Discrete Mathematics: $x\leq y+\epsilon \implies x\leq y$

Solution 1:

The claim is equivalent to showing that if $\omega\leq \epsilon$ for each $\epsilon >0$, then $\omega\leq 0$.

But, if $\omega>0$, then $\epsilon=\frac \omega 2 >0$ and $\omega \leq \frac\omega 2$ does not hold. Having proven the contrapositive, we can assert that $$(\forall\epsilon >0\;;\;\omega\leq\epsilon )\implies \omega \leq 0$$

Now let $\omega =x-y$.

Pictorially If for any $\color{green}{\epsilon >0}$ we choose, $\omega$ is to the left of $\color{green}{\epsilon}$, it must be the case $\omega$ is to the left of the green bar, that is, on the red side $\color{red}{\omega <0}$ (strictly negative numbers) or that it is on the breaking point, that is $\color{orange}{\omega=0}$.

ADD The contrapositive of the assertion is $$\omega >0\implies (\exists \epsilon >0:\omega\not\leq \epsilon)$$ or, which is the same, $$\omega >0\implies (\exists \epsilon >0:\omega> \epsilon)$$

We proved the contrapositive with $\epsilon =\omega /2$.

Solution 2:

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.