Proof Writing for Abbott Exercise

Question

I've just picked up Abbott's analysis book, and I am faced with the following problem. While it's quite simple, I am not very familiar with formal proof writing, which is what I am finding difficult. The problem is the following:

Prove or disprove: Two real numbers that satisfy $a\leq b$ if and only if $a<b+\epsilon$ for any $\epsilon>0$.

Here is what I have tried so far:

Let $a,b\in\mathbb{R}$ such that $a\leq b$. Suppose that there exists some number $\epsilon_0>0$ such that $a\geq b+\epsilon_0$. Then, as $a\leq b$, we have $b+\epsilon_0\leq b$ resulting in $\epsilon_0\leq 0$, which gives a contradiction.

Next, let $a,b\in\mathbb{R}$ such that $a<b+\epsilon$ for any $\epsilon>0$. Then we have that $a-b<\epsilon$ for every $\epsilon>0$, which admits the possiblity of $a=b$, thus giving a contradiction.

Here are some of the questions I have:

$1)$ How can I improve this proof overall, in terms of wording and structure?

$2)$ How can I do this without contradiction?

$3)$ I feel like the last part of my proof isn't really a solid statement, should it be fixed?

Answer 1

Direct proof for the forward part: $$a\leq b<b+\epsilon \text{ for all } \epsilon>0$$

Yes, your last part is incorrect.

Let $a<b+\epsilon$ for all $\epsilon >0.$ We need to show $a\leq b.$ If not, then we have $a>b.$ Let $\epsilon=a-b>0.$ Then $a=b+\epsilon.$ This contradicts that $a<b+\epsilon.$

To prove this directly:

Since, $a<b+\epsilon$ for all $\epsilon >0.$ Therefore, $$a<b+\frac 1n$$ for all $n \in \mathbb N.$ Thus, $$\lim_{n\to \infty}a \leq \lim_{n\to \infty}\left(b+\frac 1n\right).$$ Thus, $a\leq b.$

Answer 2

The first part is correct.

The second part is incorrect because $a-b<\epsilon$ for all $\epsilon>0$ doesn't necessarily imply that $a=b$.

A way to prove this part is using the contrapositive statement, that is: suppose you have an assertion to be proved that says $A\implies B$, then proving this assertion is equivalent to prove $\lnot B\implies\lnot A$.

Then the contrapositive of the assertion $$(\forall\epsilon>0: a<b+\epsilon)\implies a\le b$$ is $$a>b\implies(\exists\epsilon_0>0:a\ge b+\epsilon_0)$$

what is easy to show choosing $\epsilon_0:=a-b$, what is positive by assumption, then $$a\ge b+(a-b)=a$$

holds.

Answer 3

If $a\le b $ then $a+\epsilon \le b+\epsilon $ for all $\epsilon $. (Axiom that $a<b\implies a+x <b+x$ for all $x $) And futhurmore if $\epsilon > 0$ then $a+\epsilon >a $ (same axiom as $0+a <\epsilon+a $) and so, by transitivity $a<a+\epsilon \le b+\epsilon $.

I could not understand your second part at all.

If $a < b+\epsilon$ for all $\epsilon >0$. Suppose $a>b $. Then let $\epsilon = a-b>0$. So $a <b+\epsilon=b+(a-b)=a $. That's a contradiction so $a\le b $.

If you want a direct proof:

If $a < b+\epsilon $ for all $\epsilon > 0$. Then if $a \ge b+k $ that would mean $k\not > 0$ so that would mean $k\le 0$. Well, $a=b+(a-b) $ so $a-b \le 0$. And so $a\le b $.

Answer 4

Your first step ($a\leq b\, \, \, \, a,b\in R\to a < b + \epsilon\, \, \, \, \forall$ (for all) $\epsilon \in R$) looks good. My only nitpick is that flipping between $\geq$ and $\leq$ reduces readability a bit.

As for your second step, I see a couple problems.

First, the logic:

You derive a contradiction by reaching $a = b$ which can only mean you assumed $a \not = b$. This suggests to me that you got a bit mixed up about what exactly you were proving.

Which brings me to point 2:

Your statement of the problem is incomplete and (more debatably) awkward. It could be more effectively written as:

$\quad$Now, suppose instead that $a,b \in R$ such that $a<b+\epsilon$ for all $ \epsilon \in R^+$

Since it looks like you understand the issue and, as you say, are struggling with the formalization, so I imagine you can finish it from here. (It is as you thought a proof by contradiction).

As to your third question, it's generally difficult to prove that things are impossible / true in all cases except by supposing otherwise and finding a contradiction. But in this case, it is possible. Consider the statement (for the first step),

Suppose $a, b, \epsilon \in R$ such that $a\leq b$ and $\epsilon > 0$. Then it must be that $a - b \leq 0$. Thus, $a - b < \epsilon$. Therefore $a < b +\epsilon$.

I can't find a proof for part 2 without using contradiction.