Real Numbers

# The Set of Real Numbers $$\mathbb{R}$$

Theorem: $$\mathbb{R}$$ is an ordered field, with the least upper bound property, and $$\mathbb{R}$$ contains $$\mathbb{Q}$$ as a subfield.

## Construction of $$\mathbb{R}$$

Dedekind: A cut $$\alpha$$ is a subset of $$\mathbb{Q}$$ satisfying

1. (nontriviality) i.e. $$\alpha \neq \emptyset, \mathbb{Q}$$,
2. (closed downwards or closed to the left) if $$p \in \alpha$$, $$q \in \mathbb{Q}$$, and $$q < p$$, then $$q \in \alpha$$, and
3. (no largest member) if $$p \in \alpha$$ then $$p < r$$ for some $$r \in \alpha$$.

For example, the set $$A = \{x : x^2 < 2\}$$ is not a cut because it does not satisfy (2). The set $$\mathbb{Q}_{-}$$ is a cut. The set $$\beta = \{x \in \mathbb{Q} : x \leq 2\}$$ is not a cut because it does not satisfy (3).

Let $$\mathbb{R} := \{\alpha : \alpha \text{ is a cut}\}$$ ($$:=$$ means defined to be).

## Properties of $$\mathbb{R}$$

### Order on $$\mathbb{R}$$

$$\alpha < \beta$$ is defined as $$\alpha \subsetneq \beta$$, or $$\alpha$$ is a proper subset of $$\beta$$.

Proof of well-definedness. This definition is well-defined because it satisfies

• trichotomy: for all cuts $$\alpha$$ and $$\beta$$, either $$\alpha \subsetneq \beta$$, $$\alpha = \beta$$, or $$\beta \subsetneq \alpha$$, and thus either $$\alpha < \beta$$, $$\alpha = \beta$$, or $$\beta < \alpha$$.
• transitivity: let $$\alpha, \beta, \gamma$$ be cuts, $$\alpha < \beta$$, and $$\beta < \gamma$$. Since $$\alpha < \beta$$, $$\alpha \subsetneq \beta$$. Similarly, $$\beta \subsetneq \gamma$$. Thus, $$\alpha \subsetneq \gamma$$, due to transitivity of inclusion. Hence, $$\alpha < \gamma$$. QED.

### Operations on $$\mathbb{R}$$

Let $$\alpha + \beta := \{r + s : r \in \alpha, s \in \beta\}$$.

Proof of well-definedness. Let $$\alpha$$ and $$\beta$$ be cuts. $$\alpha + \beta$$ is a cut because

• it is nontrivial: since $$\alpha$$ is a cut, there exists $$r \in \alpha$$. Similarly, there exists $$s \in \beta$$. So, there exists $$r + s$$. Hence, $$\alpha + \beta \neq \emptyset$$. Since $$\alpha$$ is a cut, there exists $$p \in \mathbb{Q} \setminus \alpha$$ such that $$p > r$$ for all $$r \in \alpha$$. Similarly, there exists $$q \in \mathbb{Q} \setminus \beta$$ such that $$q > s$$ for all $$s \in \beta$$. Thus, $$p + q > r + s$$ for all $$r \in \alpha$$ and $$s \in \beta$$. Hence, $$\alpha + \beta \neq \mathbb{Q}$$.
• it is closed downwards: let $$p \in \alpha + \beta$$, and $$q < p$$. Since $$p \in \alpha + \beta$$, there exist $$r \in \alpha$$ and $$s \in \beta$$ such that $$p = r + s$$. Since $$q < p = r + s$$, $$q - s < r$$, so $$q - s \in \alpha$$. Then $$q = (q - s) + s$$, as desired.
• it has no largest member: Fix arbitrary $$p \in \alpha$$ and $$q \in \beta$$. Since $$\alpha$$ is a cut, there exists $$r \in \alpha$$ such that $$p < r$$. Similarly, there exists $$s \in \beta$$ such that $$q < s$$. Hence, $$p + q < r + s \in \alpha + \beta$$, as desired. QED.

Additive identity $$0^\star := \mathbb{Q}_{-}$$.

Proof of well-definedness.

$$0^\star$$ is a cut because

• it is nontrivial: $$0^\star \neq \emptyset$$ since there exist negative rationals. $$0^\star \neq \mathbb{Q}$$ since there exist positive rationals.
• it is closed downward: let $$p \in 0^\star$$ and $$p \in \mathbb{Q}$$ with $$q < p$$. Since $$p \in 0^\star$$, it is less than $$0$$. Hence, $$q < 0$$ and thus $$q \in 0^\star$$.
• it has no largest element: Let $$p \in 0^\star$$. Consider $$\frac{p}{2}$$. We have $$p < \frac{p}{2} < 0$$, as desired.

$$\alpha + 0^\star = \alpha$$ because

• $$\alpha + 0^\star \subset \alpha$$: let $$p \in \alpha + 0^\star$$. Hence, there exist $$r \in \alpha$$ and $$s \in \beta$$ such that $$p = r + s$$. We have $$p = r + s < r + 0 < r \in \alpha$$, as desired.
• $$\alpha \subset \alpha + 0^\star$$: let $$p \in \alpha$$. Since $$\alpha$$ is a cut, there exists $$q \in \alpha$$ such that $$q > p$$. Let $$r = p - q$$. $$r \in 0^\star$$ since $$r < 0$$. We have $$p = q + r$$, as desired. QED.

Additive inverse for $$\alpha$$ is $$\beta := \{p : \exists r > 0 \text { such that } -p - r \notin \alpha\}$$.

Definition of multiplication: if $$\alpha, \beta \in \mathbb{R}_{+}$$ ($$\alpha, \beta > 0^\star$$), then $$\alpha\beta := \{p : p < rs \text{ for some } r \in \alpha, s \in \beta, r > 0, s > 0\}$$.

Multiplicative identity $$1^\star := \{q \in \mathbb{Q} : q < 1\}$$.

Given a set of cuts $$A$$, let $$\gamma = \bigcup \{x : x \in \alpha\}$$. $$\gamma$$ is a cut and is $$\sup A$$.

### $$\mathbb{R}$$ is an Extension of $$\mathbb{Q}$$

$$\mathbb{R}$$ contains $$\mathbb{Q}$$ as a subfield. Associate $$q \in \mathbb{Q}$$ with the cut $$q^\star = \{r \in \mathbb{Q} : r < q\}$$.

Formally, $$f : \mathbb{Q} \rightarrow \mathbb{R}$$ is $$f(q \in \mathbb{Q}) = q^\star$$, where $$q^\star = \{r \in \mathbb{Q} : r < q\}$$.

Then $$\mathbb{Q}' = \{q^\star : q \in \mathbb{Q}\}$$ is a subfield of $$\mathbb{R}$$.

Notice “$$\sqrt{2}$$” sits in $$\mathbb{R}$$ as the cut $$\gamma = \{q : q^2 < 2 \text{ or } q < 0\}$$.

### $$\mathbb{R}$$ has the Least Upper Bound Property

Let $$A$$ contain cuts with upper bound $$\beta$$, and $$\gamma = \bigcup\{\alpha : \alpha \in A\}$$, a subset of $$\mathbb{Q}$$.

Check that $$\gamma$$ is a cut and $$\gamma = \sup A$$.

Proof. $$\gamma$$ is a cut because

• it is nontrivial since it contains cuts that are bounded above,
• it is closed downwards since it is a union of cuts which are closed downwards, so for all $$q \in \gamma$$, there exists $$\alpha \in A$$ such that $$\gamma \in \alpha$$ and for all $$r < q$$, $$r \in \alpha$$, hence $$r \in \gamma$$, and
• it has no largest member since for all $$q \in \gamma$$, there exists $$\alpha \in A$$ such that $$\gamma \in \alpha$$ but $$\alpha$$ has no largest member so there exists $$r \in \alpha$$ such that \$q < r hence $$r \in \gamma$$.

$$\gamma$$ is an upper bound because $$\gamma$$ contains all $$\alpha \in A$$ and order on cuts is defined by inclusion.

$$\gamma$$ is the least upper bound because for all $$\delta < \gamma$$, there exists $$x \in \gamma$$ such that $$x \notin \delta$$, so there exists $$\alpha \in A$$ such that $$x \in \alpha$$, hence $$\delta$$ is not an upper bound for $$\alpha$$ and thus $$\delta$$ is not an upper bound for $$A$$. QED.

$$\mathbb{R}$$ is the only ordered field with the least upper bound property.

Consequence: the length “$$\sqrt{2}$$” can be thought of as $$\sup\{1, 1.4, 1.41, 1.414, 1.4142, \dots\}$$, whose shorthand is $$1.4142135\dots$$.

More generally, $$a^{\frac{1}{n}} := \sup\{r \in \mathbb{Q} : r^n < a\}$$, i.e. roots exist.

### Consequences

Archimedean property: if $$x, y \in \mathbb{R}$$ and $$x > 0$$, then there exists $$n \in \mathbb{N}$$ such that $$nx > y$$.

Proof (by contradiction). Consider $$A = \{nx : n \in \mathbb{N}\}$$. If $$A$$ is bounded by $$y$$ (i.e. $$nx < y$$ for all $$n$$) then $$A$$ has a least upper bound $$\alpha$$ (by the least upper bound property). Then $$\alpha - x$$ is not an upper bound for $$A$$. Hence, $$\alpha - x < mx$$ for some $$m \in \mathbb{N}$$. Then $$\alpha < (m + 1)x$$. Thus $$\alpha$$ is not an upper bound for $$A$$, a contradiction! Therefore, if $$x, y \in \mathbb{R}$$ and $$x > 0$$, then there exists $$n \in \mathbb{N}$$ such that $$nx > y$$. QED.

Corollary. If $$x > 0$$ then there exists $$n \in \mathbb{N}$$ such that $$\frac{1}{n} < x$$.

Theorem ($$\mathbb{Q}$$ is dense in $$\mathbb{R}$$). Between $$x, y \in \mathbb{R}$$, and $$x < y$$, there is a $$q \in \mathbb{Q}$$ such that $$x < q < y$$.

Proof. Choose $$n$$ such that $$\frac{1}{n} < y - x$$ (by Archimedean property). Consider multiples of $$\frac{1}{n}$$. These are unbounded (by the Archimedean property). Choose the first multiple such that $$\frac{m}{n} > x$$. If $$\frac{m}{n}$$ is not less than $$y$$ then $$\frac{m - 1}{n} < x$$ and $$\frac{m}{n} > y$$, which implies $$\frac{1}{n} > y - x$$, a constradiction. So, $$\frac{m}{n} < y$$. QED.