Lectures

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 xx is an element of a set AA, we write xAx\in A. If xx is not an element of AA, we write xAx\notin A. For example, if A={1,2,3}A=\{1,2,3\}, then 1A1\in A and 4A4\notin A.

There are multiple ways to describe a set. One is to list its elements, as in A={1,2,3}A=\{1,2,3\}. Another is to describe a property that characterizes its elements, as in B={x:x>0}B=\{x\in\mathbb R:x>0\}, which is the set of positive real numbers.

Some examples of sets include:

  • The empty set \emptyset.

  • The set of integers ={,2,1,0,1,2,}\mathbb Z=\{\ldots,-2,-1,0,1,2,\ldots\}.

  • The set of real numbers \mathbb R.

  • The set of natural numbers ={1,2,3,}\mathbb N=\{1,2,3,\ldots\}.

  • The set of rational numbers ={p/q:p,q\{0}}\mathbb Q=\{p/q:p\in\mathbb Z,q\in\mathbb Z\setminus\{0\}\}.

  • The set of all subsets of a finite set SS, called the power set and denoted 2S2^S.

Definition 1.1 (Inclusion and equality). For sets AA and BB, write ABA\subseteq B (AA is a subset of BB) if every element of AA belongs to BB. The sets are equal if ABA\subseteq B and BAB\subseteq A. Write ABA\subset B (AA is a proper subset of BB) if ABA\subseteq B and ABA\neq B.

For example, the statements {1,2}{1,2,3},{1,2,3}={3,2,1},,\{1,2\}\subseteq\{1,2,3\}, \qquad \{1,2,3\}=\{3,2,1\}, \qquad \mathbb Z\subset\mathbb R, \qquad \mathbb N\subset\mathbb Z are all true. Furthermore, for any set SS, we have S\emptyset\subseteq S and SSS\subseteq S.

Definition 1.2 (Set operations). For sets AA and BB, the union, intersection, and set difference are AB={x:xA or xB},AB={x:xA and xB},A\B={xA:xB}.\begin{aligned} A\cup B&=\{x:x\in A\text{ or }x\in B\},\\ A\cap B&=\{x:x\in A\text{ and }x\in B\},\\ A\setminus B&=\{x\in A:x\notin B\}. \end{aligned} The complement of a set AA relative to a universal set UU is Ac={x:xU and xA}=U\AA^c=\{x:x\in U\text{ and }x\notin A\}=U\setminus A.

These operations are called union, intersection, and set difference. They are typically illustrated by Venn diagrams:

Figure 1.1. Union, intersection, set difference, and complement.

Of course, we can also define the union and intersection of more than two sets. Given a collection of sets {Ai}iI\{A_i\}_{i\in I} indexed by a set II, we define iIAi={x:xAi for some iI},iIAi={x:xAi for every iI}.\bigcup_{i\in I}A_i=\{x:x\in A_i\text{ for some }i\in I\}, \qquad \bigcap_{i\in I}A_i=\{x:x\in A_i\text{ for every }i\in I\}.

The following theorem summarizes some properties of aforementioned set operations. Its proof is left as an exercise.

Theorem 1.3 (Set algebra). For sets AA, BB, and CC,

  1. AB=BAA\cup B=B\cup A, AB=BAA\cap B=B\cap A (commutativity),

  2. A(BC)=(AB)CA\cup(B\cup C)=(A\cup B)\cup C, A(BC)=(AB)CA\cap(B\cap C)=(A\cap B)\cap C (associativity),

  3. A(AB)=AA\cup(A\cap B)=A, A(AB)=AA\cap(A\cup B)=A (absorption),

  4. A\B=ABcA\setminus B=A\cap B^c (set difference),

  5. (AB)c=AcBc(A\cup B)^c=A^c\cap B^c, (AB)c=AcBc(A\cap B)^c=A^c\cup B^c (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 SS is a collection of nonempty, pairwise disjoint subsets of SS whose union is SS. For example, the sets of even and odd integers form a partition of \mathbb Z.

1.2 Cartesian Product and n\mathbb R^n

Definition 1.4. A Cartesian product of two sets AA and BB is the set A×B={(a,b):aA,bB}.A\times B=\{(a,b):a\in A, b\in B\}.

For example, if A={1,2}A=\{1,2\} and B={x,y}B=\{x,y\}, then A×B={(1,x),(1,y),(2,x),(2,y)}.A\times B=\{(1,x),(1,y),(2,x),(2,y)\}. The product A1××AnA_1\times\cdots\times A_n consists of tuples (a1,,an)(a_1,\ldots,a_n) satisfying aiAia_i\in A_i for every ii. An nn-fold product of a set AA with itself is denoted An=A××AA^n=A\times\cdots\times A.

The most commonly used in economics example of an nn-fold product is n\mathbb R^n. In this course, we treat elements of n\mathbb R^n as column vectors: x=(x1xn)n,x=\begin{pmatrix}x_1\\\vdots\\x_n\end{pmatrix}\in\mathbb R^n, but that convention is not important until we discuss linear algebra.

Sometimes we need to compare vectors in n\mathbb R^n. While not all vectors are comparable, we can compare them coordinatewise.

Definition 1.5 (Coordinatewise comparisons). For x,ynx,y\in\mathbb R^n, write xy if xiyi for every i,x>y if xy and xy,xy if xi>yi for every i.\begin{aligned} x\geq y &\text{ if } x_i\geq y_i \text{ for every } i,\\ x>y &\text{ if } x\geq y \text{ and } x\neq y,\\ x\gg y &\text{ if } x_i>y_i \text{ for every } i. \end{aligned} The nonnegative and positive orthants are +n={xn:x0},++n={xn:x0}.\mathbb R^n_+=\{x\in\mathbb R^n:x\geq0\}, \qquad \mathbb R^n_{++}=\{x\in\mathbb R^n:x\gg0\}.

2 Functions

2.1 Vocabulary and properties

Definition 2.1 (Function). Let XX and YY be two sets. A function f:XYf: X \to Y is a rule that assigns to each xXx\in X exactly one element f(x)Yf(x)\in Y. We also call ff a map, or a mapping, from XX to YY.

For a function f:XYf:X\to Y, we call XX the domain and YY the codomain. For a subset AXA \subseteq X of the domain, the image of AA is the set f(A)={f(x):xA}Yf(A)=\{f(x):x\in A\}\subseteq Y of values taken by ff on AA. In particular, we call f(X)f(X) the range of ff. The preimage of a subset BYB\subseteq Y of the codomain is the set f1(B)={xX:f(x)B}Xf^{-1}(B)=\{x\in X:f(x)\in B\}\subseteq X of points in the domain that map into BB. The graph of ff is the set of pairs {(x,f(x)):xX}X×Y.\{(x,f(x)):x\in X\}\subseteq X\times Y.

Definition 2.2 (Injective, surjective, bijective). A function f:XYf:X\to Y is

  1. injective, or one-to-one, if f(x)=f(x)f(x)=f(x') implies x=xx=x' for all x,xXx,x'\in X;

  2. surjective, or onto, if its range is all of YY, that is, f(X)=Yf(X)=Y;

  3. bijective if it is both injective and surjective.

Injectivity says that distinct points of the domain have distinct values, and surjectivity says that every yYy\in Y equals f(x)f(x) for at least one xXx\in X. The two conditions are checked in opposite directions: injectivity starts from two points of XX and compares their values, while surjectivity starts from a point of YY and asks for a point of XX that reaches it.

Note that whether a function is surjective depends on the declared domain and codomain. For example, the function f(x)=x2f(x) = x^2 is not surjective if f:f:\mathbb R\to\mathbb R but surjective if f:+f:\mathbb R\to \mathbb{R}_{+}. On the other hand, the function g(x)=x3g(x)=x^3 is surjective if g:g: \mathbb{R} \to \mathbb{R} or g:++g: \mathbb{R}_+ \to \mathbb{R}_+.

Definition 2.3 (Composition). Let f:XYf:X\to Y and g:YZg:Y\to Z, so that the codomain of ff is the domain of gg. The composition of ff and gg is the function gf:XZg\circ f:X\to Z defined by (gf)(x)=g(f(x))for every xX.(g\circ f)(x)=g(f(x))\quad\text{for every }x\in X.

2.2 Inverse Function

Let f:XYf:X\to Y be a function and let yYy\in Y. Recalling the definition of a preimage, let us consider the set f1({y})f^{-1}(\{y\}), which contains all elements of XX that map to the point yy. In general, this set may be empty, contain one point, or contain multiple points. By Definition 2.2, if ff is surjective, then f1({y})f^{-1}(\{y\}) contains at least one point for every yYy\in Y, and if ff is injective, then it contains at most one point. Consequently, if ff is bijective, then it contains exactly one point. This allows us to slightly abuse notation and define f1(y)f^{-1}(y) to be the unique point of XX that maps to yy, so that f1({y})={f1(y)}f^{-1}(\{y\})=\{f^{-1}(y)\}.

Definition 2.4 (Inverse). Let f:XYf:X\to Y be a bijective function. The inverse of ff is the function f1:YXf^{-1}:Y\to X that assigns to each yYy\in Y the unique xXx\in X with f(x)=yf(x)=y.

The function ff and its inverse f1f^{-1} undo each other, in the sense that f1(f(x))=x for every xX,f(f1(y))=y for every yY.f^{-1}(f(x))=x\ \text{ for every }x\in X, \qquad f(f^{-1}(y))=y\ \text{ for every }y\in Y.

Figure 2.1. Applying f and then f^{-1} returns every point of X to itself; applying f^{-1} and then f returns every point of Y to itself.

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 f:XYf:X\to Y, the following are equivalent.

  1. ff is bijective.

  2. There is a function g:YXg:Y\to X with (gf)(x)=x(g\circ f)(x)=x for every xXx\in X and (fg)(y)=y(f\circ g)(y)=y for every yYy\in Y.

Moreover, the function gg in (2) is unique and equals f1f^{-1}.

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 ff is bijective. Then f1:YXf^{-1}:Y\to X is defined by Definition 2.4, and we take g=f1g=f^{-1}. Let xXx\in X and put y=f(x)y=f(x). By Definition 2.4, f1(y)f^{-1}(y) is the unique point of XX that ff maps to yy; since f(x)=yf(x)=y, that point is xx. Using Definition 2.3 to expand the composition, (f1f)(x)=f1(f(x))=f1(y)=x.(f^{-1}\circ f)(x)=f^{-1}(f(x))=f^{-1}(y)=x. Now let yYy\in Y and put x=f1(y)x=f^{-1}(y), so that f(x)=yf(x)=y by Definition 2.4. Then (ff1)(y)=f(f1(y))=f(x)=y.(f\circ f^{-1})(y)=f(f^{-1}(y))=f(x)=y. Both equations hold for arbitrary xXx\in X and yYy\in Y, so g=f1g=f^{-1} is a function of the kind required in (2).

(2) implies (1). Assume such a gg exists. To show that ff is injective, let x,xXx,x'\in X satisfy f(x)=f(x)f(x)=f(x'). Applying gg to both sides and using Definition 2.3, x=(gf)(x)=g(f(x))=g(f(x))=(gf)(x)=x,x=(g\circ f)(x)=g(f(x))=g(f(x'))=(g\circ f)(x')=x', where the first and last equalities are the hypothesis on gg. So ff is injective in the sense of Definition 2.2. To show that ff is surjective, let yYy\in Y and put x=g(y)x=g(y), which lies in XX because g:YXg:Y\to X. Then f(x)=f(g(y))=(fg)(y)=y,f(x)=f(g(y))=(f\circ g)(y)=y, so yf(X)y\in f(X). Since yy was an arbitrary point of YY, this gives Yf(X)Y\subseteq f(X), and the reverse inclusion f(X)Yf(X)\subseteq Y holds because ff takes its values in YY. Hence f(X)=Yf(X)=Y, which is surjectivity in Definition 2.2. Being both injective and surjective, ff is bijective.

Uniqueness. Let gg be any function satisfying (2). By the previous paragraph ff is bijective, so f1f^{-1} is defined. Let yYy\in Y and put x=f1(y)x=f^{-1}(y), so that f(x)=yf(x)=y by Definition 2.4. Then g(y)=g(f(x))=(gf)(x)=x=f1(y).g(y)=g(f(x))=(g\circ f)(x)=x=f^{-1}(y). The functions gg and f1f^{-1} have the same domain YY and the same codomain XX, and we have just shown that they agree at every yYy\in Y. Hence g=f1g=f^{-1}. ◻

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, “2+2=42+2=4” is a true statement and “2+2=52+2=5” is a false one.

Given a statement PP, its negation ¬P\lnot P is true if and only if PP is false. Given two statements PP and QQ, their conjunction PQP\wedge Q is true if and only if both PP and QQ are true, while their disjunction PQP\vee Q is true if and only if at least one of PP or QQ is true. We typically illustrate these logical operations with truth tables: P¬PTFFTPQPQTTTTFFFTFFFFPQPQTTTTFTFTTFFF\begin{array}{c|c} P&\lnot P\\ \hline T&F\\ F&T \end{array} \qquad \begin{array}{cc|c} P&Q&P\wedge Q\\ \hline T&T&T\\ T&F&F\\ F&T&F\\ F&F&F \end{array} \qquad \begin{array}{cc|c} P&Q&P\vee Q\\ \hline T&T&T\\ T&F&T\\ F&T&T\\ F&F&F \end{array}

3.2 Implication and equivalence

Perhaps the most important logical operation is implication. We define it via a truth table as follows: PQPQTTTTFFFTTFFT\begin{array}{cc|c} P&Q&P\Rightarrow Q\\ \hline T&T&T\\ T&F&F\\ F&T&T\\ F&F&T \end{array}

The implication PQP\Rightarrow Q is read “PP implies QQ,” or “if PP, then QQ,” or “PP is sufficient for QQ,” or “QQ is necessary for PP.” Its converse is QPQ\Rightarrow P, and its contrapositive is ¬Q¬P\lnot Q\Rightarrow\lnot P. If PQP\Rightarrow Q and QPQ\Rightarrow P both hold, we write PQP\Leftrightarrow Q and say that PP and QQ are equivalent, or PP if and only if (IFF) QQ.

Our first proof involves showing that an implication and its contrapositive are equivalent.

Theorem 3.1 (Contrapositive law). (PQ)(¬Q¬P).(P\Rightarrow Q)\iff(\lnot Q\Rightarrow\lnot P).

Proof. The truth tables for PQP\Rightarrow Q and ¬Q¬P\lnot Q\Rightarrow\lnot P are PQPQTTTTFFFTTFFTPQ¬Q¬P¬Q¬PTTFFTTFTFFFTFTTFFTTT\begin{array}{cc|c} P&Q&P\Rightarrow Q\\ \hline T&T&T\\ T&F&F\\ F&T&T\\ F&F&T \end{array} \qquad \begin{array}{cc|cc|c} P&Q&\neg Q&\neg P&\neg Q\Rightarrow\neg P\\ \hline T&T&F&F&T\\ T&F&T&F&F\\ F&T&F&T&T\\ F&F&T&T&T \end{array}

Since the truth tables are the same, the statements are equivalent. ◻

Note that an implication PQP\Rightarrow Q and its converse QPQ\Rightarrow P 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 PP and QQ, ¬(PQ)¬P¬Q,¬(PQ)¬P¬Q,¬(PQ)P¬Q.\begin{aligned} \lnot(P\wedge Q)&\iff \lnot P\vee\lnot Q,\\ \lnot(P\vee Q)&\iff \lnot P\wedge\lnot Q,\\ \lnot(P\Rightarrow Q)&\iff P\wedge\lnot Q. \end{aligned}

3.3 Properties and quantifiers

Given a set XX, a property on XX is a sentence P(x)P(x) that becomes either true or false (hence a statement) once a particular xXx \in X is specified. For instance, if X=X = \mathbb{Z} and P(x)P(x) is “xx is even,” then P(2)P(2) is true and P(3)P(3) is false.

To turn a property into a statement, we use quantifiers. The universal quantifier \forall means “for every” and the existential quantifier \exists means “there exists.” Continuing the example above, the statement x,P(x)\forall x\in\mathbb Z,\ P(x) is false, since 33 is not even, while the statement x,P(x)\exists x\in\mathbb Z,\ P(x) is true, since 22 is even.

Theorem 3.3 (Negation rules for quantifiers). For a property P(x)P(x) on a set XX, ¬(xX,P(x))xX,¬P(x),¬(xX,P(x))xX,¬P(x).\begin{aligned} \lnot(\forall x\in X,\ P(x))&\iff \exists x\in X,\ \lnot P(x),\\ \lnot(\exists x\in X,\ P(x))&\iff \forall x\in X,\ \lnot P(x). \end{aligned}

For example, ABA\subseteq B means x(xAxB)\forall x\,(x\in A\Rightarrow x\in B). Its negation is x(xA and xB).\exists x\,(x\in A\text{ and }x\notin B).

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 PQP\Rightarrow Q, we assume PP and derive QQ. In a direct proof of a universal statement xX,P(x)\forall x\in X,\ P(x), we let xXx\in X be arbitrary, meaning that nothing is assumed about xx beyond xXx\in X, and derive P(x)P(x); since the argument used no property of xx, it applies to every element of XX.

Below is one example of a direct proof. It uses the following vocabulary: an integer nn is even if n=2an=2a for some aa\in\mathbb Z, and odd if n=2a+1n=2a+1 for some aa\in\mathbb Z. Every integer is even or odd, and no integer is both; we take that as given.

Lemma 4.1. If nn is an even integer, then 5n3-5n-3 is an odd integer.

Proof. Let nn be an arbitrary even integer. Then n=2an=2a for some aa\in\mathbb Z, so 5n3=5(2a)3=10a3=2(5a2)+1.-5n-3=-5(2a)-3=-10a-3=2(-5a-2)+1. Now set b=5a2b=-5a-2. Since aa is an integer, so is bb, and we obtain 5n3=2b+1-5n-3=2b+1. Hence 5n3-5n-3 is odd. ◻

A proof by contrapositive of PQP\Rightarrow Q is a direct proof of ¬Q¬P\lnot Q\Rightarrow\lnot P: we assume ¬Q\lnot Q and derive ¬P\lnot P. 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 5n3-5n-3 is not odd, hence even, and its conclusion is that nn is not even, hence odd.

Second proof of Lemma 4.1. We prove the contrapositive: if 5n3-5n-3 is an even integer, then nn is an odd integer. Let nn be an integer for which 5n3-5n-3 is even, so that 5n3=2b-5n-3=2b for some bb\in\mathbb Z. Adding 6n+36n+3 to both sides gives n=(5n3)+6n+3=2b+6n+3=2(b+3n+1)+1.n=(-5n-3)+6n+3=2b+6n+3=2(b+3n+1)+1. Now set c=b+3n+1c=b+3n+1. Since bb and nn are integers, so is cc, and the display reads n=2c+1n=2c+1. So nn 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 SS, we assume that SS is false, and from that assumption we derive some statement CC together with its negation ¬C\lnot C. Whatever CC may be, the statement C¬CC\wedge\lnot C is false, so the assumption cannot hold and SS 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 S(¬S(C¬C)),S\iff\left(\lnot S\Rightarrow(C\wedge\lnot C)\right), which holds for every statement CC: showing that ¬S\lnot S leads to a contradiction is the same as showing that SS 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 PQP\Rightarrow Q, its negation is P¬QP\wedge\lnot Q by Theorem 3.2, so the proof begins by assuming both PP and ¬Q\lnot Q. Which statement CC 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 xx\in\mathbb R. If xεx\leq\varepsilon for every real number ε>0\varepsilon>0, then x0x\leq0.

Proof. The hypothesis PP is that xεx\leq\varepsilon for every ε>0\varepsilon>0, and the conclusion QQ is that x0x\leq0. By Theorem 3.2, to assume the implication false is to assume PP and ¬Q\lnot Q. Therefore, we start by assuming that xε for every ε>0,andx>0.x\leq\varepsilon\ \text{ for every }\varepsilon>0, \qquad\text{and}\qquad x>0. By the assumption on the right, x/2>0x/2 > 0. The assumption on the left holds for every ε>0\varepsilon > 0, so it holds for ε=x/2\varepsilon = x/2, giving xx2.x\leq\frac{x}{2}. Subtracting x/2x/2 from both sides gives x/20x/2\leq0, and multiplying by 22 gives x0x\leq0. The statement CC is x0x\leq0: we have just derived it, and its negation x>0x>0 is one of our assumptions. So CC is both true and false, a contradiction. ◻

4.4 Proof by induction

Some statements come indexed by the natural numbers: for each nn\in\mathbb N there is a statement P(n)P(n), and the claim is that P(n)P(n) holds for every nn. The following theorem reduces that claim, which involves infinitely many statements, to two checks.

Theorem 4.3 (Principle of mathematical induction). For each nn\in\mathbb N, let P(n)P(n) be a statement. Suppose that

  1. P(1)P(1) is true, and

  2. for every nn\in\mathbb N, P(n)P(n+1)P(n)\Rightarrow P(n+1).

Then P(n)P(n) is true for every nn\in\mathbb N.

Condition (1) is the base step and condition (2) is the induction step; the assumption P(n)P(n) 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 nn\in\mathbb N be arbitrary, assume P(n)P(n), and derive P(n+1)P(n+1).

Lemma 4.4 (Bernoulli’s inequality). Let xx\in\mathbb R with x>1x>-1. Then for every nn\in\mathbb N, (1+x)n1+nx.(1+x)^n\geq1+nx.

Proof. Fix x>1x>-1 and let P(n)P(n) be the displayed inequality.

Base step. For n=1n=1 both sides equal 1+x1+x, so P(1)P(1) holds.

Induction step. Let nn\in\mathbb N be arbitrary and assume P(n)P(n), that is, (1+x)n1+nx(1+x)^n\geq1+nx (induction hypothesis). Multiply the induction hypothesis by 1+x1+x, which is positive because x>1x>-1, to obtain (1+x)n+1=(1+x)n(1+x)(1+nx)(1+x)=1+(n+1)x+nx2.(1+x)^{n+1}=(1+x)^n(1+x)\geq(1+nx)(1+x)=1+(n+1)x+nx^2. Since nx20nx^2\geq0, we have (1+x)n+11+(n+1)x(1+x)^{n+1}\geq1+(n+1)x, which is P(n+1)P(n+1). Thus P(n)P(n+1)P(n)\Rightarrow P(n+1) for every nn\in\mathbb N.

By Theorem 4.3, P(n)P(n) holds for every nn\in\mathbb N. ◻