Real Multiplication is Closed

Theorem
The operation of multiplication on the set of real numbers $$\R$$ is closed.

That is:
 * $$x \in \R, y \in \R \implies x y \in \R$$

Proof
From the definition, the real numbers are the set of all equivalence classes $$\left[\!\left[{\left \langle {x_n} \right \rangle}\right]\!\right]$$ of Cauchy sequences of rational numbers.

Let $$x = \left[\!\left[{\left \langle {x_n} \right \rangle}\right]\!\right], y = \left[\!\left[{\left \langle {y_n} \right \rangle}\right]\!\right]$$, where $$\left[\!\left[{\left \langle {x_n} \right \rangle}\right]\!\right]$$ and $$\left[\!\left[{\left \langle {y_n} \right \rangle}\right]\!\right]$$ are such equivalence classes.

From the definition of real multiplication, $$x \times y$$ is defined as $$\left[\!\left[{\left \langle {x_n} \right \rangle}\right]\!\right] \times \left[\!\left[{\left \langle {y_n} \right \rangle}\right]\!\right] = \left[\!\left[{\left \langle {x_n \times y_n} \right \rangle}\right]\!\right]$$.

We have that $$\forall i \in \N: x_i \in \Q, y_i \in \Q$$, therefore $$x_i \times y_i \in \Q$$.

So it follows that $$\left[\!\left[{\left \langle {x_n \times y_n} \right \rangle}\right]\!\right] \in \R$$.