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:

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

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:

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

That is:

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

Let $\epsilon > 0$, such that:

• $\exists N_1: \forall i, j \ge N_1: \left\vert{x_i - x'_j}\right\vert < \epsilon / 2$;
• $\exists N_2: \forall i, j \ge N_2: \left\vert{y_i - y'_j}\right\vert < \epsilon / 2$.

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

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

But:

 $\displaystyle \epsilon$ $>$ $\displaystyle \left\vert{x_i - x'_j}\right\vert + \left\vert{y_i - y'_j}\right\vert$ $\displaystyle$ $>$ $\displaystyle \left\vert{\left({x_i - x'_j}\right) + \left({y_i - y'_j}\right)}\right\vert$ Triangle Inequality $\displaystyle$ $=$ $\displaystyle \left\vert{\left({x_i + y_i}\right) - \left({x'_j + y'_j}\right)}\right\vert$

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

Hence the result.

$\blacksquare$