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 $\eqclass {\sequence {x_n} } {}$ of Cauchy sequences of rational numbers.

Let $x = \eqclass {\sequence {x_n} } {}, y = \eqclass {\sequence {y_n} } {}$, where $\eqclass {\sequence {x_n} } {}$ and $\eqclass {\sequence {y_n} } {}$ are such equivalence classes.

From the definition of real multiplication, $x \times y$ is defined as $\eqclass {\sequence {x_n} } {} \eqclass {\sequence {y_n} } {} = \eqclass {\sequence {x_n \times y_n} } {}$.

We need to show that:


 * $\sequence {x_n}, \sequence {x'_n} \in \eqclass {\sequence {x_n} } {}, \sequence {y_n}, \sequence {y'_n} \in \eqclass {\sequence {y_n} } {} \implies \sequence {x_n \times y_n} = \sequence {x'_n \times y'_n}$

That is:
 * $\forall \epsilon > 0: \exists N: \forall i, j > N: \size {\paren {x_i \times y_i} - \paren {x'_j \times y'_j} } < \epsilon$

As $\eqclass {\sequence {x_n} } {}$ and $\eqclass {\sequence {y_n} } {}$ are Cauchy sequences, from Cauchy Sequence is Bounded they are bounded.

Let $B_x = 2 \map \sup {\sequence {x_n} }$ and $B_y = 2 \map \sup {\sequence {y_n} }$.

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

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

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

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

So:

Hence the result.