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

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

It remains to show that $\left\langle{x_n y_n}\right\rangle$ 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 $\alpha, \beta \in \R$.

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

By the definition of real multiplication, it suffices to consider the case where $\alpha > 0^*$ and $\beta \ge 0^*$.

We can choose $p_0 \in \alpha$ such that $p_0 \notin 0^*$.

Since $\alpha$ is a Dedekind cut, we can choose $p \in \alpha$ such that $p > p_0$.

It follows that $0 \le p_0 < p$.

Since $\beta$ is non-empty, it follows that $\alpha \beta$ is non-empty.

Since $\alpha$ and $\beta$ are proper subsets of $\Q$, we can choose $p', q_0 \in \Q$ such that $p' \notin \alpha$ and $q_0 \notin \beta$.

Let $q' = \max {\left\{{q_0, 1}\right\}}$.

Using a proof by contradiction, we show that $p'q' \notin \alpha \beta$.

If $p'q' \in \alpha \beta$, then there exist $p$ and $q$ such that $p \in \alpha$, $p > 0$, $q \in \beta$, and $pq = p'q'$.

We have that $p < p'$, $q < q_0 \le q'$, and $q' \ge 1 > 0$.

It follows that $pq < pq' < p'q'$, a contradiction.

Hence, $\alpha \beta \ne \Q$.

Let $p \in \alpha$, $p > 0$, $q \in \beta$, $r \in \Q$, $r < pq$.

Then $p^{-1} r < q$, and so $p^{-1} r \in \beta$.

Hence, $r = p \left({p^{-1} r}\right) \in \alpha \beta$.

We can choose $s \in \beta$ such that $q < s$.

It follows that $pq < ps$ and $ps \in \alpha \beta$.

Hence, $\alpha \beta$ is a Dedekind cut of $\left({\Q, \le}\right)$.