Real Addition is Well-Defined

Theorem
The operation of addition 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:


 * $$\left[\!\left[{\left \langle {x_n} \right \rangle}\right]\!\right] \equiv \left[\!\left[{\left \langle {y_n} \right \rangle}\right]\!\right] \iff \forall \epsilon > 0: \exists n \in \N: \forall i, j > n: \left|{x_i - y_j}\right| < \epsilon$$.

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 addition, $$x + y$$ is defined as $$\left[\!\left[{\left \langle {x_n} \right \rangle}\right]\!\right] + \left[\!\left[{\left \langle {y_n} \right \rangle}\right]\!\right] = \left[\!\left[{\left \langle {x_n + y_n} \right \rangle}\right]\!\right]$$.

We need to show that:


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

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

Let $$\epsilon > 0$$, such that:
 * $$\exists N_1: \forall i, j \ge N_1: \left|{x_i - x'_j}\right| < \epsilon / 2$$;
 * $$\exists N_2: \forall i, j \ge N_2: \left|{y_i - y'_j}\right| < \epsilon / 2$$.

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

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

But:

$$ $$ $$

So $$\forall i, j \ge N: \left|{\left({x_i + y_i}\right) - \left({x'_j + y'_j}\right)}\right| < \epsilon$$.

Hence the result.