User:Abcxyz/Sandbox/Real Numbers/Real Multiplication is Closed

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

Let $\times$ denote multiplication on $\R$.

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

Proof 1
Let $\left({\R, +, \times, \le}\right)$ denote the real numbers, as axiomatically defined as a Dedekind complete totally ordered field.

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

Proof 2
Let $\R$ denote the set of real numbers, as constructed from Cauchy sequences.

Let $\times$ denote multiplication on $\R$.

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

From Rational Multiplication is Closed, we have that:
 * $\forall n \in \N: x_n \times y_n \in \Q$

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

Let $\epsilon \in \Q_{>0}$ be a strictly positive rational number.

Since a Cauchy sequence is bounded, there exist $A, B \in \Q$ such that, for all $n \in \N$:
 * $\left\vert{x_n}\right\vert \le A$


 * $\left\vert{y_n}\right\vert \le B$

There exist $N_1, N_2 \in \N$ such that, for all $m, n \in \N$:
 * $m, n \ge N_1 \implies \left\vert{x_n - x_m}\right\vert < \dfrac \epsilon {A + B + 1}$


 * $m, n \ge N_2 \implies \left\vert{y_n - y_m}\right\vert < \dfrac \epsilon {A + B + 1}$

Let $N = \max {\{{N_1, N_2}\}} \in \N$.

Then, if $m, n \in \N$, $m, n \ge N$:

Proof 3
Let $\R$ denote the set of real numbers, as constructed from Dedekind cuts.

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

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

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