Modifying a Textbook Theorem in Real Analysis In my real analysis book, there is a theorem which states that if we let $x,y \in \mathbb{R}$ such that $x \leq y + \epsilon$ for all $\epsilon > 0$, then $x \leq y$. 
My question is, if we instead have $k\cdot \epsilon$ where $k$ is a nonzero constant, will the inequality still be true? That is, if $x \leq y+k\cdot \epsilon$, then $x \leq y$? Or must we have $k\cdot \epsilon > 0$?
 A: Yes, if $k$ is assumed positive.
Consider fixed but arbitrary $x,y \in \mathbb{R}$ and $k>0$. Suppose
$$\mathop{\forall}_{\varepsilon>0} x \leq y+k\cdot\varepsilon$$
Then using the positivity of $k$, we deduce:
$$\mathop{\forall}_{\varepsilon>0} x \leq y+k\cdot(\varepsilon \cdot k^{-1})$$
So $$\mathop{\forall}_{\varepsilon>0} x \leq y+\varepsilon$$
Hence by the quoted theorem, we have $$x \leq y.$$
We conclude that:
$$\left(\mathop{\forall}_{\varepsilon>0} x \leq y+k \cdot \varepsilon\right) \rightarrow x \leq y$$
In summary:

Proposition.
$$\mathop{\forall}_{k>0}\,\,\mathop{\forall}_{x,y \in \mathbb{R}}\left[\left(\mathop{\forall}_{\varepsilon>0} x \leq y+k \cdot \varepsilon\right) \rightarrow x \leq y\right]$$

Remark. I think there's something wrong with present-day mathematical notation. Notice that its kind of unclear how positivity was used. A good notation would not suffer these kinds of problems.
A: $x \leq y+k\cdot \epsilon$ iff $\dfrac xk \leq \dfrac yk+\epsilon$ and so, by what you know, $\dfrac xk \leq \dfrac yk$. Now multiply by $k$.
We need to assume that $k>0$.
