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:


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

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

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

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

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

So:

$$ $$ $$ $$ $$ $$

Hence the result.