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 real numbers.

That is:
 * $\epsilon > 0$

Then:
 * $a \le b$

Proof
$a > b$.

Then:
 * $a - b > 0$

By hypothesis, we have:
 * $\forall \epsilon > 0: a < b + \epsilon$

Let $\epsilon = a - b$.

Then:
 * $a < b + \paren {a - b} \implies a < a$

The result follows by Proof by Contradiction.