Real Plus Epsilon

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 $$a \le b$$.

Proof
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.