Real Plus Epsilon

[edit] Theorem

Let $a, b \in \R$ , such that:

$\forall \epsilon \in \R^*_+: a < b + \epsilon$

where $\R^*_+$ is the set of strictly positive reals, i.e. $\epsilon > 0$ .

Then $a \le b$ .

Suppose $a > b$ . Then $a - b > 0$ .

But $\forall \epsilon > 0: a < b + \epsilon$ (by hypothesis).

Let $\epsilon = a - b$ . Then $a < b + \left({a - b}\right) \implies a < a$ .

The result follows by Proof by Contradiction.

$\blacksquare$