## Section1.1Basic properties

Note: 1.5 lectures

The main object we work with in analysis is the set of real numbers. As this set is so fundamental, often much time is spent on formally constructing the set of real numbers. However, we take an easier approach here and just assume that a set with the correct properties exists. We start with the definitions of those properties.

### Definition1.1.1.

An ordered set is a set $$S$$ together with a relation $$<$$ such that

1. (trichotomy) For all $$x, y \in S\text{,}$$ exactly one of $$x < y\text{,}$$ $$x=y\text{,}$$ or $$y < x$$ holds.

2. (transitivity) If $$x,y,z \in S$$ are such that $$x < y$$ and $$y < z\text{,}$$ then $$x < z\text{.}$$

We write $$x \leq y$$ if $$x < y$$ or $$x=y\text{.}$$ We define $$>$$ and $$\geq$$ in the obvious way.

The set of rational numbers $$\Q$$ is an ordered set by letting $$x < y$$ if and only if $$y-x$$ is a positive rational number, that is if $$y-x = \nicefrac{p}{q}$$ where $$p,q \in \N\text{.}$$ Similarly, $$\N$$ and $$\Z$$ are also ordered sets.

There are other ordered sets than sets of numbers. For example, the set of countries can be ordered by landmass, so India $$>$$ Lichtenstein. A typical ordered set that you have used since primary school is the dictionary. It is the ordered set of words where the order is the so-called lexicographic ordering. Such ordered sets often appear, for example, in computer science. In this book we will mostly be interested in ordered sets of numbers.

### Definition1.1.2.

Let $$E \subset S\text{,}$$ where $$S$$ is an ordered set.

1. If there exists a $$b \in S$$ such that $$x \leq b$$ for all $$x \in E\text{,}$$ then we say $$E$$ is bounded above and $$b$$ is an upper bound of $$E\text{.}$$

2. If there exists a $$b \in S$$ such that $$x \geq b$$ for all $$x \in E\text{,}$$ then we say $$E$$ is bounded below and $$b$$ is a lower bound of $$E\text{.}$$

3. If there exists an upper bound $$b_0$$ of $$E$$ such that whenever $$b$$ is an upper bound for $$E$$ we have $$b_0 \leq b\text{,}$$ then $$b_0$$ is called the least upper bound or the supremum of $$E\text{.}$$ See Figure 1.1. We write

\begin{equation*} \sup\, E := b_0 . \end{equation*}
4. Similarly, if there exists a lower bound $$b_0$$ of $$E$$ such that whenever $$b$$ is a lower bound for $$E$$ we have $$b_0 \geq b\text{,}$$ then $$b_0$$ is called the greatest lower bound or the infimum of $$E\text{.}$$ We write

\begin{equation*} \inf\, E := b_0 . \end{equation*}

When a set $$E$$ is both bounded above and bounded below, we say simply that $$E$$ is bounded.

The notation $$\sup E$$ and $$\inf E$$ is justified as the supremum (or infimum) is unique (if it exists): If $$b$$ and $$b'$$ are suprema of $$E\text{,}$$ then $$b \leq b'$$ and $$b' \leq b\text{,}$$ because both $$b$$ and $$b'$$ are the least upper bounds, so $$b=b'\text{.}$$

A simple example: Let $$S := \{ a, b, c, d, e \}$$ be ordered as $$a < b < c < d < e\text{,}$$ and let $$E := \{ a, c \}\text{.}$$ Then $$c\text{,}$$ $$d\text{,}$$ and $$e$$ are upper bounds of $$E\text{,}$$ and $$c$$ is the least upper bound or supremum of $$E\text{.}$$

A supremum or infimum for $$E$$ (even if it exists) need not be in $$E\text{.}$$ The set $$E := \{ x \in \Q : x < 1 \}$$ has a least upper bound of 1, but 1 is not in the set $$E$$ itself. The set $$G := \{ x \in \Q : x \leq 1 \}$$ also has an upper bound of 1, and in this case $$1 \in G\text{.}$$ The set $$P := \{ x \in \Q : x \geq 0 \}$$ has no upper bound (why?) and therefore it cannot have a least upper bound. The set $$P$$ does have a greatest lower bound: 0.

### Definition1.1.3.

An ordered set $$S$$ has the least-upper-bound property if every nonempty subset $$E \subset S$$ that is bounded above has a least upper bound, that is $$\sup\, E$$ exists in $$S\text{.}$$

The least-upper-bound property is sometimes called the completeness property or the Dedekind completeness property 1 . As we will note in the next section, the real numbers have this property.

### Example1.1.4.

The set $$\Q$$ of rational numbers does not have the least-upper-bound property. The subset $$\{ x \in \Q : x^2 < 2 \}$$ does not have a supremum in $$\Q\text{.}$$ We will see later (Example 1.2.3) that the supremum is $$\sqrt{2}\text{,}$$ which is not rational 3 . Suppose $$x \in \Q$$ such that $$x^2 = 2\text{.}$$ Write $$x=\nicefrac{m}{n}$$ in lowest terms. So $${(\nicefrac{m}{n})}^2 = 2$$ or $$m^2 = 2n^2\text{.}$$ Hence, $$m^2$$ is divisible by 2, and so $$m$$ is divisible by 2. Write $$m = 2k$$ and so $${(2k)}^2 = 2n^2\text{.}$$ Divide by 2 and note that $$2k^2 = n^2\text{,}$$ and hence $$n$$ is divisible by 2. But that is a contradiction as $$\nicefrac{m}{n}$$ is in lowest terms.

That $$\Q$$ does not have the least-upper-bound property is one of the most important reasons why we work with $$\R$$ in analysis. The set $$\Q$$ is just fine for algebraists. But us analysts require the least-upper-bound property to do any work. We also require our real numbers to have many algebraic properties. In particular, we require that they are a field.

### Definition1.1.5.

A set $$F$$ is called a field if it has two operations defined on it, addition $$x+y$$ and multiplication $$xy\text{,}$$ and if it satisfies the following axioms:

(A1)

If $$x \in F$$ and $$y \in F\text{,}$$ then $$x+y \in F\text{.}$$

(A2)

(commutativity of addition) $$x+y = y+x$$ for all $$x,y \in F\text{.}$$

(A3)

(associativity of addition) $$(x+y)+z = x+(y+z)$$ for all $$x,y,z \in F\text{.}$$

(A4)

There exists an element $$0 \in F$$ such that $$0+x = x$$ for all $$x \in F\text{.}$$

(A5)

For every element $$x\in F\text{,}$$ there exists an element $$-x \in F$$ such that $$x + (-x) = 0\text{.}$$

(M1)

If $$x \in F$$ and $$y \in F\text{,}$$ then $$xy \in F\text{.}$$

(M2)

(commutativity of multiplication) $$xy = yx$$ for all $$x,y \in F\text{.}$$

(M3)

(associativity of multiplication) $$(xy)z = x(yz)$$ for all $$x,y,z \in F\text{.}$$

(M4)

There exists an element $$1 \in F$$ (and $$1 \not= 0$$) such that $$1x = x$$ for all $$x \in F\text{.}$$

(M5)

For every $$x\in F$$ such that $$x \not= 0$$ there exists an element $$\nicefrac{1}{x} \in F$$ such that $$x(\nicefrac{1}{x}) = 1\text{.}$$

(D)

(distributive law) $$x(y+z) = xy+xz$$ for all $$x,y,z \in F\text{.}$$

### Example1.1.6.

The set $$\Q$$ of rational numbers is a field. On the other hand $$\Z$$ is not a field, as it does not contain multiplicative inverses. For example, there is no $$x \in \Z$$ such that $$2x = 1\text{,}$$ so (M5) is not satisfied. You can check that (M5) is the only property that fails 4 .

We will assume the basic facts about fields that are easily proved from the axioms. For example, $$0x = 0$$ is easily proved by noting that $$xx = (0+x)x = 0x+xx\text{,}$$ using (A4), (D), and (M2). Then using (A5) on $$xx\text{,}$$ along with (A2), (A3), and (A4), we obtain $$0 = 0x\text{.}$$

### Definition1.1.7.

A field $$F$$ is said to be an ordered field if $$F$$ is also an ordered set such that

1. For $$x,y,z \in F\text{,}$$ $$x < y$$ implies $$x+z < y+z\text{.}$$

2. For $$x,y \in F\text{,}$$ $$x > 0$$ and $$y > 0$$ implies $$xy > 0\text{.}$$

If $$x > 0\text{,}$$ we say $$x$$ is positive. If $$x < 0\text{,}$$ we say $$x$$ is negative. We also say $$x$$ is nonnegative if $$x \geq 0\text{,}$$ and $$x$$ is nonpositive if $$x \leq 0\text{.}$$

It can be checked that the rational numbers $$\Q$$ with the standard ordering is an ordered field.

Note that iv implies in particular that $$1 > 0\text{.}$$

### Proof.

Let us prove i. The inequality $$x > 0$$ implies by item i of definition of ordered fields that $$x + (-x) > 0 + (-x)\text{.}$$ Apply the algebraic properties of fields to obtain $$0 > -x\text{.}$$ The “vice versa” follows by similar calculation.

For ii, notice that $$y < z$$ implies $$0 < z - y$$ by item i of the definition of ordered fields. Apply item ii of the definition of ordered fields to obtain $$0 < x(z-y)\text{.}$$ By algebraic properties, $$0 < xz - xy\text{.}$$ Again by item i of the definition, $$xy < xz\text{.}$$

Part iii is left as an exercise.

To prove part iv first suppose $$x > 0\text{.}$$ By item ii of the definition of ordered fields, $$x^2 > 0$$ (use $$y=x$$). If $$x < 0\text{,}$$ we use part iii of this proposition, where we plug in $$y=x$$ and $$z=0\text{.}$$

To prove part v, notice that $$\nicefrac{1}{x}$$ cannot be equal to zero (why?). Suppose $$\nicefrac{1}{x} < 0\text{,}$$ then $$\nicefrac{-1}{x} > 0$$ by i. Apply part ii (as $$x > 0$$) to obtain $$x(\nicefrac{-1}{x}) > 0x$$ or $$-1 > 0\text{,}$$ which contradicts $$1 > 0$$ by using part i again. Hence $$\nicefrac{1}{x} > 0\text{.}$$ Similarly, $$\nicefrac{1}{y} > 0\text{.}$$ Thus $$(\nicefrac{1}{x})(\nicefrac{1}{y}) > 0$$ by definition of ordered field and by part ii

\begin{equation*} (\nicefrac{1}{x})(\nicefrac{1}{y})x < (\nicefrac{1}{x})(\nicefrac{1}{y})y . \end{equation*}

By algebraic properties we get $$\nicefrac{1}{y} < \nicefrac{1}{x}\text{.}$$

Parts vi and vii are left as exercises.

The product of two positive numbers (elements of an ordered field) is positive. However, it is not true that if the product is positive, then each of the two factors must be positive. For instance, $$(-1)(-1) = 1 > 0\text{.}$$

### Proof.

We show the contrapositive: If either one of $$x$$ or $$y$$ is zero, or if $$x$$ and $$y$$ have opposite signs, then $$xy$$ is not positive. If either $$x$$ or $$y$$ is zero, then $$xy$$ is zero and hence not positive. Hence assume that $$x$$ and $$y$$ are nonzero and have opposite signs. Without loss of generality suppose $$x > 0$$ and $$y < 0\text{.}$$ Multiply $$y < 0$$ by $$x$$ to get $$xy < 0x = 0\text{.}$$

### Example1.1.10.

The reader may also know about the complex numbers, usually denoted by $$\C\text{.}$$ That is, $$\C$$ is the set of numbers of the form $$x + iy\text{,}$$ where $$x$$ and $$y$$ are real numbers, and $$i$$ is the imaginary number, a number such that $$i^2 = -1\text{.}$$ The reader may remember from algebra that $$\C$$ is also a field; however, it is not an ordered field. While one can make $$\C$$ into an ordered set in some way, it is not possible to put an order on $$\C$$ that would make it an ordered field: In every ordered field, $$-1 < 0$$ and $$x^2 > 0$$ for all nonzero $$x\text{,}$$ but in $$\C\text{,}$$ $$i^2 = -1\text{.}$$

Finally, an ordered field that has the least-upper-bound property has the corresponding property for greatest lower bounds.

### Proof.

Let $$B := \{ -x : x \in A \}\text{.}$$ Let $$b \in F$$ be a lower bound for $$A\text{:}$$ If $$x \in A\text{,}$$ then $$x \geq b\text{.}$$ In other words, $$-x \leq -b\text{.}$$ So $$-b$$ is an upper bound for $$B\text{.}$$ Since $$F$$ has the least-upper-bound property, $$c:=\sup\, B$$ exists, and $$c \leq -b\text{.}$$ As $$y \leq c$$ for all $$y \in B\text{,}$$ then $$-c \leq x$$ for all $$x \in A\text{.}$$ So $$-c$$ is a lower bound for $$A\text{.}$$ As $$-c \geq b\text{,}$$ $$-c$$ is the greatest lower bound of $$A\text{.}$$

### Subsection1.1.1Exercises

#### Exercise1.1.1.

Prove part iii of Proposition 1.1.8. That is, let $$F$$ be an ordered field and $$x,y,z \in F\text{.}$$ Prove If $$x < 0$$ and $$y < z\text{,}$$ then $$xy > xz\text{.}$$

#### Exercise1.1.2.

Let $$S$$ be an ordered set. Let $$A \subset S$$ be a nonempty finite subset. Then $$A$$ is bounded. Furthermore, $$\inf\, A$$ exists and is in $$A$$ and $$\sup\, A$$ exists and is in $$A\text{.}$$ Hint: Use induction.

#### Exercise1.1.3.

Prove part vi of Proposition 1.1.8. That is, let $$x, y \in F\text{,}$$ where $$F$$ is an ordered field, such that $$0 < x < y\text{.}$$ Show that $$x^2 < y^2\text{.}$$

#### Exercise1.1.4.

Let $$S$$ be an ordered set. Let $$B \subset S$$ be bounded (above and below). Let $$A \subset B$$ be a nonempty subset. Suppose all the $$\inf$$s and $$\sup$$s exist. Show that

\begin{equation*} \inf\, B \leq \inf\, A \leq \sup\, A \leq \sup\, B . \end{equation*}

#### Exercise1.1.5.

Let $$S$$ be an ordered set. Let $$A \subset S$$ and suppose $$b$$ is an upper bound for $$A\text{.}$$ Suppose $$b \in A\text{.}$$ Show that $$b = \sup\, A\text{.}$$

#### Exercise1.1.6.

Let $$S$$ be an ordered set. Let $$A \subset S$$ be nonempty and bounded above. Suppose $$\sup\, A$$ exists and $$\sup\, A \notin A\text{.}$$ Show that $$A$$ contains a countably infinite subset.

#### Exercise1.1.7.

Find a (nonstandard) ordering of the set of natural numbers $$\N$$ such that there exists a nonempty proper subset $$A \subsetneq \N$$ and such that $$\sup\, A$$ exists in $$\N\text{,}$$ but $$\sup\, A \notin A\text{.}$$ To keep things straight it might be a good idea to use a different notation for the nonstandard ordering such as $$n \prec m\text{.}$$

#### Exercise1.1.8.

Let $$F := \{ 0, 1, 2 \}\text{.}$$

1. Prove that there is exactly one way to define addition and multiplication so that $$F$$ is a field if $$0$$ and $$1$$ have their usual meaning of (A4) and (M4).

2. Show that $$F$$ cannot be an ordered field.

#### Exercise1.1.9.

Let $$S$$ be an ordered set and $$A$$ is a nonempty subset such that $$\sup \, A$$ exists. Suppose there is a $$B \subset A$$ such that whenever $$x \in A$$ there is a $$y \in B$$ such that $$x \leq y\text{.}$$ Show that $$\sup \, B$$ exists and $$\sup \, B = \sup \, A\text{.}$$

#### Exercise1.1.10.

Let $$D$$ be the ordered set of all possible words (not just English words, all strings of letters of arbitrary length) using the Latin alphabet using only lower case letters. The order is the lexicographic order as in a dictionary (e.g. aa $$<$$ aaa $$<$$ dog $$<$$ door). Let $$A$$ be the subset of $$D$$ containing the words whose first letter is `a' (e.g. a $$\in A\text{,}$$ abcd $$\in A$$). Show that $$A$$ has a supremum and find what it is.

#### Exercise1.1.11.

Let $$F$$ be an ordered field and $$x,y,z,w \in F\text{.}$$

1. Prove part vii of Proposition 1.1.8. That is, if $$x \leq y$$ and $$z \leq w\text{,}$$ then $$x+z \leq y+w\text{.}$$

2. Prove that if $$x < y$$ and $$z \leq w\text{,}$$ then $$x+z < y+w\text{.}$$

#### Exercise1.1.12.

Prove that any ordered field must contain a countably infinite set.

#### Exercise1.1.13.

Let $$\N_{\infty} := \N \cup \{ \infty \}\text{,}$$ where elements of $$\N$$ are ordered in the usual way amongst themselves, and $$k < \infty$$ for every $$k \in \N\text{.}$$ Show $$\N_{\infty}$$ is an ordered set and that every subset $$E \subset \N_{\infty}$$ has a supremum in $$\N_{\infty}$$ (make sure to also handle the case of an empty set).

#### Exercise1.1.14.

Let $$S := \{ a_k : k \in \N \} \cup \{ b_k : k \in \N \}\text{,}$$ ordered such that $$a_k < b_j$$ for every $$k$$ and $$j\text{,}$$ $$a_k < a_m$$ whenever $$k < m\text{,}$$ and $$b_k > b_m$$ whenever $$k < m\text{.}$$

1. Show that $$S$$ is an ordered set.

2. Show that every subset of $$S$$ is bounded (both above and below).

3. Find a bounded subset of $$S$$ that has no least upper bound.

Named after the German mathematician Julius Wilhelm Richard Dedekind 2  (1831–1916).
https://en.wikipedia.org/wiki/Richard_Dedekind
This is true for all other roots of 2, and interestingly, the fact that $$\sqrt[k]{2}$$ is never rational for $$k > 1$$ implies no piano can ever be perfectly tuned in all keys. See for example: https://youtu.be/1Hqm0dYKUx4.
An algebraist would say that $$\Z$$ is an ordered ring, or perhaps more precisely a commutative ordered ring.
For a higher quality printout use the PDF versions: https://www.jirka.org/ra/realanal.pdf or https://www.jirka.org/ra/realanal2.pdf