Real Plus Epsilon

Theorem
Let $a, b \in \R$, such that:
 * $\forall \epsilon \in \R_{>0}: a < b + \epsilon$

where $\R_{>0}$ 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.