Examining the theorem, for all $\epsilon \gt 0$, if $x \le y + \epsilon$, then $x \le y$

Question

For this theorem, I understand the case in which x $\lt y + \epsilon$. From there it follows visually that x could either be equivalent to y itself, or x could also be less than y; in either case, the conclusion holds. However, is it always the case that x $\lt y + \epsilon$? Is it even possible for $x = y + \epsilon$ for all epsilon? Because if this were true, wouldn't our conclusion that $x \le y$ be false? Wouldn't $x \gt y$?

Answer 1

I will show epsilon by E You can see it like this way, x-y<=E. It means, x-y should always be less than or equal to any positive element(E) of field. If x=y+E for all E>0 then x is variable, not fixed but we deal with fixed values of x and y which can follow the given inequality for variable E. If x is considered fixed in x=y+E, then order properties of field is ruled out! (x=y+1 and x=y+2 for fixed x and y is not possible). So we have to find the possibilities of inequalities between x and y. So possibilities are x<=y.