0.1 Course foundations and notation

Motivation

Chapter 0 establishes the language used by every later calculation. A formula is not a complete mathematical object until its allowed inputs are known; a claimed range is not proved until both containments are checked; and a change in quantifier order can turn a true assertion into a false one. These are not minor matters of presentation. They determine what a statement means.

The same discipline governs finite sums and products. An index is a bound variable, so changing its name changes nothing, while shifting it requires the bounds and the indexed term to move together. The aim of this note is therefore to make notation auditable: each symbol has a declared role, each function has a domain and codomain, and each worked computation preserves those roles.

Definitions

Sets and their descriptions

Definition

Sets, membership, subsets, and equality

A set is a collection of distinct objects called its elements or members. We write $x\in A$ when $x$ belongs to $A$, and $x\notin A$ otherwise.

The statement $A\subseteq B$ means that every element of $A$ is also an element of $B$. Two sets are equal precisely when they have the same elements.

A finite or patterned set may be given by listing its elements, such as $\{1,2,3\}$ or $\{0,1,2,3,\ldots\}$. Set-builder notation instead gives an ambient set and a selecting property:

$$\{x\in S:P(x)\}.$$

The notation says: take exactly those $x$ from $S$ for which $P(x)$ is true. The ambient set matters. For example, the solutions of $x^2-1=0$ are $1$ and $-1$ in $\mathbb R$, but $\{x\in\mathbb N:x^2-1=0\}=\{1\}$.

Definition

Standard number sets and intervals

We use

$$\begin{aligned} \mathbb N&:=\{0,1,2,3,\ldots\},\\ \mathbb Z&:=\{\ldots,-2,-1,0,1,2,\ldots\},\\ \mathbb Z^+&:=\{1,2,3,\ldots\},\\ \mathbb Q&:=\left\{\frac pq:p,q\in\mathbb Z,\ q\ne0\right\}, \end{aligned}$$

and $\mathbb R$ for the real numbers. Thus $\mathbb Z^+\subseteq\mathbb N\subseteq\mathbb Z\subseteq\mathbb Q\subseteq\mathbb R$. The symbol $:=$ means “is defined to be.”

For real $a\lt b$, the intervals $[a,b]$, $(a,b)$, and $(a,b]$ include both, neither, and only the right endpoint, respectively. Also $[a,\infty)=\{x\in\mathbb R:a\le x\}$. The empty set is $\varnothing$.

Definition

Set operations

For sets $A$ and $B$:

• $A\cap B$ contains the elements common to both sets;
• $A\cup B$ contains the elements belonging to at least one of them;
• $A\setminus B$ contains the elements of $A$ that do not belong to $B$.

These descriptions determine exactly which regions a Venn diagram should shade. Intersection keeps the overlap, union keeps every point in either circle, and $A\setminus B$ keeps only the part of $A$ outside $B$. Difference is therefore directional: $A\setminus B$ need not equal $B\setminus A$. Intervals are simply sets of real numbers, so the same operations apply to them. For instance, the real solutions of $x^2>1$ form the union of two open intervals and also the complement of the closed interval $[-1,1]$.

Logic and quantifiers

Definition

Quantified statements

The symbol $\forall$ means “for every,” $\exists$ means “there exists at least one,” and $\exists!$ means “there exists exactly one.” The implication $P\Rightarrow Q$ says that $P$ implies $Q$; $P\Leftrightarrow Q$ says that each statement implies the other.

Quantifiers are read from left to right. In $\forall x\in\mathbb Z,\exists y\in\mathbb Z, x+y=0$, the choice of $y$ may depend on $x$. In $\exists y\in\mathbb Z,\forall x\in\mathbb Z, x+y=0$, one fixed $y$ would have to work for every $x$.

Unique existence contains two claims. To prove $\exists!x\in S$ with a stated property, one must first produce an element of $S$ having the property and then show that no second element can have it. Likewise, an implication gives only one direction. The symbol $\Leftrightarrow$ is appropriate only after both the forward and reverse implications have been established.

Functions and inverses

Definition

Function, domain, codomain, and range

A function $f:A\to B$ assigns to each $a\in A$ exactly one element $f(a)$ of $B$. The set $A$ is the domain and $B$ is the codomain. The range is the set of outputs that actually occur:

$$f(A):=\{f(x)\in B:x\in A\}.$$

The formula alone does not determine a function. Domain and codomain are part of its declaration. Changing the domain can change the range. With the rule and domain fixed, changing to another codomain that contains every output leaves the range unchanged, but can change surjectivity and whether an inverse exists.

Definition

Injective, surjective, and bijective

For a function $f:A\to B$:

• $f$ is injective if $f(x_1)=f(x_2)$ implies $x_1=x_2$ for all $x_1,x_2\in A$;
• $f$ is surjective if every $y\in B$ equals $f(x)$ for at least one $x\in A$;
• $f$ is bijective if it is both injective and surjective.

Definition

Inverse function

Let $f:A\to B$. A function $g:B\to A$ is an inverse of $f$ when

$$g(f(x))=x\quad(x\in A),\qquad f(g(y))=y\quad(y\in B).$$

When it exists, the inverse is unique and is denoted by $f^{-1}$.

Finite sums and products

Definition

Summation and product symbols

Let $m,n\in\mathbb Z$ with $m\le n$, and let $a_m,a_{m+1},\ldots,a_n$ be numbers. Then

$$\sum_{i=m}^{n}a_i=a_m+a_{m+1}+\cdots+a_n, \qquad \prod_{i=m}^{n}a_i=a_m a_{m+1}\cdots a_n.$$

The letter $i$ is a dummy index: it identifies corresponding terms but has no meaning outside the indexed expression.

Consequently, $\sum_{i=m}^{n}a_i$ and $\sum_{k=m}^{n}a_k$ denote the same number. This harmless renaming differs from an index shift, which changes the labels attached to the terms and must be compensated in the bounds. The operators also encode different arithmetic: $\sum$ adds its terms, whereas $\prod$ multiplies its factors. For example, $\prod_{i=1}^{n}i=n!$, while $\sum_{i=1}^{n}i$ is an arithmetic sum. Keeping the operator visible prevents an algebraically plausible but logically wrong continuation.

Theorem / Proposition

Theorem

Double-inclusion criterion for set equality

For any sets $A$ and $B$, $A=B$ if and only if $A\subseteq B$ and $B\subseteq A$.

Theorem

Range of the square function

If $f:\mathbb R\to\mathbb R$ is defined by $f(x)=x^2$, then $f(\mathbb R)=[0,\infty)$.

Theorem

Inverse criterion

A function has an inverse if and only if it is bijective. When an inverse exists, it is unique.

Theorem

Basic laws for finite sums

For a constant $C$ and compatible finite sequences,

$$\sum_{i=m}^{n}Ca_i=C\sum_{i=m}^{n}a_i, \qquad \sum_{i=m}^{n}(a_i\pm b_i) =\sum_{i=m}^{n}a_i\pm\sum_{i=m}^{n}b_i.$$

For every integer shift $s$,

$$\sum_{i=m}^{n}a_i=\sum_{i=m+s}^{n+s}a_{i-s}.$$

Finite double sums may be expanded in either order:

$$\left(\sum_{i=m}^{n}a_i\right) \left(\sum_{j=p}^{q}b_j\right) =\sum_{i=m}^{n}\sum_{j=p}^{q}a_i b_j =\sum_{j=p}^{q}\sum_{i=m}^{n}a_i b_j.$$

Proof Sketch or Proof Idea

For the double-inclusion criterion, suppose first that $A=B$. Every element of either set is then an element of the other, so both inclusions hold. Conversely, if $A\subseteq B$ and $B\subseteq A$, an object belongs to $A$ exactly when it belongs to $B$; hence the sets have the same elements.

For the square-function range, two containments are essential. If $y\in f(\mathbb R)$, then $y=x^2\ge0$ for some real $x$, so $f(\mathbb R)\subseteq[0,\infty)$. If $y\in[0,\infty)$, choose $x=\sqrt y\in\mathbb R$. Then $f(x)=y$, proving the reverse containment.

For the inverse criterion, an inverse sends equal outputs back to equal inputs, so $f$ must be injective; and the equation $f(g(y))=y$ reaches every codomain element, so $f$ is surjective. Conversely, if $f$ is bijective, each $y\in B$ has exactly one preimage. Define $g(y)$ to be that preimage. Both composition identities follow. Any other inverse must choose the same unique preimage, which also proves uniqueness.

The finite-sum laws follow by expanding the indicated terms and using ordinary distributivity. In an index shift, the first old term $a_m$ reappears when the new index is $m+s$, and the last old term $a_n$ reappears at $n+s$. Thus the bounds and the subscript must be shifted together.

Worked Examples

Worked example

1. Describe and combine sets

The positive even integers are

$$E=\{2m:m\in\mathbb Z^+\} =\{n\in\mathbb Z^+:n=2m\text{ for some }m\in\mathbb Z^+\}.$$

If $A=\{1,2\}$, $B=\{2,3\}$, and $C=\{3\}$, then $A\cap B=\{2\}$, $A\cap C=\varnothing$, $A\cup B=\{1,2,3\}$, $A\setminus B=\{1\}$, and $B\setminus A=\{3\}$.

Finally,

$$\{x\in\mathbb R:x^2>1\} =(-\infty,-1)\cup(1,\infty) =\mathbb R\setminus[-1,1].$$

Worked example

2. Read quantifiers in order

The statement

$$\forall x\in\mathbb Z,\exists y\in\mathbb Z, x+y=0$$

is true because, after $x$ is given, we may choose $y=-x$. The reversed order $\exists y\in\mathbb Z,\forall x\in\mathbb Z, x+y=0$ is false: setting $x=0$ forces $y=0$, but then $x=1$ makes the sum nonzero.

Worked example

3. Domain changes and a piecewise endpoint

For $f:\mathbb R\to\mathbb R$, $f(x)=x^2$, the range is $[0,\infty)$. For $g:[-1,2)\to\mathbb R$, defined by $g(x)=x^2$ for $x\in[-1,2)$, the range is $[0,4)$. The value $4$ is approached as $x$ approaches $2$ but is not attained.

Now define $h:\mathbb R\to\mathbb R$ by

$$h(x)=\begin{cases}x+1,&x\ge2,\\3-x,&x\lt2.\end{cases}$$

At $x=2$ only the first branch applies, so $h(2)=3$. The second branch would give $1$, but its condition explicitly excludes the endpoint.

Worked example

4. Classify three sine functions

The function $f:\mathbb R\to\mathbb R$, $f(x)=\sin x$, is neither injective nor surjective. Restricting only the codomain gives $g:\mathbb R\to[-1,1]$, which is surjective but not injective. Restricting the domain as well gives $h:[-\pi/2,\pi/2]\to[-1,1]$, which is strictly increasing and reaches every codomain value; hence it is bijective.

Worked example

5. Exponential and logarithmic inverses

Let $f:\mathbb R\to(0,\infty)$ be $f(x)=e^x$. It is bijective, and its inverse is $f^{-1}:(0,\infty)\to\mathbb R$, $f^{-1}(y)=\ln y$. Indeed,

$$\ln(e^x)=x\quad(x\in\mathbb R), \qquad e^{\ln y}=y\quad(y>0).$$

Both identities are needed because they verify the two compositions on their respective domains.

Worked example

6. Prove a cubic function is bijective and invert it

Let $f:(0,\infty)\to(1,\infty)$ be $f(x)=2x^3+1$. If $f(x_1)=f(x_2)$, then

$$(x_1-x_2)(x_1^2+x_1x_2+x_2^2)=0.$$

The second factor is positive because $x_1,x_2>0$, so $x_1=x_2$ and $f$ is injective. Given $y>1$, choose $x=\sqrt[3]{(y-1)/2}>0$. Then $f(x)=y$, so $f$ is surjective. Therefore

$$f^{-1}(y)=\sqrt[3]{\frac{y-1}{2}},\qquad y\in(1,\infty).$$

Worked example

7. Expand a sum and telescope

Examples of direct expansion are

$$\sum_{i=1}^{n}i^2=1^2+2^2+\cdots+n^2, \qquad \sum_{i=0}^{10}2^i=1+2+2^2+\cdots+2^{10}.$$

For $n\in\mathbb Z^+$, partial fractions give

$$\begin{aligned} \sum_{k=1}^{n}\frac1{k(k+1)} &=\sum_{k=1}^{n}\left(\frac1k-\frac1{k+1}\right)\\ &=1-\frac1{n+1}. \end{aligned}$$

Every interior term cancels with its copy of opposite sign. The index is $k$ throughout, so each displayed summand is bound by the stated sum.

Worked example

8. Read products rather than sums

For $n\in\mathbb Z^+$,

$$\prod_{i=1}^{n}i=n!, \qquad \prod_{i=1}^{n}2^i =2^{1+2+\cdots+n} =2^{n(n+1)/2}.$$

Because every factor is positive,

$$\sum_{i=1}^{n}\log i =\log(1\cdot2\cdots n) =\log\left(\prod_{i=1}^{n}i\right).$$

The second identity begins with a product of powers, not a sum of powers.

Common Mistakes

Common mistake

Membership is not containment

$x\in A$ relates an element to a set; $A\subseteq B$ relates two sets. Their roles cannot be exchanged.

Common mistake

Set difference needs a set on the right

The real numbers except $2$ are $\mathbb R\setminus\{2\}$, not an expression that subtracts the number $2$ from a set.

Common mistake

Do not swap quantifiers

In $\forall x\exists y$, $y$ may depend on $x$. In $\exists y\forall x$, one fixed $y$ must work for all $x$.

Common mistake

Formula, domain, codomain, and range are different data

Writing $x^2$ does not by itself specify a function. Also, the codomain is the declared target, whereas the range consists only of attained values.

Common mistake

An inverse is not a reciprocal

$f^{-1}$ denotes the function undoing $f$; it does not generally mean $1/f$. Verify both compositions and their domains.

Common mistake

Shift all index data together

Changing a dummy letter is harmless, but an index shift must update the lower bound, upper bound, and subscript consistently. Also distinguish $\sum$, which adds terms, from $\prod$, which multiplies factors.

Summary

Sets are controlled by membership, and equality of sets is proved by two inclusions. Quantifiers must be read in order. A function includes its domain and codomain as well as its rule; injectivity controls uniqueness of preimages, surjectivity controls existence, and together they give an inverse. Finite sums and products require explicit bounds and consistent indices. These conventions make every later algebraic argument precise enough to check.

Exercises

1. Write the positive odd integers in two equivalent set-builder forms.
2. For $A=\{1,2,4\}$ and $B=\{2,3,4\}$, find $A\cap B$, $A\cup B$, $A\setminus B$, and $B\setminus A$.
3. Decide which of $\forall x\in\mathbb Z,\exists y\in\mathbb Z, y>x$ and $\exists y\in\mathbb Z,\forall x\in\mathbb Z, y>x$ is true.
4. Find the range of $g:[-2,1)\to\mathbb R$, $g(x)=x^2$.
5. Classify $f:\mathbb R\to[0,\infty)$, $f(x)=x^2$, as injective, surjective, both, or neither.
6. Show that $f:\mathbb R\to\mathbb R$, $f(x)=3x-5$, is bijective and find $f^{-1}$.
7. Shift the index in $\sum_{i=2}^{n}a_i$ so that the new lower bound is $5$.
8. Evaluate $\sum_{k=1}^{n}\frac1{(k+1)(k+2)}$ and simplify $\prod_{i=1}^{n}3^i$.

Solutions

Solution · Solution 1

The positive odd integers are $\{2m-1:m\in\mathbb Z^+\}$ and $\{n\in\mathbb Z^+:n=2m-1\text{ for some }m\in\mathbb Z^+\}$.

Solution · Solution 2

$A\cap B=\{2,4\}$, $A\cup B=\{1,2,3,4\}$, $A\setminus B=\{1\}$, and $B\setminus A=\{3\}$.

Solution · Solution 3

The first statement is true: after $x$ is given, take $y=x+1$. The second is false because no integer is greater than every integer; for any proposed $y$, the choice $x=y+1$ is a counterexample.

Solution · Solution 4

The least value is $0$, attained at $x=0$. The greatest possible square would be $4$ at $x=-2$, and that endpoint is included. Hence the range is $[0,4]$.

Solution · Solution 5

The function is surjective because every $y\ge0$ equals $(\sqrt y)^2$. It is not injective because $f(1)=f(-1)$. Thus it is surjective but not injective.

Solution · Solution 6

If $3x_1-5=3x_2-5$, then $x_1=x_2$, so $f$ is injective. Given any real $y$, take $x=(y+5)/3$; then $f(x)=y$, so it is surjective. Solving $y=3x-5$ for $x$ gives $f^{-1}(y)=(y+5)/3$.

Solution · Solution 7

Use the shift $s=3$. Then

$$\sum_{i=2}^{n}a_i=\sum_{j=5}^{n+3}a_{j-3}.$$

The first new term is $a_{5-3}=a_2$ and the last is $a_{n+3-3}=a_n$.

Solution · Solution 8

Since $\frac1{(k+1)(k+2)}=\frac1{k+1}-\frac1{k+2}$,

$$\sum_{k=1}^{n}\frac1{(k+1)(k+2)} =\frac12-\frac1{n+2}.$$

Also $\prod_{i=1}^{n}3^i=3^{1+2+\cdots+n}=3^{n(n+1)/2}$.