1.3 Sets

The most fundamental objects we will use in our studies (and really in all of math) are sets. Much of what follows might be review, but it is very important that you are fluent in the language of set theory. Most of the notation we use below is standard, although some might be a little different than what you have seen before.

For us, a set will simply be an unordered collection of objects. Two examples: we could consider the set of all actors who have played The Doctor on Doctor Who, or the set of natural numbers between 1 and 10 inclusive. In the first case, Tom Baker is an element (or member) of the set, while Idris Elba, among many others, is not an element of the set. Also, the two examples are of different sets. Two sets are equal exactly if they contain the exact same elements. For example, the set containing all of the vowels in the declaration of independence is precisely the same set as the set of vowels in the word “questionably” (namely, all of them); we do not care about order or repetitions, just whether the element is in the set or not.

Notation

We need some notation to make talking about sets easier. Consider,

$A equals open brace 1, 2, 3 close brace$

This is read, “ $A$ is the set containing the elements 1, 2 and 3.” We use curly braces “ ${,}$ ” to enclose elements of a set. Some more notation:

$a ∈ open brace a, b, c close brace$

The symbol “ $∈$ ” is read “is in” or “is an element of.” Thus the above means that $a$ is an element of the set containing the letters $a$ , $b$ , and $c$ . Note that this is a true statement. It would also be true to say that $d$ is not in that set:

$d ∉ open brace a, b, c close brace$

Be warned: we write “ $x ∈ A$ ” when we wish to express that one of the elements of the set $A$ is $x$ . For example, consider the set,

$A equals open brace 1, b, open brace x, y, z close brace, ∅ close brace$

This is a strange set, to be sure. It contains four elements: the number 1, the letter b, the set ${x, y, z}$ , and the empty set $∅ equals open brace close brace$ , the set containing no elements. Is $x$ in $A$ ? The answer is no. None of the four elements in $A$ are the letter $x$ , so we must conclude that $x ∉ A$ . Similarly, consider the set $B equals open brace 1, b close brace$ . Even though the elements of $B$ are elements of $A$ , we cannot say that the set $B$ is one of the elements of $A$ . Therefore $B ∉ A$ . (Soon we will see that $B$ is a subset of $A$ , but this is different from being an element of $A$ .)

We have described the sets above by listing their elements. Sometimes this is hard to do, especially when there are a lot of elements in the set (perhaps infinitely many). For instance, if we want $A$ to be the set of all even natural numbers, would could write,

$A equals open brace 0, 2, 4, 6, and so on close brace$

, but this is a little imprecise. A better way would be

$A = {x \in N : \exists n \in N (x = 2 n)}$

Let's look at this carefully. First, there are some new symbols to digest: “ $N$ ” is the symbol usually used to denote that natural numbers, which we will take to be the set $open brace 0, 1, 2, 3, and so on close brace$ . Next, the colon, “:”, is read such that; it separates the elements that are in the set from the condition that the elements in the set must satisfy. So putting this all together, we would read the set as, “the set of all $x$ in the natural numbers, such that there exists some $n$ in the natural numbers for which $x$ is twice $n$ .” In other words, the set of all natural numbers, that are even. Here is another way to write the same set.

$A equals open brace x ∈ N: x is even close brace$

Note: Sometimes mathematicians use $|$ or $∍$ for the “such that” symbol instead of the colon. Also, there is a fairly even split between mathematicians about whether $0$ is an element of the natural numbers, so be careful there.

This notation is usually called set builder notation. It tells us how to build a set by telling us precisely the condition elements must meet to gain access (the condition is the logical statement after the “ $:$ ” symbol). Reading and comprehending sets written in this way takes practice. Here are some more examples:

Example 1

Describe each of the following sets both in words and by listing out enough elements to see the pattern.

$open brace x: x plus 3 ∈ N close brace$ .
$open brace x ∈ N: x plus 3 ∈ N close brace$ .
$open brace x: x ∈ N ∨ minus x ∈ N close brace$ .
$open brace x: x ∈ N ∧ minus x ∈ N close brace$ .

This is the set of all numbers which are 3 less than a natural number (i.e., that if you add 3 to them, you get a natural number). The set could also be written as ${- 3, - 2, - 1, 0, 1, 2, \dots}$ (note that 0 is a natural number, so $minus 3$ is in this set because $- 3 + 3 = 0$ ).
This is the set of all natural numbers which are 3 less than a natural number. So here we just have $open brace 0, 1, 2, 3 and so on close brace$ .
This is the set of all integers (positive and negative whole numbers, written $Z$ ). In other words, ${\dots, - 2, - 1, 0, 1, 2, \dots}$ .
Here we want all numbers $x$ such that $x$ and $- x$ are natural numbers. There is only one: 0. So we have the set $open brace 0 close brace$ .

There is also a subtle variation on set builder notation. While the condition is generally given after the “such that”, sometimes it is hidden in the first part. Here is an example.

Example 2

List a few elements in the sets below and describe them in words. The set $Z$ is the set of integers; positive and negative whole numbers.

$A equals open brace x ∈ Z: x squared ∈ N close brace$
$B equals open brace x squared: x ∈ N close brace$

The set of integers that pass the condition that their square is a natural number. Well, every integer, when you square it, gives you a non-negative integer, so a natural number. Thus $A = Z = {\dots, - 2, - 1, 0, 1, 2, 3, \dots}$ .
Here we are looking for the set of all $x^{2}$ s where $x$ is a natural number. So this set is simply the set of perfect squares. $B equals open brace 0, 1, 4, 9, 16, and so on close brace$ .
Another way we could have written this set, using more strict set builder notation, would be as $B = {x \in N : x = n^{2} for some n \in N}$ .

We already have a lot of notation, and there is more yet. Below is a handy chart of symbols. Some of these will be discussed in greater detail as we move forward.

Summary

Special sets

$∅$
The empty set is the set which contains no elements.
$U$
The universe set is the set of all elements.
$N$
The set of natural numbers. That is, $N equals open brace 0, 1, 2, 3 and so on close brace$ .
$Z$
The set of integers. That is, $Z = {\dots, - 2, - 1, 0, 1, 2, 3, \dots}$ .
$Q$
The set of rational numbers.
$R$
The set of real numbers.
$P (A)$
The power set of any set $A$ is the set of all subsets of $A$ .

Summary

Set Theory Notation

${,}$
We use these braces to enclose the elements of a set. So $open brace 1, 2, 3 close brace$ is the set containing 1, 2, and 3.
$:$
$open brace x: x greater than 2 close brace$ is the set of all $x$ such that $x$ is greater than 2.
$∈$
$2 ∈ open brace 1, 2, 3 close brace$ asserts that 2 is an element of the set $open brace 1, 2, 3 close brace$ .
$∉$
$4 ∉ open brace 1, 2, 3 close brace$ because 4 is not an element of the set $open brace 1, 2, 3 close brace$ .
$⊆$
$A ⊆ B$ asserts that $A$ is a subset of $B$ : every element of $A$ is also an element of $B$ .
$⊂$
$A ⊂ B$ asserts that $A$ is a proper subset of $B$ : every element of $A$ is also an element of $B$ , but $A \neq B$ .
$\cap$
$A \cap B$ is the intersection of $A$ and $B$ : the set containing all elements which are elements of both $A$ and $B$ .
$\cup$
$A \cup B$ is the union of $A$ and $B$ : is the set containing all elements which are elements of $A$ or $B$ or both.
$\times$
$A \times B$ is the Cartesian product of $A$ and $B$ : the set of all ordered pairs $(a, b)$ with $a ∈ A$ and $b ∈ B$ .
$∖$
$A ∖ B$ is set difference between $A$ and $B$ : the set containing all elements of $A$ which are not elements of $B$ .
$\bar{A}$
The complement of $A$ is the set of everything which is not an element of $A$ .
$| A |$
The cardinality (or size) of $A$ is the number of elements in $A$ .

Investigate!

Find the cardinality of each set below.
1. $A equals open brace 3, 4, and so on, 15 close brace$ .
2. $B equals open brace n ∈ N: 2 less than n less than or equal to 200 close brace$ .
3. $C = {n \leq 100 : n \in N \land \exists m \in N (n = 2 m + 1)}$ .
Find two sets $A$ and $B$ for which $vertical bar A vertical bar equals 5$ , $vertical bar B vertical bar equals 6$ , and $vertical bar A union B vertical bar equals 9$ . What is $| A \cap B |$ ?
Find sets $A$ and $B$ with $| A | = | B |$ such that $vertical bar A union B vertical bar equals 7$ and $vertical bar A intersection B vertical bar equals 3$ . What is $| A |$ ?
Let $A equals open brace 1, 2, and so on, 10 close brace$ . Define $B_{2} = {B \subseteq A : | B | = 2}$ . Find $vertical bar B sub 2 vertical bar$ .
For any sets $A$ and $B$ , define $A B equals open brace a b: a ∈ A ∧ b ∈ B close brace$ . If $A equals open brace 1, 2 close brace$ and $B equals open brace 2, 3, 4 close brace$ , what is $| A B |$ ? What is $| A \times B |$ ?

Relationships Between Sets

We have already said what it means for two sets to be equal: they have exactly the same elements. Thus, for example,

$open brace 1, 2, 3 close brace equals open brace 2, 1, 3 close brace$

(Remember, the order the elements are written down in does not matter.) Also,

${1, 2, 3} = {1, 1 + 1, 1 + 1 + 1} = {I, I I, I I I} = {1, 2, 3, 1 + 2}$

since these are all ways to write the set containing the first three positive integers (how we write them doesn't matter, just what they are).

What about the sets $A equals open brace 1, 2, 3 close brace$ and $B equals open brace 1, 2, 3, 4 close brace$ ? Clearly $A \neq B$ , but notice that every element of $A$ is also an element of $B$ . Because of this we say that $A$ is a subset of $B$ , or in symbols $A ⊂ B$ or $A ⊆ B$ . Both symbols are read “is a subset of.” The difference is that sometimes we want to say that $A$ is either equal to or is a subset of $B$ , in which case we use $⊆$ . This is analogous to the difference between $<$ and $\leq$ .

Example 3

Let $A equals open brace 1, 2, 3, 4, 5, 6 close brace$ , $B equals open brace 2, 4, 6 close brace$ , $C equals open brace 1, 2, 3 close brace$ and $D equals open brace 7, 8, 9 close brace$ . Determine which of the following are true, false, or meaningless.

$A ⊂ B$ .
$B ⊂ A$ .
$B ∈ C$ .
$∅ ∈ A$ .
$∅ ⊂ A$ .
$A < D$ .
$3 ∈ C$ .
$3 ⊂ C$ .
$open brace 3 close brace ⊂ C$ .

False. For example, $1 ∈ A$ but $1 ∉ B$ .
True. Every element in $B$ is an element in $A$ .
False. The elements in $C$ are 1, 2, and 3. The set $B$ is not equal to 1, 2, or 3.
False. $A$ has exactly 6 elements, and none of them are the empty set.
True. Everything in the empty set (nothing) is also an element of $A$ . Notice that the empty set is a subset of every set.
Meaningless. A set cannot be less than another set.
True. $3$ is one of the elements of the set $C$ .
Meaningless. $3$ is not a set, so it cannot be a subset of another set.
True. $3$ is the only element of the set $open brace 3 close brace$ , and is an element of $C$ , so every element in $open brace 3 close brace$ is an element of $C$ .

In the example above, $B$ is a subset of $A$ . You might wonder what other sets are subsets of $A$ . If you collect all these subsets of $A$ into a new set, we get a set of sets. We call the set of all subsets of $A$ the power set of $A$ , and write it $P (A)$ .

Example 4

Let $A equals open brace 1, 2, 3 close brace$ . Find $P (A)$ .

$P (A)$ is a set of sets, all of which are subsets of $A$ . So

$P open parenthesis A close parenthesis equals open brace ∅, open brace 1 close brace, open brace 2 close brace, open brace 3 close brace, open brace 1, 2 close brace, open brace 1, 3 close brace, open brace 2, 3 close brace, open brace 1, 2, 3 close brace close brace$

Notice that while $2 ∈ A$ , it is wrong to write $2 ∈ P open parenthesis A close parenthesis$ since none of the elements in $P (A)$ are numbers! On the other hand, we do have $open brace 2 close brace ∈ P open parenthesis A close parenthesis$ because $open brace 2 close brace ⊆ A$ .

What does a subset of $P (A)$ look like? Notice that $open brace 2 close brace ⊈ P open parenthesis A close parenthesis$ because not everything in $open brace 2 close brace$ is in $P (A)$ . But we do have $open brace open brace 2 close brace close brace ⊆ P open parenthesis A close parenthesis$ . The only element of $open brace open brace 2 close brace close brace$ is the set $open brace 2 close brace$ which is also an element of $P (A)$ . We could take the collection of all subsets of $P (A)$ and call that $P (P (A))$ . Or even the power set of that set of sets of sets.

Another way to compare sets is by their size. Notice that in the example above, $A$ has 6 elements and $B$ , $C$ , and $D$ all have 3 elements. The size of a set is called the set's cardinality. We would write $vertical bar A vertical bar equals 6$ , $vertical bar B vertical bar equals 3$ , and so on. For sets that have a finite number of elements, the cardinality of the set is simply the number of elements in the set. Note that the cardinality of $open brace 1, 2, 3, 2, 1 close brace$ is 3. We do not count repeats (in fact, $open brace 1, 2, 3, 2, 1 close brace$ is exactly the same set as $open brace 1, 2, 3 close brace$ ). There are sets with infinite cardinality, such as $N$ , the set of rational numbers (written $Q$ ), the set of even natural numbers, and the set of real numbers ( $R$ ). It is possible to distinguish between different infinite cardinalities, but that is beyond the scope of this text. For us, a set will either be infinite, or finite; if it is finite, the we can determine its cardinality by counting elements.

Example 5

Find the cardinality of $A equals open brace 23, 24, and so on, 37, 38 close brace$ .
Find the cardinality of $B equals open brace 1, open brace 2, 3, 4 close brace, ∅ close brace$ .
If $C equals open brace 1, 2, 3 close brace$ , what is the cardinality of $P (C)$ ?

Since $38 - 23 = 15$ , we can conclude that the cardinality of the set is $vertical bar A vertical bar equals 16$ (you need to add one since 23 is included).
Here $vertical bar B vertical bar equals 3$ . The three elements are the number 1, the set $open brace 2, 3, 4 close brace$ , and the empty set.
We wrote out the elements of the power set $P (C)$ above, and there are 8 elements (each of which is a set). So $vertical bar P open parenthesis C close parenthesis vertical bar equals 8$ . (You might wonder if there is a relationship between $| A |$ and $| P (A) |$ for all sets $A$ . This is a good question which we will return to in Counting.)

Operations On Sets

Is it possible to add two sets? Not really, however there is something similar. If we want to combine two sets to get the collection of objects that are in either set, then we can take the union of the two sets. Symbolically,

$C = A \cup B$

, read, “ $C$ is the union of $A$ and $B$ ,” means that the elements of $C$ are exactly the elements which are either an element of $A$ or an element of $B$ (or an element of both). For example, if $A equals open brace 1, 2, 3 close brace$ and $B equals open brace 2, 3, 4 close brace$ , then $A union B equals open brace 1, 2, 3, 4 close brace$ .

The other common operation on sets is intersection. We write,

$C = A \cap B$

and say, “ $C$ is the intersection of $A$ and $B$ ,” when the elements in $C$ are precisely those both in $A$ and in $B$ . So if $A equals open brace 1, 2, 3 close brace$ and $B equals open brace 2, 3, 4 close brace$ , then $A intersection B equals open brace 2, 3 close brace$ .

Often when dealing with sets, we will have some understanding as to what “everything” is. Perhaps we are only concerned with natural numbers. In this case we would say that our universe is $N$ . Sometimes we denote this universe by $U$ . Given this context, we might wish to speak of all the elements which are not in a particular set. We say $B$ is the complement of $A$ , and write,

$B = \bar{A}$

when $B$ contains every element not contained in $A$ . So, if our universe is $open brace 1, 2, and so on, 9, 10 close brace$ , and $A equals open brace 2, 3, 5, 7 close brace$ , then $A bar equals open brace 1, 4, 6, 8, 9, 10 close brace$ .

Of course we can perform more than one operation at a time. For example, consider

$A \cap \bar{B}$

This is the set of all elements which are both elements of $A$ and not elements of $B$ . What have we done? We've started with $A$ and removed all of the elements which were in $B$ . Another way to write this is the set difference:

$A intersection B bar equals A ∖ B$

It is important to remember that these operations (union, intersection, complement, and difference) on sets produce other sets. Don't confuse these with the symbols from the previous section (element of and subset of). $A \cap B$ is a set, while $A ⊆ B$ is true or false. This is the same difference as between $3 plus 2$ (which is a number) and $3 less than or equal to 2$ (which is false).

Example 6

Let $A equals open brace 1, 2, 3, 4, 5, 6 close brace$ , $B equals open brace 2, 4, 6 close brace$ , $C equals open brace 1, 2, 3 close brace$ and $D equals open brace 7, 8, 9 close brace$ . If the universe is $U equals open brace 1, 2, and so on, 10 close brace$ , find:

$A \cup B$ .
$A \cap B$ .
$B \cap C$ .
$A \cap D$ .
$\bar{B \cup C}$ .
$A ∖ B$ .
$(D \cap \bar{C}) \cup \bar{A \cap B}$ .
$∅ union C$ .
$∅ intersection C$ .

$A \cup B = {1, 2, 3, 4, 5, 6} = A$ since everything in $B$ is already in $A$ .
$A \cap B = {2, 4, 6} = B$ since everything in $B$ is in $A$ .
$B intersection C equals open brace 2 close brace$ as the only element of both $B$ and $C$ is 2.
$A intersection D equals ∅$ since $A$ and $D$ have no common elements.
$B union C bar equals open brace 5, 7, 8, 9, 10 close brace$ . First we find that $B union C equals open brace 1, 2, 3, 4, 6 close brace$ , then we take everything not in that set.
$A ∖ B equals open brace 1, 3, 5 close brace$ since the elements 1, 3, and 5 are in $A$ but not in $B$ . This is the same as $A \cap \bar{B}$ .
$open parenthesis D intersection C bar close parenthesis union A intersection B bar equals open brace 1, 3, 5, 7, 8, 9, 10 close brace$ . The set contains all elements that are either in $D$ but not in $C$ (i.e., $open brace 7, 8, 9 close brace$ ), or not in both $A$ and $B$ (i.e., $open brace 1, 3, 5, 7, 8, 9, 10 close brace$ ).
$∅ union C equals C$ since nothing is added by the empty set.
$∅ intersection C equals ∅$ since nothing can be both in a set and in the empty set.

Having notation like this is useful. We will often want to add or remove elements from sets, and our notation allows us to do so precisely.

You might notice that the symbols for union and intersection slightly resemble the logic symbols for “or” and “and.” This is no accident. What does it mean for $x$ to be an element of $A \cup B$ ? It means that $x$ is an element of $A$ or $x$ is an element of $B$ (or both). That is,

$x ∈ A union B ⇔ x ∈ A ∨ x ∈ B$

Similarly,

$x ∈ A intersection B ⇔ x ∈ A ∧ x ∈ B$

Also,

$x ∈ A bar ⇔ ¬ open parenthesis x ∈ A close parenthesis$

. which says $x$ is an element of the complement of $A$ if $x$ is not an element of $A$ .

There is one more way to combine sets which will be useful for us: the Cartesian product, $A \times B$ . This sounds fancy but is nothing you haven't seen before. When you graph a function in calculus, you graph it in the Cartesian plane. This is the set of all ordered pairs of real numbers $(x, y)$ . We can do this for any pair of sets, not just the real numbers with themselves.

Put another way, $A \times B = {(a, b) : a \in A \land b \in B}$ . The first coordinate comes from the first set and the second coordinate comes from the second set. Sometimes we will want to take the Cartesian product of a set with itself, and this is fine: $A \times A = {(a, b) : a, b \in A}$ (we might also write $A^{2}$ for this set). Notice that in $A \times A$ , we still want all ordered pairs, not just the ones where the first and second coordinate are the same. We can also take products of 3 or more sets, getting ordered triples, or quadruples, and so on.

Example 8

Let $A equals open brace 1, 2 close brace$ and $B equals open brace 3, 4, 5 close brace$ . Find $A \times B$ and $A \times A$ . How many elements do you expect to be in $B \times B$ ?

$A \times B = {(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5)}$ .

$A \times A = A^{2} = {(1, 1), (1, 2), (2, 1), (2, 2)}$ .

$| B \times B | = 9$ . There will be 3 pairs with first coordinate $3$ , three more with first coordinate $4$ , and a final three with first coordinate $5$ .

Venn Diagrams

There is a very nice visual tool we can use to represent operations on sets. A Venn diagram displays sets as intersecting circles. We can shade the region we are talking about when we carry out an operation. We can also represent cardinality of a particular set by putting the number in the corresponding region.

Two overlapping circles enclosed in a rectangular box. Circles labeled A and B.

Three overlapping circles enclosed in a rectangle. The circle labeled A is in the top left, the circle labeled B is in the top right, the circle labeled C is in the bottom center. The circles intersect to create seven interior regions: one inside all circles, three inside just a pair of circles, and three inside only one circle.

Each circle represents a set. The rectangle containing the circles represents the universe. To represent combinations of these sets, we shade the corresponding region. For example, we could draw $A \cap B$ as:

Two overlapping circles enclosed in a rectangular box. Circles labeled A and B. The overlapping region inside both circles is shaded light gray.

Here is a representation of $A \cap \bar{B}$ , or equivalently $A ∖ B$ :

Two overlapping circles enclosed in a rectangular box. Circles labeled A and B. The region inside circle A but outside circle B is shaded light gray.

A more complicated example is $(B \cap C) \cup (C \cap \bar{A})$ , as seen below.

Notice that the shaded regions above could also be arrived at in another way. We could have started with all of $C$ , then excluded the region where $C$ and $A$ overlap outside of $B$ . That region is $(A \cap C) \cap \bar{B}$ . So the above Venn diagram also represents $C \cap \bar{((A \cap C) \cap \bar{B})}$ . So using just the picture, we have determined that

$(B \cap C) \cup (C \cap \bar{A}) = C \cap \bar{((A \cap C) \cap \bar{B})}$

Let $A equals open brace x ∈ N: 3 less than or equal to x less than or equal to 13 close brace$ , $B equals open brace x ∈ N: x is even close brace$ , and $C equals open brace x ∈ N: x is odd close brace$ .

Find $A \cap B$ .
Find $A \cup B$ .
Find $B \cap C$ .
Find $B \cup C$ .

Find an example of sets $A$ and $B$ such that $A intersection B equals open brace 3, 5 close brace$ and $A union B equals open brace 2, 3, 5, 7, 8 close brace$ .

For example, $A equals open brace 2, 3, 5, 7, 8 close brace$ and $B equals open brace 3, 5 close brace$ .

Find an example of sets $A$ and $B$ such that $A ⊆ B$ and $A ∈ B$ .

For example, $A equals open brace 1, 2, 3 close brace$ and $B equals open brace 1, 2, 3, 4, 5, open brace 1, 2, 3 close brace close brace$

Recall $Z = {\dots, - 2, - 1, 0, 1, 2, \dots}$ (the integers). Let $Z^{+} = {1, 2, 3, \dots}$ be the positive integers. Let $2 Z$ be the even integers, $3 Z$ be the multiples of 3, and so on.

Is $Z^{+} \subseteq 2 Z$ ? Explain.
Is $2 Z \subseteq Z^{+}$ ? Explain.
Find $2 Z intersection 3 Z$ . Describe the set in words, and using set notation.
Express ${x \in Z : \exists y \in Z (x = 2 y \lor x = 3 y)}$ as a union or intersection of two sets already described in this problem.

No.
No.
$2 Z intersection 3 Z$ is the set of all integers which are multiples of both 2 and 3 (so multiples of 6). Therefore $2 Z \cap 3 Z = {x \in Z : \exists y \in Z (x = 6 y)}$ .
$2 Z union 3 Z$ .

Let $A sub 2$ be the set of all multiples of 2 except for $2$ . Let $A sub 3$ be the set of all multiples of 3 except for 3. And so on, so that $A_{n}$ is the set of all multiples of $n$ except for $n$ , for any $n greater than or equal to 2$ . Describe (in words) the set $A sub 2 union A sub 3 union A sub 4 union ⋯ bar$ .

It might help to think about what the union $A sub 2 union A sub 3$ is first. Then think about what numbers are not in that union. What will happen when you also include $A sub 5$ ?

Draw a Venn diagram to represent each of the following:

$A \cup \bar{B}$
$\bar{(A \cup B)}$
$A \cap (B \cup C)$
$(A \cap B) \cup C$
$\bar{A} \cap B \cap \bar{C}$
$open parenthesis A union B close parenthesis ∖ C$

$A \cup \bar{B}$ :
$\bar{(A \cup B)}$ :
$A \cap (B \cup C)$ :
$(A \cap B) \cup C$ :
$\bar{A} \cap B \cap \bar{C}$ :
$open parenthesis A union B close parenthesis ∖ C$ :

Describe a set in terms of $A$ and $B$ (using set notation) which has the following Venn diagram:

Two overlapping circles enclosed in a rectangular box. Circles labeled A and B. The part of circle A outside circle B and the part of circle B outside circle A are both shaded light gray; the lens where the two circles overlap is left white, as is the part of the box outside both circles.

Let $A = {a, b, c, d}$ . Find $P (A)$ .

We are looking for a set containing 16 sets.

$P open parenthesis A close parenthesis equals open brace, ∅, open brace a close brace, open brace b close brace, open brace c close brace, open brace d close brace, open brace a, b close brace, open brace a, c close brace, open brace a, d close brace, open brace b, c close brace, open brace b, d close brace,; open brace c, d close brace open brace a, b, c close brace, open brace a, b, d close brace, open brace a, c, d close brace, open brace b, c, d close brace, open brace a, b, c, d close brace close brace$

Let $A equals open brace 1, 2, 3, 4, 5, 6 close brace$ . Find all sets $B ∈ P open parenthesis A close parenthesis$ which have the property $open brace 2, 3, 5 close brace ⊆ B$ .

Find an example of sets $A$ and $B$ such that $vertical bar A vertical bar equals 4$ , $vertical bar B vertical bar equals 5$ , and $vertical bar A union B vertical bar equals 9$ .

For example, $A equals open brace 1, 2, 3, 4 close brace$ and $B equals open brace 5, 6, 7, 8, 9 close brace$ gives $A union B equals open brace 1, 2, 3, 4, 5, 6, 7, 8, 9 close brace$ .

Find an example of sets $A$ and $B$ such that $vertical bar A vertical bar equals 3$ , $vertical bar B vertical bar equals 4$ , and $vertical bar A union B vertical bar equals 5$ .

Are there sets $A$ and $B$ such that $| A | = | B |$ , $vertical bar A union B vertical bar equals 10$ , and $vertical bar A intersection B vertical bar equals 5$ ? Explain.

Let $A equals open brace 2, 4, 6, 8 close brace$ . Suppose $B$ is a set with $vertical bar B vertical bar equals 5$ .

What are the smallest and largest possible values of $| A \cup B |$ ? Explain.
_____ $\leq | A \cup B | \leq$ _____
What are the smallest and largest possible values of $| A \cap B |$ ? Explain.
_____ $\leq | A \cap B | \leq$ _____
What are the smallest and largest possible values of $| A \times B |$ ? Explain.
_____ $\leq | A \times B | \leq$ _____

Let $X equals open brace n ∈ N: 10 less than or equal to n less than 20 close brace$ . Find examples of sets with the properties below and very briefly explain why your examples work.

A set $A ⊆ N$ with $vertical bar A vertical bar equals 10$ such that $X ∖ A equals open brace 10, 12, 14 close brace$ .
A set $B ∈ P open parenthesis X close parenthesis$ with $vertical bar B vertical bar equals 5$ .
A set $C ⊆ P open parenthesis X close parenthesis$ with $vertical bar C vertical bar equals 5$ .
A set $D ⊆ X times X$ with $vertical bar D vertical bar equals 5$
A set $E ⊆ X$ such that $vertical bar E vertical bar ∈ E$ .

Let $A$ , $B$ and $C$ be sets.

Suppose that $A ⊆ B$ and $B ⊆ C$ . Does this mean that $A ⊆ C$ ? Prove your answer. Hint: to prove that $A ⊆ C$ you must prove the implication, “for all $x$ , if $x ∈ A$ then $x ∈ C$ .”
Suppose that $A ∈ B$ and $B ∈ C$ . Does this mean that $A ∈ C$ ? Give an example to prove that this does NOT always happen (and explain why your example works). You should be able to give an example where $| A | = | B | = | C | = 2$ .

In a regular deck of playing cards there are 26 red cards and 12 face cards. Explain, using sets and what you have learned about cardinalities, why there are only 32 cards which are either red or a face card.

Find an example of a set $A$ with $vertical bar A vertical bar equals 3$ which contains only other sets and has the following property: for all sets $B ∈ A$ , we also have $B ⊆ A$ . Explain why your example works. (FYI: sets that have this property are called transitive.)

Consider the sets $A$ and $B$ , where $A equals open brace 3, vertical bar B vertical bar close brace$ and $B equals open brace 1, vertical bar A vertical bar, vertical bar B vertical bar close brace$ . What are the sets?

We need to be a little careful here. If $B$ contains 3 elements, then $A$ contains just the number 3 (listed twice). So that would make $vertical bar A vertical bar equals 1$ , which would make $B equals open brace 1, 3 close brace$ , which only has 2 elements. Thus $vertical bar B vertical bar not equal to 3$ . This means that $vertical bar A vertical bar equals 2$ , so $B$ contains at least the elements 1 and 2. Since $vertical bar B vertical bar not equal to 3$ , we must have $vertical bar B vertical bar equals 2$ , which agrees with the definition of $B$ .

Therefore it must be that $A equals open brace 2, 3 close brace$ and $B equals open brace 1, 2 close brace$

Explain why there is no set $A$ which satisfies $A equals open brace 2, vertical bar A vertical bar close brace$ .

It looks like you should be able to define the set $A$ like this. But consider the two possible values for $| A |$ .

Find all sets $A$ , $B$ , and $C$ which satisfy the following.

$\begin{aligned} A = & {1, | B |, | C |} \\ B = & {2, | A |, | C |} \\ C = & {1, 2, | A |, | B |} \end{aligned}$

Discrete Mathematics: An Open Introduction, 3rd edition, by Oscar Levin (discrete.openmathbooks.org), licensed under CC BY-SA 4.0; this adaptation is distributed under the same license. License: CC-BY-SA-4.0.