User:Abcxyz/Sandbox/Real Numbers

Axiomatic Definition
The real numbers are a totally ordered field $\left({\R, +, \times, \le}\right)$ such that $\left({\R, \le}\right)$ is Dedekind complete.

The axiom that $\left({\R, \le}\right)$ is Dedekind complete is known as the (Dedekind) completeness axiom.

Construction from Cauchy Sequences
Let $X$ be the set of all rational Cauchy sequences.

Let $\sim$ be the relation on $X$ defined as:
 * $\left\langle{x_n}\right\rangle \sim \left\langle{y_n}\right\rangle \iff \forall \epsilon \in \Q_{>0}: \exists N \in \N: \forall n \in \N: n > N \implies \left\vert{x_n - y_n}\right\vert < \epsilon$

By Equivalence Relation on Cauchy Sequences, $\sim$ is an equivalence relation on $X$.

The set $\R$ of real numbers is defined as the quotient set $X/{\sim}$.

Construction from Dedekind Cuts
The set $\R$ of real numbers is defined as the set of all Dedekind cuts of the totally ordered set $\left({\Q, \le}\right)$ of rational numbers.

Real Numbers as Dedekind Completion of Rational Numbers
Let $\left({\left({\R, \le}\right), \phi}\right)$ be the Dedekind completion of the ordered set $\left({\Q, \le}\right)$ of rational numbers.

Then $\left({\R, \le}\right)$ is the ordered set of real numbers.

Addition/Axiomatic Definition
Let $\left({\R, +, \times, \le}\right)$ denote the totally ordered field of real numbers.

The binary operation $+$ is called addition.

Addition/Construction from Cauchy Sequences
Let $\R$ denote the set of real numbers.

Addition, denoted $+$, is the binary operation on $\R$ defined as:
 * $\left[{\!\left[{\left\langle{x_n}\right\rangle}\right]\!}\right] + \left[{\!\left[{\left\langle{y_n}\right\rangle}\right]\!}\right] = \left[{\!\left[{\left\langle{x_n + y_n}\right\rangle}\right]\!}\right]$

Addition/Construction from Dedekind Cuts
Let $\R$ denote the set of real numbers.

Addition, denoted $+$, is the binary operation on $\R$ defined as:
 * $\alpha + \beta = \left\{{p + q: p \in \alpha, \, q \in \beta}\right\}$

Addition/Real Numbers as Dedekind Completion of Rational Numbers
Let $\left({\R, \le}\right)$ denote the ordered set of real numbers.

Let $\left({\left({\R, \le}\right), \phi}\right)$ be the Dedekind completion of the ordered set $\left({\Q, \le}\right)$ of rational numbers.

Let $\left({\Q, +}\right)$ denote the additive group of rational numbers.

We have that $\left({\Q, +, \le}\right)$ is an Archimedean ordered group.

By this theorem, there exists a unique binary operation $+$ on $\R$ such that:
 * $({1}): \quad \left({\R, +, \le}\right)$ is an ordered group
 * $({2}): \quad \phi$ is a group homomorphism from $\left({\Q, +}\right)$ to $\left({\R, +}\right)$

This binary operation $+$ is called addition.

Theorem
Let $\R$ denote the set of real numbers.

Let $+$ denote addition on $\R$.

Then $\R$ is closed under $+$.

Proof 1
Let $\left({\R, +, \times, \le}\right)$ denote the totally ordered field of real numbers.

By the field axioms, $\R$ is closed under $+$.

Proof 2
Let $\R$ denote the set of real numbers.

Let $+$ denote addition on $\R$.

Let $x, y \in \R$, $x = \left[{\!\left[{\left\langle{x_n}\right\rangle}\right]\!}\right]$, $y = \left[{\!\left[{\left\langle{y_n}\right\rangle}\right]\!}\right]$.

First, we have that $x + y$ is well-defined.

Next, from Rational Addition is Closed, we have that:
 * $\forall n \in \N: x_n + y_n \in \Q$

It remains to show that $\left\langle{x_n + y_n}\right\rangle$ is a Cauchy sequence.

Proof 3
Let $\R$ denote the set of real numbers.

Let $+$ denote addition on $\R$.

Let $\alpha, \beta \in \R$.

It is to be shown that $\alpha + \beta$ is a Dedekind cut of $\left({\Q, \le}\right)$.

We have the result Rational Addition is Closed.

It follows that $\alpha + \beta \subseteq \Q$.

By the definition of a Dedekind cut, $\alpha$ and $\beta$ are non-empty, so $\alpha + \beta$ is non-empty.

By the definition of a Dedekind cut, we can choose $r \in \Q \setminus \alpha$ and $s \in \Q \setminus \beta$.

It follows that $r + s \in \Q \setminus \left({\alpha + \beta}\right)$.

Suppose that $p \in \alpha$, $q \in \beta$.

By the definition of a Dedekind cut, we can choose $r \in \beta$ such that $q < r$.

It follows that $p + r \in \alpha + \beta$ and $p + q < p + r$.

Suppose that $s \in \Q$, $s < p + q$.

Then $-p + s < q$; therefore, by the definition of a Dedekind cut, we have that $-p + s \in \beta$.

Hence, $s = p + \left({-p + s}\right) \in \alpha + \beta$.

Proof 4
Let $\R$ denote the set of real numbers.

Let $+$ denote addition on $\R$.

By definition, $\left({\R, +}\right)$ is a group.

By the group axioms, $\R$ is closed under $+$.