
<li><m>A \times C</m>.</li>
<li><m>C \times (A \cap B)</m>.</li>
<li><m>(C \times A) \cap (C \times B)</m>.</li>
</ol></p>
</statement>
</exercise>
<exercise>
<statement>
<p>Prove or disprove: If <m>A, B \in \mathcal P(S)</m>, then <m>A \cup B \in \mathcal P(S)</m>.</p>
</statement>
</exercise>
<exercise>
<statement>
<p>Prove or disprove: If <m>A, B \in \mathcal P(S)</m>, then <m>A \cap B \in \mathcal P(S)</m>.</p>
</statement>
</exercise>
<exercise>
<statement>
<p>Is it true that <m>A \times \emptyset = \emptyset \times A = \emptyset</m>? Justify your thinking.</p>
</statement>
</exercise>
</exercises>
</chapter>