Sets

(a) $\{\text{A},\text{L},\text{G},\text{E},\text{B},\text{R}\}$   (b) $\{-3,-2,-1,0,1\}$   (c) $\{1,3,5,7,9,11\}$

(a) The letters of ALGEBRA in order are A, L, G, E, B, R, A. The letter A occurs twice, but repeating an element does not change a set, so it is written once:

Six elements, from a seven-letter word.

(b) The description restricts $x$ to integers, so only whole numbers, positive, negative or zero, qualify. The condition uses one weak inequality and one strict inequality: $-3$ is included because of the $\le$, and $2$ is excluded because of the . Walking up from $-3$:

(c) The odd positive integers are $1,3,5,\dots$, and the condition "less than $12$" stops the list at $11$:

Note that , so $11$ is included; there is no odd integer between $11$ and $12$.

(a)   (b) $\{x|x\text{ is a positive even integer}\}$, or $\{2n|n\text{ is a positive integer}\}$   (c) $\{x|x^2=9\}$

(a) Two conditions must hold at once, and they can be combined into a double inequality. "Greater than $-2$" is strict, so use ; "at most $5$" allows $5$ itself, so use $\le$:

Since no universal set is named, $x$ ranges over the real numbers, as agreed in this section.

(b) The pattern is the positive even numbers. One way is to state the property directly,

and another is to build the elements from a parameter,

read as "the set of all numbers of the form $2n$ where $n$ is a positive integer." Both describe the same set, and either is acceptable.

(c) The property is "the square equals $9$":

In roster form this is $\{-3,3\}$. Writing $\{3\}$ would be wrong, since $(-3{)}^2=9$ as well.

(a) True   (b) False   (c) False   (d) True   (e) False   (f) True

The four elements of $A$ are

(a) True. The number $4$ is on the list.

(b) False. The number $1$ is not on the list. The set $\{1\}$ is, and a set is not the same thing as the object inside it.

(c) False. The claim $\{1\}\subset A$ means every element of $\{1\}$ lies in $A$. The only element of $\{1\}$ is $1$, and by (b) it does not. What is true here is $\{1\}\in A$.

(d) True. The elements of $\{\{1\}\}$ number exactly one, namely $\{1\}$, and $\{1\}$ is an element of $A$. So the inclusion holds. Comparing (c) and (d) is the whole point of this exercise: adding a pair of braces converts a true membership statement into a true subset statement.

(e) False. The elements of $\{2,3\}$ are $2$ and $3$, and neither is an element of $A$. Again the true statement is $\{2,3\}\in A$.

(f) True. The empty set is a subset of every set, vacuously: there is no element of $\varnothing$ that could fail to lie in $A$.

(a) $0$   (b) $1$   (c) $2$   (d) $1$.   $\varnothing \in \varnothing$ is false; $\varnothing \subset \varnothing$ is true.

Count what sits between the outermost braces, treating each inner set as a single object.

(a) $\varnothing =\{\}$ has nothing between its braces, so it has $0$ elements.

(b) $\{\varnothing \}$ has one thing between its braces, namely $\varnothing$. So it has $1$ element and is therefore not empty. This is the point made in the section with the boxes: a box containing an empty box is not itself empty.

(c) $\{\varnothing ,\{\varnothing \}\}$ lists two objects, $\varnothing$ and $\{\varnothing \}$, and they are different from each other by (a) and (b). So it has $2$ elements.

(d) $\{0\}$ has one element, the number $0$. Do not confuse the number zero with the empty set; $\{0\}$ is a set with one element, while $\varnothing$ has none.

Is $\varnothing \in \varnothing$? No. Membership in $\varnothing$ would require $\varnothing$ to have at least one element, and it has none.

Is $\varnothing \subset \varnothing$? Yes. Every set is a subset of itself, and in any case the empty set is a subset of every set.

(a) $\{1,2,3,4,5,6,8\}$   (b) $\{2,4\}$   (c) $\{1,3,5\}$   (d) $\{6,8\}$   (e) $\varnothing$   (f) $\{1,2,3,4,5,7\}$

(a) Union. Collect everything that appears in $A$ or in $B$, writing each element once:

The elements $2$ and $4$ belong to both sets but are listed a single time.

(b) Intersection. Keep only what appears in both lists. Scanning $A$: $2$ is in $B$, $4$ is in $B$, and $1,3,5$ are not. So

(c) $A-B$. Start with $A$ and delete anything that also lies in $B$, that is, delete $2$ and $4$:

(d) $B-A$. Start with $B$ this time and delete $2$ and $4$:

Comparing (c) and (d) shows clearly that set difference is not symmetric.

(e) $B\cap C$. Every element of $B$ is even and every element of $C$ is odd, so no number can be in both:

The sets $B$ and $C$ are disjoint.

(f) $(A\cap B)\cup C$. Work inside the parentheses first, using (b):

It is customary, though not required, to list the elements in increasing order.

Both sides of the first identity equal $\{3,4\}$; both sides of the second equal $\{1,2,3,4,5,6\}$.

First identity, left side. Inside the parentheses,

and intersecting with $A=\{1,2,3,4\}$ keeps only $3$ and $4$:

First identity, right side. Compute each intersection separately,

and take their union:

The two sides agree.

Second identity, left side. Inside the parentheses,

so

Second identity, right side. Compute each union,

and intersect them. Every element of the first list also appears in the second, so the intersection is the first list itself:

The two sides agree again. These two rules are the distributive laws for sets: intersection distributes over union, and union distributes over intersection. Checking one numerical case does not prove them in general, but it does make them easy to remember.

Both equal $\{1,2,5,6\}$. This common value is called the symmetric difference of $A$ and $B$.

First expression. Removing from $A$ what it shares with $B$, and then removing from $B$ what it shares with $A$,

so

Second expression. The union and the intersection are

and deleting the second from the first leaves

Observation. The two results are the same. Both describe the elements that belong to exactly one of the two sets, that is, to one of them but not to both. This set is called the symmetric difference of $A$ and $B$, often written $A\mathrm{△}B$. On a Venn diagram it is the whole of both circles with the overlapping lens removed.

$\varnothing ,\{a\},\{b\},\{c\},\{a,b\},\{a,c\},\{b,c\},\{a,b,c\}$; there are $8$ of them, and a set with $n$ elements has $2^n$ subsets.

Organise the list by the number of elements, which guarantees that none is missed and none is repeated.

• No elements: $\varnothing$. That is $1$ subset.
• One element: $\{a\}$, $\{b\}$, $\{c\}$. That is $3$ subsets.
• Two elements: $\{a,b\}$, $\{a,c\}$, $\{b,c\}$. That is $3$ subsets.
• Three elements: $\{a,b,c\}$, the whole set $S$ itself. That is $1$ subset.

Altogether $1+3+3+1=8$ subsets. Two of them are easy to forget: the empty set, which is a subset of every set, and $S$ itself, since $S\subset S$.

The general count. Building a subset amounts to making one independent decision per element of $S$: put it in, or leave it out. With $n$ elements there are $n$ such decisions, each with $2$ outcomes, so the number of subsets is

For $n=3$ this gives $2^3=8$, matching the list. The collection of all subsets of $S$ is called the power set of $S$.

Both follow directly from the definitions of intersection and union.

First claim: $A\cap B\subset A$. By the definition of subset, we must show that every element of $A\cap B$ is an element of $A$. So let $x$ be an arbitrary element of $A\cap B$. By the definition of intersection,

In particular $x\in A$, which is what had to be shown. Since $x$ was arbitrary, $A\cap B\subset A$.

Second claim: $A\subset A\cup B$. Let $x$ be an arbitrary element of $A$. By the definition of union,

and the first alternative already holds. Hence $x\in A\cup B$, and since $x$ was arbitrary, $A\subset A\cup B$. $◻$

Putting the two together gives the chain

and by the same argument with the roles of $A$ and $B$ exchanged, $A\cap B\subset B\subset A\cup B$ as well. Both proofs are short because each amounts to reading the definition and keeping only the part that is needed: "and" gives away either half, while "or" is satisfied by either half.

$26$ study at least one subject; $4$ study neither.

Let $U$ be the set of all $30$ students, $M$ the set of those studying mathematics and $P$ the set of those studying physics. The data are

where $|X|$ denotes the number of elements of $X$.

At least one subject. "At least one" is the union $M\cup P$. Adding $18$ and $15$ counts the $7$ students who take both subjects twice, once in each total, so subtract that overlap once:

To find we have to subtract $|M\cap P|$ from $|M|+|P|$, to avoid double counting those who study both. This is called the inclusion-exclusion principle for two sets.

Neither subject. These students form $U-(M\cup P)$, everything in the class outside the union:

A check with the four Venn regions. Mathematics only: $18-7=11$. Physics only: $15-7=8$. Both: $7$. Neither: $4$. The four counts add to $11+8+7+4=30$, the size of the class, so nothing has been double counted or lost.

(a) True   (b) False   (c) False   (d) True

(a) True. Suppose $A\subset B$. Every element of $A\cap B$ lies in $A$, so $A\cap B\subset A$. Conversely, if $x\in A$ then $x\in B$ as well, because $A\subset B$; so $x$ lies in both sets, that is, $x\in A\cap B$. Hence $A\subset A\cap B$. Both inclusions give $A\cap B=A$.

(b) False. Take $A=\{1,2\}$ and $B=\{2,3\}$. Then

and these are different sets. The two differences are equal only in the special case $A=B$, when both are $\varnothing$.

(c) False. Disjoint sets need not be empty. Take $A=\{1\}$ and $B=\{2\}$. Then $A\cap B=\varnothing$, yet neither set is empty. The statement confuses sets with numbers: for numbers, $ab=0$ does force $a=0$ or $b=0$, but there is no analogous rule for sets.

(d) True. By definition, $x\in A\cup \varnothing$ means $x\in A$ or $x\in \varnothing$. The second alternative is impossible, so the condition reduces to $x\in A$. Hence the two sets have exactly the same elements. The empty set plays the same role for union that $0$ plays for addition.