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chapter_6/exercises.md
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@ -1390,3 +1390,309 @@ Omitted.
given set that is represented as the array $a[1], a[2], \dots, a[n]$. given set that is represented as the array $a[1], a[2], \dots, a[n]$.
Omitted. Omitted.
---
Page 427
**Exercise Set 6.2**
1.
a. To say that an element is in $A \cap (B \cup C)$ means that it is in __ (1)
__ and in __ (2) __.
b. To say that an element is in $(A \cap B) \cup C$ means that it is in __ (1)
__ or in __ (2) __.
c. To say that an element is in $A - (B \cap C)$ means that it is in __ (1) __
and not in __ (2)__.
d. To prove that $(A \cup B) \cap C \subseteq A \cup (B \cap C)$, we suppose
that $x$ is any element in __ (1) __. Then we must show that __ (2) __.
e. If $A$, $B$, and $C$ are any sets such that $B \subseteq C$, to prove that
$A \cap B \subseteq A \cap C$, we suppose that $x$ is any element in __ (1) __.
Then we must show that __ (2) __.
2. The following are two proofs that for all sets $A$ and $B$,
$A - B \subseteq A$. The first is less formal, and the second is more formal.
Fill in the blanks.
a. **Proof:** Suppose $A$ and $B$ are any sets. To show that
$A - B \subseteq A$, we must show that every element in __ (1) __ is in __ (2)
__. But any element in $A - B$ is in __ (3) __ and not in __ (4) __ (by
definition of $A - B$). In particular, such an element is in $A$.
b. **Proof:** Suppose $A$ and $B$ are any sets and $x \in A - B$. _[We must show
that __ (1) __.]_ By definition of set difference, $x \in$ __ ( 2 ) __ and
$x \notin$ __ (3) __. In particular, $x \in$ __ (4) __ _[which is what was to be
shown]._
In 3 and 4, supply explanations of the stesp in the given proofs.
3. **Theorem:** For all sets $A$, $B$, and $C$, if $A \subseteq C$,
$B \subseteq C$, then $A \subseteq C$.
**Proof:**
| Statement | Explanation |
| ------------------------------------------------------------------------------------ | ------------------------------------- |
| Suppose $A$, $B$, and $C$ are any sets such that $A \subseteq B$ and $B \subseteq C$ | starting point |
| We must show that $A \subseteq C$. | conclusion to be shown |
| Let $x$ be any element in $A$. | start of an element proof |
| Then $x$ is in $B$. | __ (a) __ |
| It follows that $x$ is in $C$. | __ (b) __ |
| Thus every element in $A$ is in $C$ | since $x$ could be any element of $A$ |
| Therefore, $A \subseteq C$ _[as was to be shown]._ | __ \(c\) __ |
4. **Theorem:** For all sets $A$ and $B$, if $A \subseteq B$, then
$A \cup B \subseteq B$.
**Proof:**
| Statement | Explanation |
| ----------------------------------------------------------------- | -------------------------------------------- |
| Suppose $A$, $B$, and $C$ are any sets such that $A \subseteq B$. | starting point |
| We must show that $A \cup B \subseteq B$ | conclusion to be shown |
| Let $x$ be any element in $A \cup B$. | start of an element proof |
| Then $x$ is in $A$ or $x$ is in $B$. | __ (a) __ |
| In case $x$ is in $A$, then $x$ is in $B$ | __ (b) __ |
| In case $x$ is in $B$, then $x$ is in $B$. | tautology ($p \to p$) |
| So in either case $x$ is in $B$. | proof by division into cases |
| Thus every element in $A \cup B$ is in $B$ | since $x$ could be any element of $A \cup B$ |
| Therefore, $A \cup B \subseteq B$ _[as was to be shown]._ | __ \(c\) __ |
5. Prove that for all sets $A$ and $B$, $(B - A) = B \cap A^c$.
6. Let $\cap$ and $\cup$ stand for the words "intersection" and "union",
respectively. Fill in the blanks in the following proof that for all sets
$A$, $B$, and $C$, $A \cap (B \cup C) = (A \cap C) \cup (A \cap C)$.
**Proof:** Suppose $A$, $B$, and $C$ are any sets.
(1) Proof that $A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C)$:
Let $x \in A \cap (B \cup C)$. _[We must show that $x \in$ __ (a) __ ]._
By definition of $\cap$, $x \in$ __ (b) __ and $x \in B \cup C$.
Thus $x \in A$ and, by definition of $\cup$, $x \in B$ or __ \(c\) __.
_Case 1 $(x \in A \text{ and } x \in B)$:_ In this case, $x \in A \cap B$ by
definition of $\cap$.
_Case 2 $(x \in A \text{ and } x \in C)$:_ IN this case, $x \in A \cap C$ by
definition of $\cap$.
By cases 1 and 2, $x \in A \cap B$ or $x \in A \cap C$, and so, by definition of
$\cup$, __ (d) __.
_[So $A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C)$ by definition of
subset.]_
(2) Proof that $(A \cap B) \cup (A \cap C) \subseteq A \cap (B \cup C)$:
Let $x \in (A \cap B) \cup (A \cap C)$. _[We must show that
$x \in A \cap (B \cup C)$.]_
By definition of $\cup$, $x \in A \cap B$ __ (a) __ $x \in A \cap C$.
_Case 1 $(x \in A \cap B)$:_ In this case, by definition of $\cap$, $x \in A$
and $x \in B$$.
Since $x \in B$, then $x \in B \cup C$ by definition of $\cup$.
_Case 2 $(x \in A \cap C)$:_ In this case, by definition of $\cap$, $x \in A$ __
(b) __ $x \in C$.
Since $x \in C$, then $x \in B \cup C$ by definition of $\cup$.
In both cases $x \in A$ and $$ix \in B \cup C, and so, by definition of $\cap$,
__ \(c\) __.
_[So $(A \cap B) \cup (A \cap C) \subseteq A \cap (B \cup C)$ by definition of
__ (d) __ .]_
(3) Conclusion: _[Since both subset relations have been proved, it follows, by
definition of set equality, that __ (a) __.]_
Use an element argument to prove each statement in 7-22. Assume that all sets
are subsets of a universal set $U$.
7. For all sets $A$ and $B$, $(A \cap B)^c = A^c \cup B^c$.
8. For all sets $A$ and $B$, $(A \cap B) \cup (A \cap B^c) = A$.
(This property is used in Section 9.9.)
9. For all sets $A$, $B$, and $C$,
$$ (A - B) \cup (C - B) = (A \cup C) - B $$
10. For all sets $A$, $B$, and $C$,
$$ (A \cup B) \cap C \subseteq A \cup (B \cap C) $$
11. For all sets $A$, $B$, and $C$,
$$ A \cap (B - C) \subseteq (A \cap B) - (A \cap C) $$
12. For all sets $A$, $B$, and $C$,
$$ (A \cup B) - C \subseteq (A - C) \cup (B - C) $$
13. For all sets $A$, $B$, and $C$,
$$ (A - B) \cap (C - B) = (A \cap C) - B $$
14. For all sets $A$ and $B$, $A \cup (A \cap B) = A$.
15. For every set $A$, $A \cup \emptyset = A$.
16. For all sets $A$, $B$, and $C$, if $A \subseteq B$ then
$A \cap C \subseteq B \cap C$.
17. For all sets $A$, $B$, and $C$, if $A \subseteq B$ then
$A \cup C \subseteq B \cup C$.
18. For all sets $A$ and $B$, if $A \subseteq B$ then $B^c \subseteq A^c$.
19. For all sets $A$, $B$, and $C$, if $A \subseteq B$ and $A \subseteq C$ then
$A \subseteq B \cap C$.
20. For all sets $A$, $B$, and $C$, if $A \subseteq C$ and $B \subseteq C$ then
$A \cup B \subseteq C$.
21. For all sets $A$, $B$, and $C$,
$$ A \times (B \cup C) = (A \times B) \cup (A \times C) $$
22. For all sets $A$, $B$, and $C$,
$$ A \times (B \cap C) = (A \times B) \cap (A \times C) $$
23. Find the mistake in the following "proof" that for all sets $A$, $B$, and
$C$, if $A \subseteq B$ and $B \subseteq C$ then $A \subseteq C$.
**Proof:** Suppose $A$, $B$, and $C$ are any sets such that $A \subseteq B$ and
$B \subseteq C$. Since $A \subseteq B$, there is an element $x$ such that
$x \in A$ and $x \in B$, and since $B \subseteq C$, there is an element $x$ such
that $x \in B$ and $x \in C$. Hence there is an element $x$ such that $x \in A$
and $x \in C$ and so $A \subseteq C$.
24. Find the mistake in the following "proof."
**Theorem:** For all sets $A$ and $B$, $A^c \cup B^c \subseteq (A \cup B)^c^c$
**Proof:** Suppose $A$ and $B$ are any sets, and $x \in A^c \cup B^c$. Then
$x \in A^c$ or $x \in B^c$ by definition of union. It follows that $x \notin A$
or $x \notin B$ by definition of complement, and so $x \notin A \cup B$ by
definition of union. Thus $x \in (A \cup B)^c$ by definition of complement, and
hence $A^c \cup B^c \subseteq (A \cup B)^c$.
25. Find the mistake in the following "proof" that for all sets $A$ and $B$,
$(A - B) \cup (A \cap B) \subseteq A$.
**Proof:** Suppose $A$ and $B$ are any sets, and suppose
$x \in (A - B) \cup (A \cap B)$. If $x \in A$ then $x \in A - B$, and so, by
definition of difference, $x \in A$ and $x \notin B$. In particular, $x \in A$,
and, therefore, $(A - B) \cup (A \cap B) \subseteq A$ by definition of subset.
26. Consider the Venn diagram below.
(See page 429 for image.)
a. Illustrate one of the distributive laws by shading in the region
corresponding to $A \cup (B \cap C)$ on one copy of the diagram and
$(A \cup B) \cap (A \cup C)$ on another.
b. Illustrate the other distributive law by shading in the region corresponding
to $A \cap (B \cup C)$ on one copy of the diagram and
$(A \cap B) \cup (A \cap C)$ on another.
c. Illustrate one of De Morgan's laws by shading in the region corresponding to
$(A \cup B)^c$ on one copy of the diagram and $A^c \cap B^c$ on the other.
(Leave the set $C$ out of your diagrams.)
d. Illustrate the other De Morgan's law by shading in the region corresponding
to $(A \cap B)^c$ on one copy of the diagram and $A^c \cup B^c$ on the other.
(Leave the set $C$ out of your diagrams.)
27. Fill in the blanks in the following proof that for all sets $A$ and $B$,
$(A - B) \cap (B - A) = \emptyset$.
**Proof:**
Let $A$ and $B$ be any sets and suppose $(A - B) \cap (B - A) \neq \emptyset$.
That is, suppose there is an element $x$ in __ (a) __. BY definition of __ (b)
__, $x \in A - B$ and $x \in$ __ \(c\) __. Then by definition of set difference,
$x \in A$ and $x \notin B$ and $x \in$ __ (d) __ and $x \notin$ __ (e) __. IN
particular $x \in A$ and $x \notin$ __ (f) __, which is a contradiction. Hence
_[the supposition that $(A - B) \cap (B - A) \neq \emptyset$ is false, and so]_
__ (g) __.
Use the element method for proving a set equals the empty set to prove each
statement in 28-38. Assume that all sets are subsets of a universal set $U$.
28. For all sets $A$ and $B$, $(A \cap B) \cap (A \cap B^c) = \emptyset$. (This
property is used in Section 9.9.)
29. For all sets $A$, $B$, and $C$,
$$ (A - C) \cap (B - C) \cap (A - B) = \emptyset $$
30. For every subset $A$ of a universal set $U$, $A \cap A^c = \emptyset$.
31. If $U$ denotes a universal set, then $U^c = \emptyset$.
32. For every set $A$, $A \times \emptyset = \emptyset$.
33. For all sets $A$ and $B$, if $A \subseteq B$ then $A \cap B^c = \emptyset$.
34. For all sets $A$ and $B$, if $B \subseteq A^c$ then $A \cap B = \emptyset$.
35. For all sets $A$, $B$, and $C$, if $A \subseteq B$ and
$B \cap C = \emptyset$ then $A \cap C = \emptyset$.
36. For all sets $A$, $B$, and $C$, if $C \subseteq B - A$, then
$A \cap C = \emptyset$.
37. For all sets $A$, $B$, and $C$, if $B \cap C \subseteq A$, then
$(C - A) \cap (B - A) = \emptyset$.
38. For all sets $A$, $B$, $C$, and $D$, if $A \cap C = \emptyset$ then
$(A \times B) \cap (C \times D) = \emptyset$.
Prove each statement in 39-44.
39. For all sets $A$ and $B$,
a. $(A - B) \cup (B - A) \cup (A \cap B) = A \cup B$
b. The sets $(A - B)$, $(B - A)$, and $(A \cap B)$ are mutually disjoint.
40. For every positive integer $n$, if $A$ and $B_1, B_2, B_3, \dots$ are any
sets, then
$$ A \cap \left(\bigcup_{i = 1}^{n}B_i\right) = \bigcup_{i = 1}^{n}(A \cap B_i) $$
41. For every positive integer $n$, if $A_1, A_2, A_3, \dots$ and $B$ are any
sets, then
$$ \bigcap_{i = 1}^{n}(A_i - B) = \left(\bigcup_{i = 1}^{n}A_i\right) - B $$
42. For every positive integer $n$, if $A_1, A_2, A_3, \dots$ and $B$ are any
sets, then
$$ \bigcap_{i = 1}^{n}(A_i - B) = \left(\bigcap_{i = 1}^{n}A_i\right) - B $$
43. For every positive integer $n$, if $A$ and $B_1, B_2, B_3, \dots$ are any
sets, then
$$ \bigcup_{i = 1}^{n}(A \times B_i) = A \times \left(\bigcup_{i = 1}^{n}B_i\right) $$
44. For every positive integer $n$, if $A$ and $B_1, B_2, B_3, \dots$ are any
sets, then
$$ \bigcap_{i = 1}^{n}(A \times B_i) = A \times \left(\bigcap_{i = 1}^{n}B_i\right) $$

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chapter_6/notes.md
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@ -170,3 +170,265 @@ $b[1], b[2], \dots, b[n]$ [a one-dimensional array representing the set $B$]_
$i := 1, \text{answer} := A \subseteq B\\ \text{\textbf{while}} (i \leq m \text{ and answer } = A \subseteq B )\\ \ \ j := 1, \text{found} := \text{"no"}\\ \ \ \text{\textbf{while }} (j \neq n \text{ and } \text{found}= \text{"no"})\\ \ \ \ \ \text{\textbf{if }} a[i] = b[j] \text{\textbf{ then }} \text{found} := \text{"yes"}\\ \ \ \ \ j := j + 1\\ \ \ \text{\textbf{end while}}\\ \ \ \text{[If found has not been given the value "yes" when execution reaches this point, then } a[i] \neq B\text{ .]}\\ \ \ \text{\textbf{if }} \text{found} = \text{"no"} \text{\textbf{ then }} \text{answer} := A \nsubseteq B\\ \ \ i := i + 1\\ \text{\textbf{end while}}$ $i := 1, \text{answer} := A \subseteq B\\ \text{\textbf{while}} (i \leq m \text{ and answer } = A \subseteq B )\\ \ \ j := 1, \text{found} := \text{"no"}\\ \ \ \text{\textbf{while }} (j \neq n \text{ and } \text{found}= \text{"no"})\\ \ \ \ \ \text{\textbf{if }} a[i] = b[j] \text{\textbf{ then }} \text{found} := \text{"yes"}\\ \ \ \ \ j := j + 1\\ \ \ \text{\textbf{end while}}\\ \ \ \text{[If found has not been given the value "yes" when execution reaches this point, then } a[i] \neq B\text{ .]}\\ \ \ \text{\textbf{if }} \text{found} = \text{"no"} \text{\textbf{ then }} \text{answer} := A \nsubseteq B\\ \ \ i := i + 1\\ \text{\textbf{end while}}$
**Output:** _answer [a string]_ **Output:** _answer [a string]_
---
Page 414
**Theorem 6.2.1 Some Subset Relations**
1. _Inclusion of Intersection:_ For all sets $A$ and $B$,
$$ \text{(a) } A \cap B \subseteq A \quad \text{ and } \quad \text{ (b) } A \cap B \subseteq B $$
2. _Inclusion in Union:_ For all sets $A$ and $B$,
$$ \text{(a) } A \subseteq A \cup B \quad \text{ and } \quad \text{ (b) } B \subseteq A \cup B $$
3. _Transitive Property of Subsets:_ For all sets $A$, $B$, $C$,
$$ \text{if } A \subseteq B \text{ and } B \subseteq C \text{, then } A \subseteq C $$
---
Page 415
**Procedural Versions of Set Definitions**
Let $X$ and $Y$ be subsets of a universal set $U$ and suppose $x$ and $y$ are
elements of $U$.
1. $x \in X \cup Y \Leftrightarrow x \in X \text{ or } x \in Y$
2. $x \in X \cap Y \Leftrightarrow x \in X \text{ and } x \in Y$
3. $x \in X - Y \Leftrightarrow x \in X \text{ and } x \notin Y$
4. $x \in X^c \Leftrightarrow x \notin X$
5. $(x, y) \in X \times Y \Leftrightarrow x \in X \text{ and } y \in Y$
---
Page 417
**Theorem 6.2.2 Set Identities**
Let all sets referred to below be subsets of a universal set $U$.
1. _Commutative Laws:_ For all sets $A$ and $B$,
$$ \text{(a) } A \cup B = B \cup A \quad \text{ and } \quad \text{ (b) } A \cap B = B \cap A $$
2. _Associative Laws:_ For all sets $A$, $B$, and $C$,
$$ \text{(a) } (A \cup B) \cup C = A \cup (B \cup C) \quad \text{ and } \quad \text{ (b) } (A \cap B) \cap C = A \cap (B \cap C) $$
3. _Distributive Laws:_ For all sets $A$, $B$, and $C$,
$$ \text{(a) } A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \quad \text{ and } \quad \text{ (b) } A \cap (B \cup C) = (A \cap B) \cup (A \cap C) $$
4. _Identity Laws:_ For every set $A$,
$$ \text{(a) } A \cup \emptyset = A \quad \text{ and } \quad \text{ (b) } A \cap U = A $$
5. _Complement Laws:_ For every set $A$,
$$ \text{(a) } A \cup A^c = U \quad \text{ and } \quad A \cap A^c = \emptyset $$
6. _Double Complement Law:_ For every set $A$,
$$ (A^c)^c = A $$
7. _Idempotent Laws:_ For every set $A$,
$$ \text{(a) } A \cup A = A \quad \text{ and } \quad \text{ (b) } A \cap A = A $$
8. _Universal Bound Laws:_ For every set $A$,
$$ \text{(a) } A \cup U = U \quad \text{ and } \quad \text{ (b) } A \cap \emptyset = \emptyset $$
9. _De Morgan's Laws:_ For all sets $A$ and $B$,
$$ \text{(a) } (A \cup B)^c = A^c \cap B^c \quad \text{ and } \quad \text{ (b) } (A \cap B)^c = A^c \cup B^c $$
10. _Absorption Laws:_ For all sets $A$ and $B$,
$$ \text{(a) } A \cup (A \cap B) = A \quad \text{ and } \quad \text{ (b) } A \cap (A \cup B) = A $$
11. _Complements of $U$ and $\emptyset$:_
$$ \text{(a) } U^c = \emptyset \quad \text{ and } \quad \text{ (b) } \emptyset^c = U $$
12. _Set Difference Law:_ For all sets $A$ and $B$,
$$ A - B = A \cap B^c $$
---
Page 418
**Basic Method for Proving That Sets Are Equal**
Let sets $X$ and $Y$ be given. To prove that $X = Y$:
1. Prove that $X \subseteq Y$.
2. Prove that $Y \subseteq X$.
---
Page 420
**Theorem 6.2.2(3)(a) A Distributive Law for Sets**
(Too lengthy, see page 420)
---
Page 422
**Theorem 6.2.2(9)(a) A De Morgan's Law for Sets**
For all sets $A$ and $B$, $(A \cup B)^c = A^c \cap B^c$.
**Proof:** Suppose $A$ and $B$ are sets.
_Proof that $(A \cup B)^c \subseteq A^c \cap B^c$:_
_[We must show that
$\forall x, \text{ if } x \in (A \cup B)^c \text{ then } x \in A^c \cap B^c$.]_
Suppose $x \in (A \cup B)^c$. _[We must show that $x \in A^c \cap B^c$.]_ By
definition of complement,
$$ x \notin A \cup B $$
Now to say that $x \notin A \cup B$ means that
it is false that ($x$ is in $A$ or $x$ is in $B$).
By De Morgan's laws of logic, this implies that
$x$ is not in $A$ and $x$ is not in $B$,
which can be written
$$ x \notin A \quad \text{ and } \quad x \notin B $$
Hence $x \in A^c$ and $x \in B^c$ by definition of complement. It follows, by
definition of intersection, that $x \in A^c \cap B^c$ _[as was to be shown]._ So
$(A \cup B)^c \subseteq A^c \cap B^c$ by definition of subset.
_Proof that $A^c \cap B^c \subseteq (A \cup B)^c$:_
_[We must show that
$\forall x, \text{ if } x \in A^c \cap B^c \text{ then } x \in (A \cup B)^c$.]_
Suppose $x \in A^c \cap B^c$. _[We must show that $x \in (A \cup B)^c$.]_ By
definition of intersection, $x \in A^c$ and $x \in B^c$, and by definition of
complement,
$$ x \notin A \quad \text{ and } \quad x \notin B $$
In other words,
$x$ is not in $A$ and $x$ is not in $B$.
By De Morgan's laws of logic this implies that
it is false that ($x$ is in $A$ or $x$ is in $B$),
which can be written
$$ x \notin A \cup B $$
by definition of union. Hence, by definition of complement, $x \in (A \cup B)^c$
_[as was to be shown]._ It follows that $A^c \cap B^c \subseteq (A \cup B)^c$ by
definition of subset.
_Conclusion:_ Since both set containments have been proved,
$(A \cup B)^c = A^c \cap B^c$ by definition of set equality.
---
Page 423
**Theorem 6.2.3 Intersection and Union with a Subset**
For any sets $A$ and $B$, if $A \subseteq B$, then
$$ \text{(a) } A \cap B = A \quad \text{ and } \quad \text{ (b) } A \cup B = B $$
**Proof:**
_Part (a):_ Suppose $A$ and $B$ are sets with $A \subseteq B$. To show part (a)
we must show both that $A \cap B \subseteq A$ and that $A \subseteq A \cap B$.
We already know that $A \cap B \subseteq A$ by the inclusion of intersection
property. To show that $A \subseteq A \cap B$, let $x$ be any element in $A$.
_[We must show that $x$ is in $A \cap B$.]_ But, because of the hypothesis that
$A \subseteq B$, we can conclude that $x$ is also in $B$ by definition of
subset. Hence
$$ x \in A \quad \text{ and } x \in B $$
and thus
$$ x \in A \cap B $$
by definition of intersection _[as was to be shown]._
**Proof:**
_Part (b):_ The proof of part (b) is left as an exercise.
---
Page 424
**Theorem 6.2.4 A Set with No Elements Is a Subset of Every Set**
If $E$ is a set with no elements and $A$ is any set, then $E \subseteq A$.
**Proof (by contradiction):**
Suppose not. _[We take the negation of the theorem and suppose it to be true.]_
Suppose there exists a set $E$ with no elements and a set $A$ such that
$E \nsubseteq A$. _[We must deduce a contradiction.]_ Then there would be an
element of $E$ that is not an element of $A$ _[by definition of subset]_. But
there can be no such element since $E$ has no elements. This is a contradiction.
_[Hence the supposition that there are sets $E$ and $A$, where $E$ has no
elements and $E \nsubseteq A$, is false, and so the theorem is true.]_
---
Page 424
**Corollary 6.2.5 Uniqueness of the Empty Set**
There is only one set with no elements.
**Proof:** Suppose $E_1$ and $E_2$ are both sets with no elements. By Theorem
6.2.4, $E_1 \subseteq E_2$ since $E_1$ has no elements. Also $E_2 \subseteq E_1$
since $E_2$ has no elements. Thus $E_1 = E_2$ by definition of set equality.
---
Page 425
**Proposition 6.2.6**
For all sets $A$, $B$, and $C$, if $A \subseteq B$ and $B \subseteq C^c$, then
$A \cap C = \emptyset$.
**Proof:**
Suppose $A$, $B$, and $C$ are sets such that $A \subseteq B$ and
$B \subseteq C^c$. We must show that $A \cap C = \emptyset$. Suppose not. That
is, suppose there is an element $x$ in $A \cap C$. By definition of
intersection, $x \in A$ and $x \in C$. Then, since $A \subseteq B$, $x \in B$ by
definition of subset. Also, since $B \subseteq C^c$, then $x \in C^c$ by
definition of subset again. It follows by definition of complement that
$x \notin C$. Thus $x \in C$ and $x \notin C$, which is a contradiction. So the
supposition that there is an element $x$ in $A \cap C$ is false, and thus
$A \cap C = \emptyset$ _[as was to be shown]_.

25
chapter_6/test_yourself.md
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@ -52,3 +52,28 @@ all $A_i$ are a subset of $A$, but are also disjoint.
$A$ is the union of all the sets $A_1, A_2, A_3, \dots$ and $A$ is the union of all the sets $A_1, A_2, A_3, \dots$ and
$A_i \cap A_j = \emptyset$ whenever $i \neq j$. $A_i \cap A_j = \emptyset$ whenever $i \neq j$.
---
Page 426
**Test Yourself**
1. To prove that a set $X$ is a subset of a set $A \cap B$, you suppose that $x$
is any element of $X$ and you show that $x \in A$ _____ $x \in B$.
2. To prove that a set $X$ is a subset of a set $A \cup B$, you suppose that $x$
is any element of $X$ and you show that $x \in A$ _____ $x \in B$.
3. To prove that a set $A \cup B$ is a subset of a set $X$, you start with any
element $x$ in $A \cup B$ and consider the two cases _____ and _____. You
then show that in either case _____.
4. To prove that a set $A \cap B$ is a subset of $X$, you suppose that _____ and
you show that _____.
5. To prove that a set $X$ equals a set $Y$, you prove that _____ and that
_____.
6. To prove that a set $X$ does not equal a set $Y$, you need to find an element
that is in _____ and not _____ or that is in _____ and not _____.