Real Analysis

Winter 2025

Part 1:

Sets

(See Chapter 1 in Book of Proof)

The central objects in mathematics are sets. A set is a collection of things. The things in a set are called elements.

Sets

Often, we denote a set by simply writing all the elements between \(\{\}\) ("curly braces") separated by commas. For example,

\[\{1,2,4,5\}\]

is the set with four elements, these elements being \(1,2,4,\) and \(5\).

We often give sets names, these will frequently be a single capital roman letter, for example, \[B = \{1,2,4,5\}.\]

To express the statement "\(x\) is an element of \(A\)" we write \(x\in A\). Thus, given the set \(B\) defined above, we see that \(1\in B\) is a true statement, and \(3\in B\) is a false statement.

We will also write \(x\notin A\) to for the statement "\(x\) is not an element of \(A\)."

Equality of sets

Two sets \(A\) and \(B\) are equal (they are the same set) if they contain exactly the same elements, that is, the following two statements hold:

If \(x\in A\), then \(x\in B\).

If \(x\in B\), then \(x\in A\).

In this case we write \(A=B\).

Important example:

Important example: Consider the sets \(A=\{1,2,2,3\}\) and \(B=\{1,2,3\}\). We can easily see that every element of \(A\) is an element of \(B\), and every element of \(B\) is an element of \(A\). Thus

\[\{1,2,2,3\}=\{1,2,3\}.\]

An element is either in a set or it is not. Sets cannot contain multiple copies of an element.

Subsets

Given two sets \(A\) and \(B\), if every element of \(A\) is also an element of \(B\), then we say that \(A\) is a subset of \(B\), and write \(A\subset B\) or \(A\subseteq B\) (these both mean the exact same thing). For example,

\[\{1,2\}\subset\{1,2,4,5\}\]

but \(\{1,2,3\}\) is not a subset of \(\{1,2,4,5\}\) since \(3\in\{1,2,3\}\) and \(3\notin\{1,2,4,5\}\)

Note that the statement "\(A=B\)" can be rewritten as "\(A\subset B\) and \(B\subset A.\)" This will be very useful later.

Ellipsis notation

The set of integers is denoted \(\mathbb{Z}\). The set of positive integers, also called the natural numbers or the naturals, is denoted \(\mathbb{N}\) or \(\mathbb{Z}^{+}\). 

We will sometimes write sets like \(\mathbb{N}\) and \(\mathbb{Z}\) with ellipsis notation, as follows,

\[\N=\{1,2,3,\ldots\}\qquad \Z = \{\ldots,-2,-1,0,1,2,3,\ldots\}\]

In this notation, we list the first several elements of a set, until there is an apparent pattern, then write an ellipsis (...). After the ellipsis we write the final element in the pattern. If no element is written after the ellipsis, then the list of elements goes on without end.

Examples.

• \(\{2,4,6,8,\ldots\}\) is the set of even positive integers.
• \(\{3,7,11,15,\ldots,43\}=\{3,5,7,11,15,19,23,27,31,35,39,43\}\)

Problems with ellipsis notation

Consider the set

\[C:=\{2,4,8,\ldots\}.\]

Questions:

Is \(16\in C\)?    Is \(14\in C\)?

What is the pattern? Is the \(n\)th term in the sequence \(2^{n}\) or \(n^2-n+2\) or something else?

It's not clear!

Suppose \(N\in\mathbb{N}\) and define the set

\[D:=\{2,4,6,\ldots,2N\}.\] If \(N=5\), then

\[D = \{2,4,6,\ldots,10\}=\{2,4,6,8,10\}\]

but if \(N=2\), then

\[D = \{2,4,6,\ldots,4\}(?)\]

We will interpret this set as \(\{2,4\}\). 

Set builder notation

Let \(A\) be a set, and \(S(x)\) denote a statement whose truth value depends on \(x\in A\). The set of \(x\in A\) such that \(S(x)\) is true is denoted

\[\{x\in A : S(x)\}\quad\text{or}\quad \{x\in A\mid  S(x)\}\quad\text{or}\quad \{x : x\in A\text{ and }S(x)\}.\]

Both the colon and the vertical line are read as "such that" or "with the property that."

Example. Consider the set of even positive integers:

\[\{x\in\mathbb{N} : x\text{ is even}\}.\]

This would be read:

"The set of \(x\) in the natural numbers such that \(x\) is even."

Using the notation from the above definition we have \(A=\N\) and \(S(x)\) is the statement "\(x\) is even"

Set builder notation

More generally, let \(A\) be a set. For each \(x\in A\) let \(f(x)\) be an expression that depends on \(x\) and \(S(x)\) denote a statement whose truth value depends on \(x\in A\). The set of all elements of the form \(f(x)\) with the property that \(x\in A\) and \(S(x)\) is true is denoted

\[\{f(x) : x\in A\text{ and }S(x)\}.\]

Example. Consider the set of even positive integers:

\[\{2x: x\in \N\}.\]

In the notation of the above definition we have  \(A=\N\), \(f(x) = 2x\), and \(S(x)\) is true for all \(x\in\N\).

Using set builder notation we can disambiguate the sets from two slides prior:

\[\{2^{n} : n\in\mathbb{N}\}\quad\text{and}\quad\{n^2-n+2 : n\in\mathbb{N}\}.\]

Examples. 

• \(\{2n-1 : n\in\N\text{ and }n\leq 5\} = \{1,3,5,7,9\}\)
• \(\{25-4n : n\in\N\text{ and }n\leq 10\text{ and } n>5\} = \{1,-3,-7,-11,-15\}\)

Consider the set 

\[\{x\in\mathbb{R} : x^{2}+1 = 0\}.\]

There does not exist a real number \(x\) such that \(x^{2}+1=0\), and hence this set does not contain any elements!

The set containing no elements is called the empty set. The empty set is denoted \(\varnothing\) or \(\{\}\).

Part 2:

Proofs and Logic

(See Chapters 2,4,5 & 6 in Book of Proof)

Proposition 1. \(\{n\in\mathbb{Z} : n+1\text{ is odd}\} = \{2n : n\in\mathbb{Z}\}.\)

Proving sets are equal.

Proof. First we will show that \(\{n\in\mathbb{Z} : n+1\text{ is odd}\} \subset \{2n : n\in\mathbb{Z}\}.\) Let \(m\in\{n\in\mathbb{Z} : n+1\text{ is odd}\}\) be arbitrary. This implies that \(m+1\) is odd, that is, there exists \(k\in\Z\) such that \(m+1=2k+1.\) Subtracting \(1\) from both sides we deduce \(m=2k\), and hence \(m\in \{2n : n\in\Z\}.\)

Next, we will show \(\{2n:n\in\mathbb{Z}\}\subset\{n\in\mathbb{Z} : n+1\text{ is odd}\}\).

Let \(p\in\{2n:n\in\mathbb{Z}\}\) be arbitrary. Then, there exists \(q\in\mathbb{Z}\) such that \(p=2q\), and hence \[p+1 = 2q+1.\] Since \(q\in\mathbb{Z}\), we conclude that \(p+1\) is odd, and hence

\(p\in\{n\in\mathbb{Z} : n+1\text{ is odd}\}.\) \(\Box\)

Definition. An integer \(n\) is even if there exists an integer \(k\) such that \(n=2k\). An integer \(n\) is odd if there exists an integer \(p\) such that \(n=2p+1\).

Proposition 2. If \(n\in\Z\) is odd, then \(n^{2}\) is odd.

Proving implications (Direct proof)

\(A : n\) is odd

\(B : n^{2}\) is odd

We wish to prove the implication \(A\Rightarrow B\). We will use a direct proof, that is, we will assume \(A\), argue directly to \(B\).

Proof. Suppose \(n\in\Z\) is odd. By definition, this means that there exists \(k\in\Z\) such that \(n=2k+1\). This implies

\[n^{2} = (2k+1)^{2} = 4k^{2}+4k+1 = 2(2k^{2}+2k)+1.\]

Since \(2k^{2}+2k\) is an integer, we conclude that \(n^{2}\) is odd. \(\Box\)

Proposition 3. Suppose \(n\in\Z\). If \(n^{2}\) is even, then \(n\) is even.

Proving implications (Contrapositive proof)

Proposition 4. Suppose \(a,b,c\in\Z\). If \(a^{2}+b^{2}=c^{2}\), then either \(a\) is even or \(b\) is even.

Proving implications (Proof by contradiction)

Proposition 4. Suppose \(a,b,c\in\Z\). If \(a^{2}+b^{2}=c^{2}\), then either \(a\) is even or \(b\) is even.

Proposition 3. Suppose \(n\in\Z\). If \(n^{2}\) is even, then \(n\) is even.

Contrapositive vs. Contradiction

Proof. Assume toward a contradiction that \(n\) is not even and \(n^{2}\) is even. This implies that \(n\) is odd, that is \(n=2k+1\) for some \(k\in\Z\). Squaring, we obtain

\[n^{2} = 4k^{2}+4k+1=2(2k^{2}+2k)+1.\]

Since \(2k^{2}+2k\) is an integer, this shows that \(n^{2}\) is odd. This contradicts the fact that \(n^{2}\) is even.\(\Box\)

Proof by contradiction: \(A\wedge(\sim\!B)\Rightarrow C\wedge(\sim\!C)\)

Contrapositive: \((\sim\!B)\Rightarrow(\sim\!A)\)

Fake proof by contradiciton: \(A\wedge(\sim\!B)\Rightarrow A\wedge(\sim\!A)\)

Moral: If your proof by contradiction concludes with \(\sim\!A\), then you should use proof by contrapositive instead!

Universal quantifier

\(A : n^{2}\) is even

\(B : n\) is even

Note that \(A\) and \(B\) below are not statements. They're truth value depends on \(n\).

These are more precisely called open sentences (or predicates or propositional functions). They should really have variables

\(A(n) : n^{2}\) is even

\(B(n) : n\) is even

Proposition 3. Suppose \(n\in\Z\). If \(n^{2}\) is even, then \(n\) is even.

Recall the proposition:

Now, Proposition 3 can be more precisely written as \[\forall\,n\in\Z,\ A(n)\Rightarrow B(n).\]

Where the symbol \(\forall\) stands for "for all" or "for every," and is called the universal quantifier.

Note that \(A(n) : n^{2}\) is even, is false unless \(n\in\Z\).

Recall that for false statements \(A\), the statement \(A\Rightarrow B\) is true for any statement \(B\). (In this case \(A\Rightarrow B\) is called a vacuously true statement.)

Thus, in the statement \(\forall\,n\in\Z,\ A(n)\Rightarrow B(n)\) the universal quantifier isn't really necessary. It would be very common to see Proposition 3 written as

Proposition 3. If \(n^{2}\) is even, then \(n\) is even.

The universal quantifier "For all \(n\in\Z\)" is there, it's just not written.

Moral: In math we often omit the universal quantifier before an implication.

Existential quantifier

To say that an open sentence \(A(x)\) is true for all \(x\) in a set \(X\) we use the universal quantifier:

\[\forall\,x\in X,\ A(x).\]

In math, we would write this using words:

"For all \(x\in X\), \(A(x)\) is true."

To say that an open sentence \(A(x)\) is true for at least one \(x\) in a set \(X\) we use the existential quantifier \[\exists\, x\in X,\ A(x).\] In words:

"There exists \(x\in X\) such that \(A(x)\) is true."

Universally quantified statements may be more common, but here is an example of a statement that is not:

"The polynomial \(x^{5}+5x^3+10\) has a real root."

Do you see that this statement is of the form \(\exists\, x\in X,\ A(x)\). What are the set \(X\) and the open sentence \(A(x)\)?

Order of quantifiers

Consider the following two statements:

\(A:\ \) For every even \(n\in\N\) there exists an odd \(m\in\N\) such that \(m>n\).

\(B:\ \) There exists odd \(m\in\N\) such that for all even \(n\in\N\), it is true that \(m>n\).

Because it sounds better in English, we will often write \(B\) as

\(B:\ \) There exists odd \(m\in\N\) such that \(m>n\) for all even \(n\in\N\).

Let \(E\) denote the set of even natrual numbers, and let \(D\) denote the set of odd natural numbers. Using symbols, we can write \(A\) and \(B\) as follows: \[A: \forall\, n\in E,\ \exists\,m\in D,\ m>n\]

\[B: \exists\,m\in D,\ \forall\, n\in E,\ m>n\]

Moral: \(A\) is true, and \(B\) is false. The order of the quantifiers matters. You must think hard to figure out the meaning of a statement, there is no shortcut!

Proving universally quantified statements

Every universally quantified statement can be converted to an implication:

\[\forall\, x\in X,\ A(x)\]

is the same as

\[(x\in X)\Rightarrow A(x)\]

Thus, to prove this statement we take an "arbitrary" element \(x\in X\), then reason that \(A(x)\) is true for that \(x\). The proof looks like

Proof. Let \(x\in X\) be arbitrary.

\[\vdots\]

\[\vdots\]

Therefore, \(A(x)\). \(\Box\)

For example, let's say we want to prove the following statement:

Proposition. For every even number \(n\in\N\), the number \(n^2+1\) is odd.

For example, let's say we want to prove the following statement:

Proposition. For every even number \(n\in\N\), the number \(n^2+1\) is odd.

Proof. Let \(n\in\N\) be an arbitrary even number. This implies that \(n=2k\) for some integer \(k\), and thus

\[n^2+1 = (2k)^{2}+1 = 4k^{2}+1 = 2(2k^{2})+1.\]

Since \(2k^2\) is an integer, we conclude that \(n^2+1\) is odd. \(\Box\)

Proving quantified statements

Theorem 1.18. For every positive real number \(x\) there exists a natural number \(n\) such that \(n>x.\)

Proposition. For every \(\varepsilon>0\) there exists \(m\in\N\) such that \(\dfrac{1}{m}<\varepsilon\).

The following is an important property of the natural numbers and the real numbers. We will prove it later, but for now we will assume it is true.

Proving quantified statements

Proposition. For every \(\varepsilon>0\) there exists \(m\in\N\) such that \(\dfrac{1}{m}<\varepsilon\).

What would it look like if we tried to prove this statement by contradiction?

Statement: \(\forall\,\varepsilon>0,\ \exists\,m\in\N,\ \dfrac{1}{m}<\varepsilon.\)

Negation: \(\exists\,\varepsilon>0,\ \forall\,m\in\N,\ \dfrac{1}{m}\geq\varepsilon.\)

In words, the negation might read:

There exists \(\varepsilon>0\) such that \(\varepsilon\leq \dfrac{1}{m}\) for all \(m\in\mathbb{N}\).

Part 3:

Properties of \(\mathbb{R}\)

(See Chapter 1 in Spivak)

The algebraic properties of the real numbers \(\mathbb{R}\):

(P1) (Associativelaw for addition) \[\forall\,a,b,c\in\mathbb{R},\ a+(b+c) = (a+b)+c.\]

(P2) (Existence of an additive idenity) \[\exists\, 0\in\mathbb{R},\ \forall a\in\mathbb{R},\ a+0=0+a=a.\]

(P3) (Existence of additive inverses) \[\forall\, a\in\mathbb{R},\ \exists -a\in\mathbb{R},\ a+(-a) = (-a)+a = 0.\]

(P4) (Commutative law for addition) \(\forall\, a,b\in\mathbb{R},\ a+b=b+a.\)

(P5) (Associative law for multiplication)  \(\forall\,a,b,c\in\mathbb{R},\ a\cdot(b\cdot c) = (a\cdot b)\cdot c.\)

(P6) (Existence of a multiplicative identity) \[\exists\, 1\in\mathbb{R}\setminus\{0\}, \forall\, a\in\mathbb{R},\ a\cdot 1 = 1\cdot a = a.\]

(P7) (Existence of multiplicative inverses) \[\forall\, a\in\mathbb{R}\setminus\{0\},\ \exists\, a^{-1}\in\mathbb{R},\ a\cdot a^{-1} = a^{-1}\cdot a = 1.\]

(P8) (Commutative law for multiplication) \(\forall\, a,b\in\mathbb{R},\ a\cdot b = b\cdot a.\)

(P9) (Distributive law) \(\forall\,a,b,c\in\mathbb{R},\ a\cdot(b+ c) = a\cdot b + a\cdot c.\)

Notation

Some consequences:

Proposition 1. The additive identity in \(\mathbb{R}\) is unique.

Proposition 3. If \(a\in\mathbb{R}\), then \(0\cdot a = 0\).

Proof. Let \(a\in\mathbb{R}\) be arbitrary. Note that

\[0\cdot a = 0\cdot a + 0\cdot a - 0\cdot a = (0+0)\cdot a - 0\cdot a = 0\cdot a - 0\cdot a = 0.\ \Box\]