Constructing the Reals

Now that we have motivated the need to explore beyond the rationals, let’s construct the reals. Intuitively, they correspond to points on a number line. Yet, however powerful it may be, geometric intuition can be rather limited from a logical perspective. It will surely help us solve many problems, once we ascertain its legitimate existence.

Let’s first make an observation.

Theorem 1. Let $\mathcal P(\mathbb Q)$ denote the collection of subsets of $\mathbb Q$ . Then the map $\iota : \mathbb Q \to \mathcal P(\mathbb Q)$ defined by $\iota(r) := \{x \in \mathbb Q : x < r\}$ is injective.

Our aim is to define the real numbers $\mathbb R$ as a collection of subsets of $\mathbb Q$ (i.e. $\mathbb R \subseteq \mathcal P(\mathbb Q)$ ), and then embed $\mathbb Q \longleftrightarrow \iota(\mathbb Q) \subseteq \mathbb R$ .

We need $\mathbb R$ to be an ordered field, and satisfy the special completeness property i.e. that any nonempty $K \subseteq \mathbb R$ that is bounded above has a supremum. To that end, we will construct $\mathbb R$ using the method of Dedekind cuts.

Definition 1. A Dedekind cut is a subset $\alpha \in \mathcal P(\mathbb Q)$ of the rationals with the following properties:

$\emptyset \subsetneq \alpha \subsetneq \mathbb Q$
For any $x \in \alpha, y \in \mathbb Q$ , $y < x$ implies $y \in \alpha$ .
For any $x \in \alpha$ , there exists $y \in \alpha$ such that $x < y$ .

Example 1. For any $r \in \mathbb Q$ , $\iota(r)$ is a Dedekind cut. Thus, we will eventually be able to identify $\mathbb Q \longleftrightarrow \iota(\mathbb Q) \subseteq \mathbb R$ and write $\mathbb Q \subseteq \mathbb R$ without ambiguity.

Example 2. For any prime number $p \in \mathbb Q$ , $\{x \in \mathbb Q : x^2 < p\} \cup \mathbb Q^-$ is a Dedekind cut.

We are now ready to define a real number. The set of real numbers $\mathbb R$ is the set of Dedekind cuts with several operations.

Lemma 1. For $\alpha,\beta \in \mathbb R$ , we define

$\alpha \leq \beta \quad \iff \quad \alpha \subseteq \beta.$

Then $\leq$ on $\mathbb R$ is a total order.

Proof. We observe that $\leq$ is a partial order since $\subseteq$ is a partial order on $\mathcal P(\mathbb Q)$ . All we need to do is to show that for any $\alpha, \beta \in \mathbb R$ , $\alpha \leq \beta$ or $\beta \leq \alpha$ .

Suppose $\neg(\alpha \leq \beta)$ . Then $\alpha \not\subseteq \beta$ means there exists $x \in \alpha$ such that $x \notin \beta$ . We aim to prove that $\beta \subseteq \alpha$ . Fix $y \in \beta$ . Since $x = y \in \beta$ is a contradiction, we have $x \neq y$ .

Since $\leq$ is a total order on $\mathbb Q$ , $y < x$ or $x < y$ . If $x < y$ , then $\beta$ being a Dedekind cut implies $x \in \beta$ , a contradiction. Thus, $y < x$ . Then $\alpha$ is a Dedekind cut implies $y \in \alpha$ , as required.

Lemma 2. Define real number addition by

$\alpha + \beta := \{x + y : x \in \alpha, y \in \beta\}$

Then $\mathbb R$ forms an abelian group under addition with additive identity $0 = \iota(0)$ and a suitably defined additive inverse $-\alpha$ for any $\alpha \in \mathbb R$ .

Proof. To check for closure, let $\alpha,\beta$ be Dedekind cuts. We need to prove that $\alpha + \beta$ is a Dedekind cut. It is clearly a nonempty proper subset of $\mathbb Q$ . For the second property, Fix $r+s \in \alpha +\beta$ and $y \in \mathbb Q$ . If $y < r+ s$ , then

$y -s < r \quad \Rightarrow \quad y-s \in \alpha \quad \Rightarrow \quad y = (y-s) + s \in \alpha + \beta.$

For the third property, fix $r+s \in \alpha+\beta$ . Since $\alpha,\beta$ are Dedekind cuts, find $u \in \alpha,v \in \beta$ such that $r < u$ and $s < v$ . Then $u + v \in \alpha +\beta$ and $r + s < u + v$ .

For the algebraic properties, it is relatively obvious that $(\alpha +\beta) + \gamma = \alpha + (\beta + \gamma)$ due to the associativity of rational number addition, and $\alpha + \beta = \beta + \alpha$ due to the commutativity of rational number addition.

The additive identity holds since $x = r + (x-r) \in \alpha + 0$ implies $\alpha \subseteq \alpha + 0$ and $x+r < x$ implies $\alpha + 0 \subseteq \alpha$ .

For the additive inverse, let $\alpha$ be a Dedekind cut. We need to construct a Dedekind cut $\beta$ such that $\alpha +\beta = 0$ . This means that we require elements of $\alpha + \beta$ to be of the form $x + y$ , where $x \in \alpha$ and $x + y < 0\iff x < -y$ . This means that there should exist $t \in \mathbb Q^+$ such that $x = -y-t$ . The elements of $\beta$ then ought to be the $y$ such that there exists $t \in \mathbb Q^+$ with $x < -y-t$ or in other words, $-y-t \notin \alpha$ .

With that in mind, we define $\beta := \{y \in \mathbb Q : (\exists t \in \mathbb Q^+ : -y-t \notin \alpha )\}$ . We need to prove that $\beta$ is a Dedekind cut, and that $\alpha + \beta = 0$ . For the first property, it is clear that $\emptyset \subsetneq \beta \subsetneq \mathbb Q$ .

For the second property, fix $u \in \beta, v \in \mathbb Q$ . Since $u \in \beta$ , find $t \in \mathbb Q^+$ such that $-u-t \notin \alpha$ . Suppose $v < u$ . Then $v = u - s$ for some $s \in \mathbb Q^+$ , which implies

$-v - s = -u > -u-t.$

Since $-u - t \notin \alpha$ , $-v-s \notin \alpha$ so that $v \in \beta$ , as required.

For the third property, fix $u \in \beta$ . Find $t \in \mathbb Q^+$ such that $-u-t \notin \alpha$ . Then $-(u+t/2)-t/2 = -u-t \notin\alpha$ too, so that $v := u + t/2 > u$ belongs to $\beta$ , as required.

To prove that $\alpha + \beta = \iota(0)$ , we need to prove in two directions. For the first direction $(\subseteq)$ , fix $u+v \in \alpha + \beta$ and find $t \in \mathbb Q^+$ such that $-v -t \notin \alpha$ . Then $-v \notin \alpha$ implies $u < -v$ so that $u + v < 0$ and hence $u +v \in \iota(0)$ .

For the second direction $(\supseteq)$ , fix $x \in \iota(0)$ i.e. $x < 0$ . Define $\epsilon := -x/2 > 0$ . Use the Archimedean property of $\mathbb Q$ to find $n \in \mathbb N$ such that $n\epsilon \in \alpha$ but $(n+1)\epsilon \notin \alpha$ . This implies $y:=-(n+2)\epsilon$ belongs to $\beta$ , since $-y-\epsilon = -(n+1)\epsilon \notin \alpha$ , so that $x = -2\epsilon = n\epsilon + y \in \alpha + \beta$ , as required.

Since additive inverse are unique, we can denote $\beta := -\alpha$ without ambiguity.

Corollary 1. The ordering $\leq$ is compatible with addition, i.e. for real numbers $\alpha,\beta,\gamma$ , $\alpha \leq \beta$ implies $\alpha + \gamma \leq \beta + \gamma$ .

Lemma 3. For multiplication, we first define $0 := \iota(0) = \mathbb Q^-$ . Then, for $\alpha > 0,\beta > 0$ , we define

$\alpha \cdot \beta := \{r \in \mathbb Q : (\exists s \in \alpha \cap \mathbb Q^+, t \in \beta \cap \mathbb Q^+ : r < s \cdot t) \}.$

Furthermore, we stipulate $0\cdot \alpha = 0 = \alpha \cdot 0$ and

$\displaystyle \alpha \cdot \beta := \begin{cases}(-\alpha)\cdot (-\beta) & \quad \text{if}\ \alpha < 0, \beta < 0, \\ -((-\alpha) \cdot \beta) & \quad \text{if}\ \alpha < 0, \beta > 0, \\ -(\alpha\cdot (-\beta)) & \quad \text{if}\ \alpha > 0, \beta < 0,\end{cases}$

Then $\mathbb R \backslash \{ 0 \}$ forms an abelian group under multiplication with multiplicative identity $1$ .

Proof. Exercise.

Lemma 4. For real numbers $\alpha,\beta,\gamma$ ,

$\begin{aligned} \alpha(\beta + \gamma) &= \alpha \beta + \alpha \gamma,\\ (\alpha + \beta)\gamma &= \alpha \gamma + \beta \gamma. \end{aligned}$

Proof. The second identity follows the first via commutativity under $+$ and $\times$ , so it suffices to prove the first identity. In fact, the first identity is a matter of tedious case-splitting after we establish it for $\alpha > 0,\beta > 0,\gamma >0$ , which works because we have $x(y+z) = xy + xz$ in the rationals.

We need to check that our definitions agree with the existing definitions for the rational numbers, summarised in the lemma below.

Lemma 5. For $x, y \in \mathbb Q$ ,

$\iota(x+y) = \iota(x)+\iota(y),\quad \iota(xy) = \iota(x)\iota(y),\quad x \geq 0 \Rightarrow \iota(x) \geq \iota(0).$

Proof. For addition, $(\supseteq)$ is obvious. For $(\subseteq)$ , fix $u \in \iota(x+y)$ so that $u < x + y$ . This means $u - x < y$ so that there exists $\epsilon > 0$ such that $(u-x) + \epsilon < y$ . In particular, $x-\epsilon < x$ so that $u = (x - \epsilon) + (u - x) +\epsilon \in \iota(x) + \iota(y)$ .

For multiplication, for simplicity, we will establish the identity for $x > 0, y > 0$ . Just like addition, $(\supseteq)$ is obvious. For $(\subseteq)$ , fix $u \in \iota(xy)$ so that $u < xy$ . Since $x > 0$ and $y > 0$ , we have $u/x < y \iff u/y < x$ . Set $\epsilon := (x-u/y)/2 \in (0, x)$ so that $u/y < x-\epsilon \iff u/(x-\epsilon) < y$ . Then

$u = (x - \epsilon) \cdot (u/(x - \epsilon)) \in \iota(x)\iota(y).$

Finally for ordering, $x > 0$ implies that $u < 0\Rightarrow u < x$ so that $\iota(0) \leq \iota(x)$ .

Finally, we need to check that $\mathbb R$ satisfies the completeness property.

Lemma 6. For any nonempty subset $K \subseteq \mathbb R$ that is bounded above, $K$ has a least upper bound in $\mathbb R$ , denoted $\sup K \in \mathbb R$ .

Proof. Fix nonempty $K \subseteq \mathbb R$ that is bounded above. Define the set $\beta \in \mathcal P(\mathbb Q)$ by

$\displaystyle \beta := \bigcup_{\alpha \in K} \alpha.$

We leave it as an exercise to prove that $\sup K = \beta \in \mathbb R$ .

In fact, the nontrivial property to prove is that $\beta$ is a proper subset of $\mathbb Q$ . Since $K$ is nonempty, there exists $\alpha \in \mathbb R$ such that $\alpha \in K$ . Since $\alpha \in \mathbb R$ , $\alpha \neq \emptyset$ . Since $\emptyset \subsetneq \alpha \subseteq \beta$ , $\beta \neq \emptyset$ .

Furthermore, since $K$ is bounded above, there exists $\gamma \in \mathbb R$ such that for any $\alpha \in K$ , $\alpha \leq \gamma \iff \alpha \subseteq \gamma$ . Since $\gamma \in \mathbb R$ , there exists $y \in \mathbb Q$ such that $y \notin \gamma$ . Hence, $y \notin \alpha$ for any $\alpha \subseteq \gamma$ . Thus, $y \notin \beta$ so that $\beta \neq \mathbb Q$ .

Thus, we have constructed the real numbers $\mathbb R$ as an ordered field that satisfies the completeness property.

—Joel Kindiak, 19 Dec 24, 1448H

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Constructing the Reals

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