Proof that if $X>Y$ then there exists some $\epsilon>0$ such that $x-\epsilon>y+\epsilon$ I am trying to prove that if $x>y$, then there exists some $\epsilon>0$ such that $x-\epsilon>y+\epsilon$. So far I have:
Suppose, for a contradiction, that no such $\epsilon$ exists. That is, $\forall \epsilon>0$, suppose that $x-\epsilon \leq y+\epsilon$.
Beyond this, I am not at all sure how to proceed. Must one make reference to the fact that limits preserve weak inequalities?
Thank you.
 A: You could use proof by contradiction, but a direct proof is preferable. Draw a number line and mark $x$ and $y$ at points such that $x>y$. From this, it should be geometrically obvious that
$$
x-\frac{x-y}{3}>y+\frac{x-y}{3} \, ,
$$
and so you can take $\varepsilon=\frac{x-y}{3}$. Of course, something being geometrically obvious is not a proof, but this steers us in the right direction. It is then simple to come up with an algebraic proof:
\begin{align}
x>y &\implies 2x+y>2y+x \\[5pt]
&\implies3x-(x-y)>3y+(x-y)\\[5pt]
&\implies x-\frac{x-y}{3}>y+\frac{x-y}{3} \, .
\end{align}
A: If $x > y$ then by subtracting $y$ from both sides we have $x - y$ > 0. Then we divide both sides by two to get $\left(x - y\right) / 2 > 0$.
Choose any $\epsilon$ such that $0 < \epsilon < \left(x - y\right) / 2$. Then multiply through by 2 to get $0 < 2 \epsilon < x - y$. Then add through by $y - \epsilon$ to get $y - \epsilon < y + \epsilon < x - \epsilon$; QED.
A: Try to develop some intuition for this problem:
Picture (or draw) a number line with $x$ to the right of $y$.
You want to add a little bit to $y$ (to get $y+\epsilon$) and subtract the same little bit from $x$ (to get $x-\epsilon$) and have  $y+\epsilon$ still be to the left of $x-\epsilon$.
With this picture, we see that we can take $\epsilon$ to be any positive value less than $\frac{x-y}{2}$. To be particular, we might, for instance, take $\epsilon = \frac{x-y}{3}$.
Then try to work out the algebra to show that this actually works (so that you have a complete proof).
A: -Here's how I would do it, a direct proof, not indirect- since x> y, x- y> 0.  Let delta= x- y and epsilon any number less than delta/2. Then y+ epsilon< y+ delta/2= y+ (x- y)/2= y+ x/2- y/2= x/2+ y/2 while x- epsilon> x- delta/2= x- (x- y)/2= x- x/2+ y/2= x/2+ y/2.
So we have y+ epsilon< x/2+ y/2< x- epsilon.
