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.

• Hint: Rearrange the inequality $x-\epsilon>y+\epsilon$ to give a bound for $\epsilon$ Jul 28, 2021 at 15:13
• Note that if $x>y$, then $x-y$ is a positive real. Jul 28, 2021 at 15:14
• If you want to continue with your approach, take the limit for $\epsilon\to0$, which gives $x\leq y$, a contradiction. Jul 28, 2021 at 15:14

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}

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).

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.

-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.