§6 ★ A Development of ℝ

Enrichment section

This section is optional. It sketches one way to construct $\mathbb{R}$ from $\mathbb{Q}$, instead of taking $\mathbb{R}$ and its completeness as given.

To motivate the construction, recall from Exercise 4.16 that

$$a = \sup\{r \in \mathbb{Q} : r < a\} \quad\text{for each } a \in \mathbb{R}.$$

The relationship is intimate: $a \le b$ if and only if $\{r \in \mathbb{Q} : r < a\} \subseteq \{r \in \mathbb{Q} : r < b\}$, and $a = b$ if and only if these sets are equal. Sets of the form $\alpha = \{r \in \mathbb{Q} : r < a\}$ satisfy:

1. $\alpha \ne \mathbb{Q}$ and $\alpha$ is not empty;
2. if $r \in \alpha$, $s \in \mathbb{Q}$ and $s < r$, then $s \in \alpha$;
3. $\alpha$ contains no largest rational.

Conversely, every subset $\alpha$ of $\mathbb{Q}$ with (i)–(iii) has this form, with $a = \sup \alpha$.

Intuition

A cut is “all the rationals to the left of a point.” Over $\mathbb{Q}$ we cannot point at $\sqrt2$, but we can still describe the set $\{r \in \mathbb{Q} : r \le 0 \text{ or } r^2 < 2\}$ using rationals alone. The construction simply declares that set to be the new number. Every gap in $\mathbb{Q}$ is filled by the cut that ends at it.

Common pitfallwhy property (iii)?

Without (iii), the sets $\{r \in \mathbb{Q} : r < \tfrac12\}$ and $\{r \in \mathbb{Q} : r \le \tfrac12\}$ would both be cuts, giving two different “real numbers” at the same place on the line. Forbidding a largest element picks one of them for each rational.

Dedekind cut.

A Dedekind cut is a subset $\alpha$ of $\mathbb{Q}$ satisfying (i)–(iii) above.

The remarks so far used our knowledge of $\mathbb{R}$, including completeness. But they suggest a construction that assumes only the ordered field $\mathbb{Q}$ and its Archimedean property:

• Define $\mathbb{R}$ to be the set of all Dedekind cuts. So a real number is a certain subset of $\mathbb{Q}$.
• Embed $\mathbb{Q}$: each rational $s$ corresponds to the cut $s^* = \{r \in \mathbb{Q} : r < s\}$, and we identify $\mathbb{Q}$ with $\mathbb{Q}^* = \{s^* : s \in \mathbb{Q}\}$.
• Order: for cuts $\alpha, \beta$, define $\alpha \le \beta$ to mean $\alpha \subseteq \beta$. Properties O1, O2, O3 hold.
• Add: $\alpha + \beta = \{r_1 + r_2 : r_1 \in \alpha,\ r_2 \in \beta\}$. This is again a cut, and A1–A4 hold.
• Multiply: this is tedious and must first be done for cuts $\ge 0^*$. A naive attempt is Exercise 6.4.

Once multiplication is defined and the remaining ordered-field properties are checked, the resulting $\mathbb{R}$ is complete. In this approach the Completeness Axiom 4.4 becomes a theorem.

Intuitionwhy completeness comes for free

Since the order is inclusion, a least upper bound of a family of cuts should be the smallest set containing all of them, namely their union. If a nonempty family of cuts is bounded above by a cut $\beta$, the union is again a cut: it is nonempty, it is not all of $\mathbb{Q}$ because it sits inside $\beta$, it is closed downward, and it has no largest element because none of its members does. Every upper bound of the family contains each member, hence contains the union. So in this model the supremum of a bounded set is simply a union of sets. That is the heart of the proof that the cuts satisfy the Completeness Axiom.

Common pitfallmultiplication

Addition can be defined elementwise, but multiplication cannot. Every cut contains all sufficiently negative rationals, so the set of products $r_1 r_2$ would contain arbitrarily large positive rationals and could not be a cut. That is why multiplication is defined first for cuts $\ge 0^*$ (Exercise 6.4 explores the naive attempt).

$\mathbb{R}$ can also be built from Cauchy sequences of rationals, and a full development starting from Peano’s axioms is also possible. See the book’s bibliography for both.

Try it: probe a cut

Membership in each set below is decided with integer arithmetic on $p$ and $q$ alone, so no real numbers are used. Test rationals, look for larger members of $\alpha$ (property (iii)) and smaller non-members (which would break property (ii)). The last two sets are not cuts.

Rational r = p/q/

Find a larger rational in αTry smaller rationalsReset

α (a hollow end is not in the set)tested r ∈ αtested r ∉ α

r = 1 ≈ 1.00000000 is in α.

A cut whose sup, √2, is not rational. This cut is the real number √2 in the construction.

Press "Find a larger rational" repeatedly: the rationals climb toward sup α but never reach a largest one, which is property (iii).

Reading alongside Abbott

Abbott constructs $\mathbb{R}$ from Dedekind cuts at the very end of the book, in Section 8.6, leaving most of the verifications as guided exercises. Like Ross, Abbott first develops analysis assuming completeness, and only then shows that a complete ordered field really exists.

Check your understanding

Take the quiz (3 questions)

Which subset of $\mathbb{Q}$ is a Dedekind cut?

$\{r \in \mathbb{Q} : r < 2\}$$\{r \in \mathbb{Q} : r \le 2\}$$\{r \in \mathbb{Q} : r^2 < 2\}$

Which cut represents $\sqrt 2$?

$\{r \in \mathbb{Q} : r < 0 \text{ or } r^2 < 2\}$$\{r \in \mathbb{Q} : r^2 < 2\}$$\{r \in \mathbb{Q} : r^2 \le 2\}$

For cuts $\alpha$ and $\beta$, what does $\alpha \le \beta$ mean?

$\alpha \subseteq \beta$$\sup \alpha \le \sup \beta$Every element of $\alpha$ is less than every element of $\beta$

Exercises

Exercise 6.1

Consider $s, t \in \mathbb{Q}$. Show

1. $s \le t$ if and only if $s^* \subseteq t^*$;
2. $s = t$ if and only if $s^* = t^*$;
3. $(s + t)^* = s^* + t^*$. (Note that $s^* + t^*$ is a sum of Dedekind cuts.)

Hint

For (a) “$\Leftarrow$”: if $t < s$, then $t \in s^*$ but $t \notin t^*$. For (c) “$\subseteq$”: if $r < s + t$, write $r = (s - \delta) + (t - \delta')$ with $\delta, \delta' > 0$ rational, for example $\delta = \delta' = \frac{s + t - r}{2}$.

Solution

Recall $s^* = \{r \in \mathbb{Q} : r < s\}$.

1. ($\Rightarrow$) If $s \le t$ and $r < s$ then $r < t$, so $s^* \subseteq t^*$. ($\Leftarrow$) Suppose $s^* \subseteq t^*$ but $t < s$. Then $t \in s^*$ but $t \notin t^*$ (as $t \not< t$), contradicting the inclusion. So $s \le t$.
2. $s = t$ iff $s \le t$ and $t \le s$ iff (by (a)) $s^* \subseteq t^*$ and $t^* \subseteq s^*$ iff $s^* = t^*$.
3. Here $s^* + t^* = \{r_1 + r_2 : r_1 < s,\ r_2 < t\}$. If $r_1 < s$ and $r_2 < t$ then $r_1 + r_2 < s + t$, so $s^* + t^* \subseteq (s+t)^*$. Conversely, if $r < s + t$ let $\delta = \frac{s + t - r}{2} > 0$. Then $s - \delta < s$, $t - \delta < t$ and $r = (s - \delta) + (t - \delta)$, so $r \in s^* + t^*$.

Tags: StandardFrom the definitions

Exercise 6.2

Show that if $\alpha$ and $\beta$ are Dedekind cuts, then so is $\alpha + \beta = \{r_1 + r_2 : r_1 \in \alpha \text{ and } r_2 \in \beta\}$.

Hint

For (i), if $s_1 \notin \alpha$ and $s_2 \notin \beta$, show $s_1 + s_2 \notin \alpha + \beta$. For (ii), if $s < r_1 + r_2$, then $s = (s - r_2) + r_2$ with $s - r_2 < r_1$. For (iii), use that neither $\alpha$ nor $\beta$ has a largest element.

Solution

We check the three properties of a cut for $\alpha + \beta$. First note: if $r \in \alpha$ and $s \notin \alpha$ then $r < s$, since $s \le r$ would put $s$ in $\alpha$ by property (ii) (or equal $r$).

(i) Both cuts are nonempty, so $\alpha + \beta$ is nonempty. Choose $s_1 \notin \alpha$ and $s_2 \notin \beta$. If $s_1 + s_2 = r_1 + r_2$ with $r_1 \in \alpha$, $r_2 \in \beta$, then $r_1 < s_1$ and $r_2 < s_2$ give $r_1 + r_2 < s_1 + s_2$, a contradiction. So $s_1 + s_2 \notin \alpha + \beta$, and $\alpha + \beta \ne \mathbb{Q}$.

(ii) Let $r_1 + r_2 \in \alpha + \beta$ with $r_1 \in \alpha$, $r_2 \in \beta$, and let $s < r_1 + r_2$. Then $s - r_2 < r_1$, so $s - r_2 \in \alpha$ by (ii) for $\alpha$, and $s = (s - r_2) + r_2 \in \alpha + \beta$.

(iii) Let $r_1 + r_2 \in \alpha + \beta$. Since $\alpha$ has no largest element there is $r_1' \in \alpha$ with $r_1' > r_1$, and then $r_1' + r_2 \in \alpha + \beta$ exceeds $r_1 + r_2$. So $\alpha + \beta$ has no largest element.

Tags: ChallengeFrom the definitions

Exercise 6.3

1. Show $\alpha + 0^* = \alpha$ for all Dedekind cuts $\alpha$.
2. We claimed, without proof, that addition of Dedekind cuts satisfies A4: for each cut $\alpha$ there is a cut $-\alpha$ with $\alpha + (-\alpha) = 0^*$. How would you define $-\alpha$?

Solution

(b) $-\alpha = \{r \in \mathbb{Q} : -r - \delta \notin \alpha \text{ for some rational } \delta > 0\}$. Without the $\delta$ the set could have a largest element, for instance when $\alpha = s^*$.

Tags: ChallengeFrom the definitions

Exercise 6.4

Let $\alpha$ and $\beta$ be Dedekind cuts and define the “product” $\alpha \cdot \beta = \{r_1 r_2 : r_1 \in \alpha \text{ and } r_2 \in \beta\}$.

1. Calculate some “products” of Dedekind cuts using the cuts $0^*$, $1^*$ and $(-1)^*$.
2. Discuss why this definition of “product” is totally unsatisfactory for defining multiplication in $\mathbb{R}$.

Solution

Every cut contains arbitrarily negative rationals, so products of two negative elements give arbitrarily large positive rationals. For example, $0^* \cdot 0^* = \{r_1 r_2 : r_1, r_2 < 0\} = \{r \in \mathbb{Q} : r > 0\}$, which is not even a cut, since it fails (ii). Likewise $1^* \cdot 1^*$ contains every rational: large positive ones come from two very negative factors and negative ones from mixed signs. So it is all of $\mathbb{Q}$ and fails (i).

Tags: ChallengeFrom the definitionsExamples & counterexamples

Exercise 6.5

1. Show $\{r \in \mathbb{Q} : r^3 < 2\}$ is a Dedekind cut, but $\{r \in \mathbb{Q} : r^2 < 2\}$ is not.
2. Does the Dedekind cut $\{r \in \mathbb{Q} : r^3 < 2\}$ correspond to a rational number in $\mathbb{R}$?
3. Show $0^* \cup \{r \in \mathbb{Q} : r \ge 0 \text{ and } r^2 < 2\}$ is a Dedekind cut. Does it correspond to a rational number in $\mathbb{R}$?

Solution

(a) $\{r^2 < 2\}$ fails (ii): $1$ is in it but $-2 < 1$ is not. (b) No: it corresponds to $\sqrt[3]2$, which is irrational by Theorem 2.2. (c) It is the cut for $\sqrt2$, which is not rational.

Tags: ChallengeFrom the definitionsExamples & counterexamples