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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.

K. Gibson
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  • $\epsilon_0$ is a particular value you can plug in the statement about $\epsilon$ to get a contradiction – J. W. Tanner Jun 20 '19 at 21:51
  • $y+\varepsilon_0=x$ because of the stipulation $\varepsilon_0=x-y$ – J. W. Tanner Jun 20 '19 at 22:36
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    When you said "Notice that $y+\varepsilon=x$", did you mean $y+\varepsilon_\color{red}0=x$? – J. W. Tanner Jun 20 '19 at 22:37
  • Why did you write "Suppose to the contrary that $x\lt y+\epsilon$ but $x\gt y$" instead of "Suppose to the contrary that $x\lt y+\epsilon$ for all $\epsilon\gt0$ but $x\gt y$"? Was that a typo? – bof Jun 20 '19 at 23:32
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    Mathematicians yous indexes just to keep multiple variables apart. They don't have anything to do with initial values. If $x > y$ we know $x - y > 0$. We assign the variable $\epsilon_0 = x-y$. Now we have a statement that is about all positive numbers. And $\epsilon_0$ is a positive number. So it is true for $\epsilon_0$. In making our statement we use $\epsilon$ to be about any positive number. $\epsilon_0=x-y$ is a specific positive number. We are using it to be a value of $\epsilon$ but the variable $\epsilon$ still refers to any positive value. – fleablood Jun 21 '19 at 05:57
  • Does this answer your question? Intuition: If $a\leq b+\epsilon$ for all $\epsilon>0$ then $a\leq b$? – user1110341 Oct 01 '23 at 06:56

2 Answers2

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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 $x \ge y + \epsilon$.

Lee Mosher
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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.

========

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 $\epsilon > 0$.

++++++++

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.

fleablood
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