Math Camp
Lecture 1: Sets, Functions, Logic, and Proofs
1 Sets
1.1 Definition and Operations
A set is a collection of objects which we call elements. We typically denote sets by capital letters and elements by lowercase letters. If is an element of a set , we write . If is not an element of , we write . For example, if , then and .
There are multiple ways to describe a set. One is to list its elements, as in . Another is to describe a property that characterizes its elements, as in , which is the set of positive real numbers.
Some examples of sets include:
The empty set .
The set of integers .
The set of real numbers .
The set of natural numbers .
The set of rational numbers .
The set of all subsets of a finite set , called the power set and denoted .
Definition 1.1 (Inclusion and equality). For sets and , write ( is a subset of ) if every element of belongs to . The sets are equal if and . Write ( is a proper subset of ) if and .
For example, the statements are all true. Furthermore, for any set , we have and .
Definition 1.2 (Set operations). For sets and , the union, intersection, and set difference are The complement of a set relative to a universal set is .
These operations are called union, intersection, and set difference. They are typically illustrated by Venn diagrams:
Of course, we can also define the union and intersection of more than two sets. Given a collection of sets indexed by a set , we define
The following theorem summarizes some properties of aforementioned set operations. Its proof is left as an exercise.
Theorem 1.3 (Set algebra). For sets , , and ,
, (commutativity),
, (associativity),
, (absorption),
(set difference),
, (De Morgan’s laws).
We say that sets are disjoint if their intersection is empty. For example, the sets of even and odd integers are disjoint. A partition of a set is a collection of nonempty, pairwise disjoint subsets of whose union is . For example, the sets of even and odd integers form a partition of .
1.2 Cartesian Product and
Definition 1.4. A Cartesian product of two sets and is the set
For example, if and , then The product consists of tuples satisfying for every . An -fold product of a set with itself is denoted .
The most commonly used in economics example of an -fold product is . In this course, we treat elements of as column vectors: but that convention is not important until we discuss linear algebra.
Sometimes we need to compare vectors in . While not all vectors are comparable, we can compare them coordinatewise.
Definition 1.5 (Coordinatewise comparisons). For , write The nonnegative and positive orthants are
2 Functions
2.1 Vocabulary and properties
Definition 2.1 (Function). Let and be two sets. A function is a rule that assigns to each exactly one element . We also call a map, or a mapping, from to .
For a function , we call the domain and the codomain. For a subset of the domain, the image of is the set of values taken by on . In particular, we call the range of . The preimage of a subset of the codomain is the set of points in the domain that map into . The graph of is the set of pairs
Definition 2.2 (Injective, surjective, bijective). A function is
injective, or one-to-one, if implies for all ;
surjective, or onto, if its range is all of , that is, ;
bijective if it is both injective and surjective.
Injectivity says that distinct points of the domain have distinct values, and surjectivity says that every equals for at least one . The two conditions are checked in opposite directions: injectivity starts from two points of and compares their values, while surjectivity starts from a point of and asks for a point of that reaches it.
Note that whether a function is surjective depends on the declared domain and codomain. For example, the function is not surjective if but surjective if . On the other hand, the function is surjective if or .
Definition 2.3 (Composition). Let and , so that the codomain of is the domain of . The composition of and is the function defined by
2.2 Inverse Function
Let be a function and let . Recalling the definition of a preimage, let us consider the set , which contains all elements of that map to the point . In general, this set may be empty, contain one point, or contain multiple points. By Definition 2.2, if is surjective, then contains at least one point for every , and if is injective, then it contains at most one point. Consequently, if is bijective, then it contains exactly one point. This allows us to slightly abuse notation and define to be the unique point of that maps to , so that .
Definition 2.4 (Inverse). Let be a bijective function. The inverse of is the function that assigns to each the unique with .
The function and its inverse undo each other, in the sense that
The theorem below shows that if we tried to define an “inverse” for a function that is not bijective, we would fail.
Theorem 2.5 (Existence of an inverse). For a function , the following are equivalent.
is bijective.
There is a function with for every and for every .
Moreover, the function in (2) is unique and equals .
Proof (optional). An equivalence asserts two implications, so we prove that (1) implies (2) and that (2) implies (1). The uniqueness claim is proved last.
(1) implies (2). Assume is bijective. Then is defined by Definition 2.4, and we take . Let and put . By Definition 2.4, is the unique point of that maps to ; since , that point is . Using Definition 2.3 to expand the composition, Now let and put , so that by Definition 2.4. Then Both equations hold for arbitrary and , so is a function of the kind required in (2).
(2) implies (1). Assume such a exists. To show that is injective, let satisfy . Applying to both sides and using Definition 2.3, where the first and last equalities are the hypothesis on . So is injective in the sense of Definition 2.2. To show that is surjective, let and put , which lies in because . Then so . Since was an arbitrary point of , this gives , and the reverse inclusion holds because takes its values in . Hence , which is surjectivity in Definition 2.2. Being both injective and surjective, is bijective.
Uniqueness. Let be any function satisfying (2). By the previous paragraph is bijective, so is defined. Let and put , so that by Definition 2.4. Then The functions and have the same domain and the same codomain , and we have just shown that they agree at every . Hence . ◻
3 Logic
3.1 Logical Statements and Operations
A statement (also referred to as a proposition or an assertion) is a sentence that is either true or false. For example, “” is a true statement and “” is a false one.
Given a statement , its negation is true if and only if is false. Given two statements and , their conjunction is true if and only if both and are true, while their disjunction is true if and only if at least one of or is true. We typically illustrate these logical operations with truth tables:
3.2 Implication and equivalence
Perhaps the most important logical operation is implication. We define it via a truth table as follows:
The implication is read “ implies ,” or “if , then ,” or “ is sufficient for ,” or “ is necessary for .” Its converse is , and its contrapositive is . If and both hold, we write and say that and are equivalent, or if and only if (IFF) .
Our first proof involves showing that an implication and its contrapositive are equivalent.
Theorem 3.1 (Contrapositive law).
Proof. The truth tables for and are
Since the truth tables are the same, the statements are equivalent.∎ ◻
Note that an implication and its converse are not equivalent. Showing that is left as an exercise, along with proving the following negation rules for statements.
Theorem 3.2 (Negation rules for statements). For statements and ,
3.3 Properties and quantifiers
Given a set , a property on is a sentence that becomes either true or false (hence a statement) once a particular is specified. For instance, if and is “ is even,” then is true and is false.
To turn a property into a statement, we use quantifiers. The universal quantifier means “for every” and the existential quantifier means “there exists.” Continuing the example above, the statement is false, since is not even, while the statement is true, since is even.
Theorem 3.3 (Negation rules for quantifiers). For a property on a set ,
For example, means . Its negation is
4 Proofs
4.1 What a proof is
A theorem is a statement that has been shown to be true, and a proof is the argument that shows it. Hammack’s Book of Proof describes a proof as follows:
“A proof of a theorem is a written verification that shows that the theorem is definitely and unequivocally true. A proof should be understandable and convincing to anyone who has the requisite background and knowledge.”
A proof is typically written in ordinary prose plus mathematical notation. However, every step in a proof must be justified by something already accepted: a hypothesis of the theorem, a definition, a result established earlier, or a rule of logic. Proofs are often accompanied by pictures, but a picture usually does not replace a formal argument.
4.2 Direct proof
In a direct proof of an implication , we assume and derive . In a direct proof of a universal statement , we let be arbitrary, meaning that nothing is assumed about beyond , and derive ; since the argument used no property of , it applies to every element of .
Below is one example of a direct proof. It uses the following vocabulary: an integer is even if for some , and odd if for some . Every integer is even or odd, and no integer is both; we take that as given.
Lemma 4.1. If is an even integer, then is an odd integer.
Proof. Let be an arbitrary even integer. Then for some , so Now set . Since is an integer, so is , and we obtain . Hence is odd. ◻
A proof by contrapositive of is a direct proof of : we assume and derive . By Theorem 3.1 the two implications are equivalent, so proving either one proves the other. It is worth switching when the negations are the easier statements to work with.
Here is a second proof of Lemma 4.1, this time by contrapositive. Its hypothesis is that is not odd, hence even, and its conclusion is that is not even, hence odd.
Second proof of Lemma 4.1. We prove the contrapositive: if is an even integer, then is an odd integer. Let be an integer for which is even, so that for some . Adding to both sides gives Now set . Since and are integers, so is , and the display reads . So is odd, and therefore not even. By Theorem 3.1, this proves Lemma 4.1. ◻
4.3 Proof by contradiction
A proof by contrapositive requires the statement to be an implication. A proof by contradiction applies to a statement of any shape. To prove a statement , we assume that is false, and from that assumption we derive some statement together with its negation . Whatever may be, the statement is false, so the assumption cannot hold and is true. The method is also called reductio ad absurdum: supposing the opposite of what we want reduces us to an absurdity.
The proof by contradiction is justified by the equivalence which holds for every statement : showing that leads to a contradiction is the same as showing that is true. Verifying the equivalence with a truth table, as in the proof of Theorem 3.1, is left as an exercise.
When the statement to be proved is an implication , its negation is by Theorem 3.2, so the proof begins by assuming both and . Which statement will do the job is usually not known at the outset: we assume the negation, derive consequences, and stop once some statement and its negation are both on the page. Often the statement contradicted is one assumed at the start, a hypothesis or the negated conclusion, and sometimes it is a fact known independently.
Lemma 4.2. Let . If for every real number , then .
Proof. The hypothesis is that for every , and the conclusion is that . By Theorem 3.2, to assume the implication false is to assume and . Therefore, we start by assuming that By the assumption on the right, . The assumption on the left holds for every , so it holds for , giving Subtracting from both sides gives , and multiplying by gives . The statement is : we have just derived it, and its negation is one of our assumptions. So is both true and false, a contradiction. ◻
4.4 Proof by induction
Some statements come indexed by the natural numbers: for each there is a statement , and the claim is that holds for every . The following theorem reduces that claim, which involves infinitely many statements, to two checks.
Theorem 4.3 (Principle of mathematical induction). For each , let be a statement. Suppose that
is true, and
for every , .
Then is true for every .
Condition (1) is the base step and condition (2) is the induction step; the assumption made while proving the induction step is the induction hypothesis. The two conditions act like a line of dominoes: the first one falls by (1), and each one that falls knocks over the next by (2), so all of them fall.
The induction step is itself a universal implication, so it is proved directly: let be arbitrary, assume , and derive .
Lemma 4.4 (Bernoulli’s inequality). Let with . Then for every ,
Proof. Fix and let be the displayed inequality.
Base step. For both sides equal , so holds.
Induction step. Let be arbitrary and assume , that is, (induction hypothesis). Multiply the induction hypothesis by , which is positive because , to obtain Since , we have , which is . Thus for every .
By Theorem 4.3, holds for every . ◻