Problems 1-23
1 $-3 < 2$
2 $4 > -6$
3 $\frac{\pi}{3} > 1$
4 $\sqrt{2} < \frac{1}{4}$
5 $\sqrt{8} < 3.1$
6 $-10 < -3\pi$
7
…
8 If $A$ is the set of positive integers, $B$ negative integers and $C$ multiples of $3$, then (a) $A \cup B$ is the set of all integers excluding $0$. (b) $A \cup B \cup C$ is the set of all integers including $0$, because $3 \cdot 0$ is an element of $C$, (c) $B \cap C$ is the set of all negative multiples of $3$, i.e. $\{ 3n: n$ is an integer $< 0 \}$, (d) $A \cap B$ is an empty set, because the intersection of these sets does not contain any elements, and (e) $\overline{C}$ $=$ $\mathbb{Z} \setminus \overline{C}$, that is the set of all integers not multiple of $3$. For example, sample around the integer $0$ is $\{$ $\cdots$, $-7$, $-5$,$-4$, $-2$, $-1$, $1$, $2$, $4$, $5$, $7$, $\cdots$ $\}$.
9 If $A$ is the set of multiples if $2$, $B$ the set of multiple of $3$ and the set $C$ the set of multiples of $5$, then (a) $A$ $\cup$ $C$ is the set of multiples of $2$, multiples of $5$ or both, such as $2$ and $10$, (b) $B$ $\cap$ $C$ is the set of numbers that are multiples of both $3$ and $5$ at the same time, like $15$ or $30$, etc., (c) $A$ $\cup$ $B$ $\cup$ $C$ is the set of numbers that are multiples of $2$, $3$, $5$ or a multiple of any possible multiplication combination of these numbers, such as $2 \cdot 3$ $=$ $6$, or $6 \cdot 5$ $=$ $30$.
10 If $A$ $=$ $(-\infty, 5)$ and $B$ $=$ $[-1,10]$, then $A \cup B$ is the set $(\infty, 10]$, and $A \cap B$ are all of the real values in the range $[-1,5)$.
11 The relation $\overline{A \cap B}$ means that arbitrary $x$ is not in the intersection of these two sets, or in other words, $x$ is not in $A$ or $x$ is not in $B$, so $x$ must be in the complement of $A$ or $B$, that is $x$ is in $\overline{A}$ $\cup$ $\overline{B}$. Because of this $\overline{A \cap B}$ $=$ $\overline{A}$ $\cup$ $\overline{B}$.
12 If arbitrary $x$ is not in $A$ or not in $B$ then it is in the complement of $A$ and $B$, but on the other hand, if arbitrary $x$ is not complement of $A$ and not in the complement of $B$, then $x$ must be in the complement of $A$ and $B$, which is the same set, so $\overline{A \cup B}$ $=$ $\overline{A}$ $\cap$ $\overline{B}$.
13 (a) The range $(1,1000)$ is bounded below by $1$ and above by $1000$, so the range is bounded. (b) The range $[-10^{37},19^{37}]$ is also bounded above and below, by $-10^{37}$ and $10^{37}$, respectively, so it’s a bounded range. (c) The range $[4, \infty]$ is bounded below by $4$, but not bounded above because positive $-\infty$ does not have a boundary, so this set is not bounded. (d) Same goes for the set $(-\infty,6)$ — The set is not bounded below because of the negative infinity, but is bounded above by $6$. (e) The set of positive odd integers is bounded below by $1$, but it’s not bounded above. (f) The set of negative rationals larger than $-20$ is bounded bounded below by $-20$ and above by a negative rational less than $0$, so this set is bounded.
14 If $A$ $=$ $[1,3]$ and $B$ $=$ $[2,4]$, then (a) $A$ $-$ $B$ $=$ $[1,2)$ and (b) $B$ $-$ $A$ $=$ $(3,4]$.
15 For the sets defined in 8: (a) $A - B$ $=$ $A$, (b) $B - A$ $=$ $B$, (c) $A - C$ $=$ $\{$ The set of positive integers not multiple of $3$ $\}$ and (d) $C - B$ $=$ $\{$ The set of positive multiples of $3$ $\}$.
16 For the sets defined in 9: (a) $A - C$ $=$ $\{$ The set of number multiple of $2$, but not multiple of $5$ $\}$, (b) $B - A$ $=$ $\{$ The set of number multiple of $3$, but not multiple of $2$ and (c) $B - C$ $=$ $\{$ The set of number multiple of $3$, but not multiple of $3$ $\}$.
17 For the intervals of 10: (a) $A - B$ $=$ $(-\infty, -1)$ and (b) $B - A$ $=$ $(5,10]$.
18 When $A$ $=$ $(-\infty,-5)$ $\cup$ $(5,\infty)$ and $B$ $=$ $[-10,3)$, (a) $A$ $-$ $(A - B)$ $=$ $A$ $-$ $((\infty,-10)$ $\cup$ $(5,\infty))$ $=$ $(\infty,-10)$ $\cup$ $(5,\infty)$. On the other hand, (b) $A$ $-$ $(B - A)$ $=$ $A$ $-$ $[-5,3))$ $=$ $A$.
19 If $A$ $\Delta$ $B$ $=$ $(A - B)$ $\cup$ $(B - A)$ and $A$ $=$ $[1,3)$ and $B$ $=$ $(2,5]$, then $A$ $\Delta$ $B$ $=$ $([1,2]) \cup ([3,5])$. This is shown in the Figure 1.
20 When $A$ $=$ $[0,5]$, $B$ $=$ $(3,6]$, and $C$ $=$ $(2,8)$, (a) $(A \ \Delta \ B) \ \Delta C$ $=$ $[0,2]$ $\cup$ $(3,5]$ $\cup$ $(6,8)$. This is illustrated in Figure 2:
(b) The operation $A \ \Delta \ (B \ \Delta \ C)$ leads to same solution as $(A \ \Delta \ B) \ \Delta \ C$, that is $A \ \Delta \ (B \ \Delta \ C)$ $=$ $[0,2]$ $\cup$ $(3,5]$ $\cup$ $(6,8)$ $=$ $(A \ \Delta \ B) \ \Delta \ C$.
21 (a) When $x$ is in $(A \cup B)$, by definition of the set union, $x$ is in $A$ or $x$ is in $B$. If $A$ and $B$ are not disjoint, $x$ can also be in $(A \cap B)$. If the latter set is removed from the former, $x$ can only be either in $A$ or $B$, but never in the intersection of $A$ and $B$, because it’s removed, so therefore $x$ is in $(A \cup B)$ $-$ $(A \cap B)$.
On the other hand, if $B$ is removed from $A$ and vice versa, $x$ is then either in the first reduced set or it is in the second reduced set, but $x$ can not be in their intersection. When considering the union of these sets, the $x$ is in $(A - B)$ $\cup$ $(B - A)$.
In any case, these sets equal each other, so by definition of symmetric difference, $A \ \Delta \ B$ $=$ $(A - B)$ $\cup$ $(B - A)$, which is also equal to $(A \cup B)$ $-$ $(A \cap B)$.
(b) See (a) and (b) of 20.
22 When $A$ and $B$ are completely overlapping each other, $A \cup B$ $=$ $A \cap B$, and when the sets are disjoint $A \cup B$ $=$ $\varnothing$ $=$ $A \cap B$.
23 (a) Relation $\varnothing$ $=$ $0$, if false because $\varnothing$ is an empty set, that is, $\varnothing$ $=$ $\{\}$, but on the other hand $0$, is something, that is, it’s a set denoting nothing, but it is not an empty set, so $\varnothing$ $=$ $\{\}$ $\neq$ $0$. (b) The relation $\varnothing$ $=$ $\{\varnothing\}$, is also false. The left hand-side of the relation is an empty set, while the right hand-side is a set containing empty set. (c) The relation $\varnothing$ $\in$ $\{\varnothing\}$ is true, because empty set, that is, the set $\varnothing$ indeed is a set of $\{\varnothing\}$. (d) The relation $\varnothing$ $\subseteq$ $\{\varnothing\}$ is also true, because $\varnothing$ is a subset of the set $\{\varnothing\}$. (e) The relation $\varnothing$ $\in$ $\varnothing$ is false, because empty set does not contain anything, or in other words, empty set is a set holding no elements, so $\varnothing$ can not be in $\varnothing$, that is, empty set can not be an element of a (another) empty set. (f) The relation $\varnothing$ $\subseteq$ $\varnothing$ is true, because empty set is an element of any set, by definition.