TITLE: Discrete Mathematics: $x\leq y+\epsilon \implies x\leq y$
QUESTION [5 upvotes]: Let $x$ and $y$ be real numbers. Prove that if $x\leq y + \epsilon$ for every positive real number $\epsilon$, then $x\leq y$.

I would like a hint as to how to prove this. Thank you. Pictorial proof would be nice too.

So this is how I word-smithed the answer given by the author of my accepted answer:
We will proceed by showing the contrapositve. We are given that
$$x\leq y+\epsilon \implies x\leq y,$$
so the contrapositive is resolved as
$$x > y \implies x>y+\epsilon,$$
or by substituting $x-y = \omega$ simply
$$\omega > 0 \implies \omega > \epsilon,$$
but if $\omega >0$, then $\epsilon = \frac{\omega}{2} > 0$; however, $\omega \leq \frac{\omega}{2}$ is false. Therefore, we can assert that for all $\epsilon > 0$
$$\omega\leq\epsilon \implies \omega \leq 0;$$
quod erat demonstrandum.
Here is my pictorial representation:

REPLY [2 votes]: Consider $x,y$ be universally quantified, then
$$\forall \epsilon\ (x\leq y+\epsilon) \implies x\leq y$$
$$\neg (x\leq y)\implies \neg\forall \epsilon\ (x\leq y+\epsilon)$$
$$\neg (x\leq y)\implies \exists \epsilon\ \neg(x\leq y+\epsilon)$$
$$x>y\implies \exists \epsilon\ (x>y+\epsilon)$$
For $x>y$, we know $x>\tfrac{x+y}{2}=y+\tfrac{x-y}{2}$ and $\tfrac{x-y}{2}>0$, so it fits the bill.