Real Multiplication is Well-Defined

Theorem
The operation of multiplication on the set of real numbers $\R$ is well-defined.

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 need to show that:


 * $\left \langle {x_n} \right \rangle, \left \langle {x'_n} \right \rangle \in \left[\!\left[{\left \langle {x_n} \right \rangle}\right]\!\right], \left \langle {y_n} \right \rangle, \left \langle {y'_n} \right \rangle \in \left[\!\left[{\left \langle {y_n} \right \rangle}\right]\!\right] \implies \left \langle {x_n \times y_n} \right \rangle = \left \langle {x'_n \times y'_n} \right \rangle$.

That is:
 * $\forall \epsilon > 0: \exists N: \forall i, j > N: \left\vert{\left({x_i \times y_i}\right) - \left({x'_j \times y'_j}\right)}\right\vert < \epsilon$.

As $\left[\!\left[{\left \langle {x_n} \right \rangle}\right]\!\right]$ and $\left[\!\left[{\left \langle {y_n} \right \rangle}\right]\!\right]$ are Cauchy, they are bounded.

Let $B_x = 2 \sup \left({\left \langle {x_n} \right \rangle}\right)$ and $B_y = 2 \sup \left({\left \langle {y_n} \right \rangle}\right)$.

Let $B = \max \left\{{B_x, B_y}\right\}$.

Now let $\epsilon > 0$. Then:
 * $\exists N_1: \forall i, j > N_1: \left\vert{B}\right\vert \left\vert{x_i - x'_j}\right\vert < \epsilon / 2$;
 * $\exists N_2: \forall i, j > N_2: \left\vert{B}\right\vert \left\vert{y_i - y'_j}\right\vert < \epsilon / 2$.

Now let $N = \max \left\{{N_1, N_2}\right\}$.

Then we have $\forall i, j \ge N: \left\vert{B}\right\vert \left\vert{x_i - x'_j}\right\vert + \left\vert{B}\right\vert \left\vert{y_i - y'_j}\right\vert < \epsilon$.

So:

Hence the result.