Can someone clarify the therorem $x \lt y + \epsilon$ iff $x \leq y$ 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.
 A: 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 $x \ge y + \epsilon$.
A: They way you transcribed this proof it has problems.
First of all it is not a proof of: $x < y+ \epsilon$ for all $\epsilon > 0$ if and only if $x \le y$.  
It is a proof only of: If $x < y+\epsilon$ for all $\epsilon > 0$ then $x \le y$.  
Proof only goes in one direction.  You will need to do another proof for:  If $x \le y$ then $x < y+\epsilon$ for all $\epsilon > 0$.
Now to the proof:

Suppose to the contrary

This means suppose $x< y + \epsilon$ for all $\epsilon > 0 \not \implies x\le y$.
This means suppose there are an $x$ and $y$ so that $x < y+\epsilon$ for all $\epsilon > 0$--
(In the proof as you wrote it you left at that this needs to be true for all $\epsilon$ each and every one, not just one in particular.)
But somehow for this $x$ and $y$, it happens that $x \not \le y$.  That would mean that $x > y$ even though $x < y + \epsilon$ for all possible $\epsilon > 0$.
Since $x > y$ that mean $x-y > 0$.  Let's give $x-y$ a name.  I'll call it $BIPPYBOO$.
So $BIPPYBOO = x-y > 0$.  
The proof you quoted called it $\epsilon_0$.  But I'm calling it $BIPPYBOO$.

Notice that y+ϵ=x

That was simply a typo.  They meant  $y + BIPPYBOO = x$.
Which is obvious.  If $BIPPYBOO =x-y$ then $y + BIPPYBOO = x$.

This contradicts the hypothesis ϵ=ϵ0.

This left out a very important word.

This contradicts the hypothesis (that $x < y + \epsilon$ for all $\epsilon > 0$) if we let $\epsilon = BIPPYBOO$.

Our hypothesis is that $x < y + \epsilon$ for any $\epsilon > 0$.  But $BIPPYBOO > 0$.  So this should be true if we set $epsilon = BIPPYBOO = x-y$.  We should have $x < y + BIPPYBOO$. 
In other words we should have $x < y + (x-y)$ because $x-y > 0$ and for ANY $\epsilon > 0$ we have $x < y + \epsilon$. 
But obvious $x < y+(x-y) = x$ is not true.
So our assumption, $x < y+\epsilon$ for all $\epsilon > 0$ means that $x > y$ is impossible  (because that would mean $x  < x +(y-x) = x$ which is just plain nutty).
So our assumption, $x < y+\epsilon$ for all $\epsilon > 0$ implies $x \le y$.
That's what the proof says.
And that's half the proof.
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Now we have to prove if $x \le y$ then for any $\epsilon > 0$ we will always have $x < y+\epsilon$.
......
Which is easy.
If $\epsilon > 0$ then $y < y + \epsilon$.  But $x \le y$ and $y < y + \epsilon$ so by transitivity $x \le y + \epsilon$ and that is true for any $\epsilon > 0$.
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So if $x < y$ for any $\epsilon > 0$ then we know $x \le y$.
And if $x \le y$ the we know that $x < y+ \epsilon$ for any $\epsilon > 0$.
So that proves our theorem.
