Theorem:
$ x \lt y + \epsilon$ for all $\epsilon \gt 0$ if and only if $x \leq y$
Suppose to the contrary that $x \lt y + \epsilon$ but $x \gt y$
Set $\epsilon_0 = x – y \gt 0$
Notice that $y + \epsilon = x$ Hence by the trichotomy property This contradicts the hypothesis $\epsilon = \epsilon_0$. Thus $x \leq y$
I would like some clarification. I am wondering why this is proven by contradiction and what is $ \epsilon_0$ I know that it stands for the initial value of $ \epsilon$. It appears that the initial value is set to $0$ How does $y + \epsilon_0 = x$? I am guessing that $y + \epsilon$ cannot be be greater than x because it is like saying that x can be greater than itself which is false.
Best Answer
This isn't about "initial values". This is about logic and about methods of proof.
In this part of the proof, you are trying to prove that the statement "for all $\epsilon > 0$, $x < y+\epsilon$" implies the statement "$x \le y$". There are of course many methods for proving that "P implies Q", and sometimes one method works better than another.
The method being used here is a proof by contrapositive: assume that "$x \le y$" is false, and so it's negation "$x > y$" is true; then use that to prove that the statement "for all $\epsilon > 0$, $x < y+\epsilon$" is false, i.e. to prove that its negation "there exists $\epsilon > 0$ such that $x \ge y + \epsilon$" is true.
To prove an existence statement, the method is what I refer to as The Hunt. You must use what you know to hunt down an appropriate value of $\epsilon > 0$, to find the right value. Finding that value requires some mathematical imagination. And once you've found the right value of $\epsilon$, you use it to prove the inequality $x \ge y + \epsilon$.
So, what value of $\epsilon$ shall we use? What do we know about $x$ and $y$? Well.... we are assuming that $x > y$. It follows that $x - y > 0$... Could that be it?
Yes! Eureka! We've found it!
Let $\epsilon = x - y$.
From this we conclude that $x = y + \epsilon$, and therefore $x \ge y + \epsilon$.