# Discrete Mathematics: $x\leq y+\epsilon \implies x\leq y$

Let $x$ and $y$ be real numbers. Prove that if $x\leq y + \epsilon$ for every positive real number $\epsilon$, then $x\leq y$.

I would like a hint as to how to prove this. Thank you. Pictorial proof would be nice too.

So this is how I word-smithed the answer given by the author of my accepted answer:

We will proceed by showing the contrapositve. We are given that $$x\leq y+\epsilon \implies x\leq y,$$ so the contrapositive is resolved as $$x > y \implies x>y+\epsilon,$$ or by substituting $x-y = \omega$ simply $$\omega > 0 \implies \omega > \epsilon,$$ but if $\omega >0$, then $\epsilon = \frac{\omega}{2} > 0$; however, $\omega \leq \frac{\omega}{2}$ is false. Therefore, we can assert that for all $\epsilon > 0$ $$\omega\leq\epsilon \implies \omega \leq 0;$$ quod erat demonstrandum.

Here is my pictorial representation: • Suppose it weren't so, i.e. $y < x$. Look for an $\epsilon$ violating the constraints. Jul 7, 2013 at 22:55
– Pedro
Jul 7, 2013 at 22:57
• It is from a discrete mathematics textbook. Should a student studying discrete mathematics also have this problem in their discrete mathematics textbook, then they will find this of interest. Jul 7, 2013 at 22:57
• @Trancot I have seen textbooks for linear give a definition and some properties of injective functions, should a question of this kind have a title "linear algebra:If we have an injective function bla bla"? Jul 7, 2013 at 23:01

The claim is equivalent to showing that if $$\omega\leq \epsilon$$ for each $$\epsilon >0$$, then $$\omega\leq 0$$.

But, if $$\omega>0$$, then $$\epsilon=\frac \omega 2 >0$$ and $$\omega \leq \frac\omega 2$$ does not hold. Having proven the contrapositive, we can assert that $$(\forall\epsilon >0\;:\;\omega\leq\epsilon )\implies \omega \leq 0$$

Now let $$\omega =x-y$$.

Pictorially If for any $$\color{green}{\epsilon >0}$$ we choose, $$\omega$$ is to the left of $$\color{green}{\epsilon}$$, it must be the case $$\omega$$ is to the left of the green bar, that is, on the red side $$\color{red}{\omega <0}$$ (strictly negative numbers) or that it is on the breaking point, that is $$\color{orange}{\omega=0}$$. ADD The contrapositive of the assertion is $$\omega >0\implies (\exists \epsilon >0:\omega\not\leq \epsilon)$$ or, which is the same, $$\omega >0\implies (\exists \epsilon >0:\omega> \epsilon)$$

We proved the contrapositive with $$\epsilon =\omega /2$$.

• Should not $\omega \leq 0$ be $\omega < 0$? Actually, no I guess not. Jul 7, 2013 at 23:08
• The claim is true if $\omega =0$. See how the proof of the contrapositive breaks down in such case, @Trancot.
– Pedro
Jul 7, 2013 at 23:09
• See my most recent edit. Jul 7, 2013 at 23:26
• Fancy answer...+1. Jul 7, 2013 at 23:34
– Pedro
Jul 7, 2013 at 23:36

Consider $$x,y$$ be universally quantified, then

$$\forall \epsilon\ (x\leq y+\epsilon) \implies x\leq y$$ $$\neg (x\leq y)\implies \neg\forall \epsilon\ (x\leq y+\epsilon)$$ $$\neg (x\leq y)\implies \exists \epsilon\ \neg(x\leq y+\epsilon)$$ $$x>y\implies \exists \epsilon\ (x>y+\epsilon)$$

For $$x>y$$, we know $$x>\tfrac{x+y}{2}=y+\tfrac{x-y}{2}$$ and $$\tfrac{x-y}{2}>0$$, so it fits the bill.

• @Angelo I edited it back out of principle. Can you please not randomly go and edit my answer without a warning? Yes, OP speaks of the situation the quantifier over $\epsilon$ goes over just positive numbers. May 21 at 14:58
• Really I do not understand what you mean. Anyway, your answer is correct and I like it, in fact I upvoted it. May 21 at 17:45
• @Angelo I don't think it's a common habit on this site to edit other people's words, unless it's a buggy or unreadable opening question May 22 at 16:32
• @Angelo Grumpy un-vote now? Well it's not a Wiki with textbook information, it's people posting text in their own name. May 22 at 17:03
• I just wrote that $\varepsilon>0\;$ and $\;\varepsilon=\frac{x-y}2$. I apologise if it bothered you so much. I did not have the slightest intention of annoying you. Sorry ! May 22 at 17:27