# How to restrict this inequality only to where it is true for all x < y?

While doing some research in game theory modeling, I've come across the following inequality:

$$\frac{x - \epsilon(1+a)}{a} \leq y \leq \frac{x + \epsilon(1+a)}{a}$$

Where $x$ and $y$ are some variables and $a \in (0,1], \epsilon \in (0,1]$ are some known constants. In addition, $0 \leq x \leq y \leq 1$.

I'd like to graph the region where the above inequality is true for all values of $y$. That is, the only pairs of $x,y$ in the solution set should be where the inequality holds not only for that pair $x,y$, but also all other pairs $x < y, y$ (graphically, no point should be in the solution set unless all points to its left up to the y-axis are also in the set and all points to its right up to y = x are also in the set).

For example: this plot shows the intersection of the inequality above (where $a = .02, \epsilon = .05$) and $0 \leq x \leq y \leq 1$. The sloped lower bound is the line $y = x$. I would like to exclude all of the points with y-value above the intersection of the lower bound with the right bound, since those points have y-values for which not every point ($x \leq y, y$) are in the solution set. That is, there are points to the left of $y=x$ but to the right of them that are not in the set.

How can I express this mathematically? Thanks!

Mathematically, the much better way to write this is $$P(y): (\forall x\in [0,y))\left(\frac{x - \epsilon(1+a)}{a} \leq y \leq \frac{x + \epsilon(1+a)}{a}\right)$$