Intuition of $a\le b$ if $a\le b+\epsilon$

Question

Intuition: If $a\leq b+\epsilon$ for all $\epsilon>0$ then $a\leq b$?
Here is the link that similar, but I am still confusing.
$a\le b+\epsilon$ means $a$ is less or equal to $b+\epsilon$. This inequality is not strict. I can understand that $a\le b$ if $a < b+\epsilon$. However, $a\le b+\epsilon$ confuses me since I need to consider $a=b+\epsilon$ in this case. How to explain $a=b+\epsilon$ in an intuitive way? In my opinion, there is no such expression $a=b+\epsilon$ since $a,b$ are fixed. If so, why this inequality not strict?

"If so, why this inequality not strict?" If the inequality to which you are referring is the conclusion that $a\leq b$, are you familiar with the fact that $a\leq a+\epsilon$ for all $\epsilon>0$? If the inequality were strict, then would that not imply that $a<a$? If instead your complaint is about the inequality in $a\leq b+\epsilon$, you could replace this by $a<b+\epsilon$ and get the same result. $a\leq b+\epsilon$ for all $\epsilon>0$ is true if and only if $a<b+\epsilon$ for all $\epsilon>0$, so it doesn't matter if you used a strict inequality vs a nonstrict one here. — JMoravitz, Dec 12 '19 at 18:44
It has nothing to do with strictness. Suppose I tell you that I have a number $x$ that satisfies $x \le \epsilon$ for all $\epsilon>0$. What can I say about $x$? I can't set $\epsilon=0$, but I can let it be arbitrarily small. So, intuitively I expect that $x \le 0$. This is straightforward to prove using contradiction. — copper.hat, Dec 12 '19 at 18:47
@JMoravitz so $a=b+\epsilon$ does not have any meaning right? — iefjkfdhfure, Dec 12 '19 at 19:00
@jf1997 it would if it were for a specific $\epsilon$ at a time. However, it is impossible for it to simultaneously be true that $a = b+e$ for some particular $e>0$ while also being true that $a\leq b+\epsilon$ for all $\epsilon >0$ since in particular when $\epsilon=\frac{e}{2}$ you would simultaneously have $a\leq b+\frac{e}{2}$ and $a=b+e$, a contradiction, since that would have implied that $a$ is less than itself. — JMoravitz, Dec 12 '19 at 19:02
@JMoravitz Can you show me if $a\leq b+\epsilon$ then $a<b+\epsilon$? I just try to prove the if and only if statement you just said. Thanks. — iefjkfdhfure, Dec 12 '19 at 19:49
@jf1997 suppose that $a\leq b + \epsilon$ is true for all $\epsilon >0$. Then by a change of variable name, setting $\frac{e}{2} = \epsilon$ it is also true that $a\leq b +\frac{e}{2}$ is true for all $e>0$. Now, notice that $a\leq b+\frac{e}{2}<b+e$ and so we have $a<b+e$ is true for all $e>0$. A final change of variable name back to $\epsilon$ completes the proof. — JMoravitz, Dec 12 '19 at 20:17
@jf1997 the "for all $\epsilon>0$" is a crucial part of the statement that I claimed. It is not true that given a specific $a,b,\epsilon$ that $a\leq b+\epsilon$ implies that $a<b+\epsilon$ if the hypothesis is only true for a particular $\epsilon$. For example, with $a=2$ and $b=e=1$ we have $a\leq b+e$ is true but $a<b+e$ is not. What is critical here is that not only is it true for a particular $\epsilon$ but it is true also for smaller $\epsilon$. — JMoravitz, Dec 12 '19 at 20:20
Does this answer your question? Intuition: If $a\leq b+\epsilon$ for all $\epsilon>0$ then $a\leq b$? — user1110341, Oct 01 '23 at 06:57

score 2 · Accepted Answer · answered Dec 13 '19 at 02:30

I think you are interpreting the $\leq$ sign as saying more than it really says.

If I write $x < y,$ I am saying that $x$ is strictly less than $y$. I am implying that it cannot be equal.

If I write $x \leq y,$ what I am saying is that

either $x$ is strictly less than $y$ or $x$ is equal to $y$
and I am not telling you which one is true!

When I write $x \leq y$ I am not saying $x$ is sometimes equal to $y$, or that $x$ might be equal to $y$ under special circumstances. I am just giving a little less information than I do when I write $x < y.$ When I write $x < y$ I have told you that $x \neq y$; when I write $x \leq y$ I have not told you that $x \neq y.$

Note that the mere fact that I have not told you something is not generally a reason to say that the thing is false.

Regarding the statement of the theorem,

$a \leq b$ if $a\leq b+\epsilon$ for all $\epsilon >0$,

the thing we often try to do in mathematics is to give as little information as we can in the premises of a theorem, and then conclude as much as we can from that little bit of information.

You can certainly see intuitively that if $a< b+\epsilon$ for all $\epsilon >0$ (strict inequality) then $a \leq b.$ But what if you only know that $a\leq b+\epsilon$ for all $\epsilon >0$? That's less information than if the inequality were strict. Do we really need someone to tell us that the comparison between $a$ and $b + \epsilon$ is a strictly "less than" inequality, or is $\leq$ enough?

You are concerned about the possibility that $a = b + \epsilon$.

We certainly do not have to consider the possibility that $a = b + \epsilon$ for all $\epsilon >0$. Aside from the fact that this is clearly impossible, it is enough to merely consider the possibility that $a = b + \epsilon$ for some $\epsilon >0$.

Let's consider that possibility. Is it possible that $a = b + \epsilon$ for some $\epsilon >0$ -- let's say, $a = b + \epsilon_1,$ where $\epsilon_1$ is some particular positive number?

OK, suppose $a = b + \epsilon_1,$ where $\epsilon_1 > 0.$ Now go back to the premise of the theorem: $a\leq b+\epsilon$ for all $\epsilon >0$. Let $\epsilon_2 = \frac{\epsilon_1}{2}$. That means $\epsilon_2 > 0,$ so $\epsilon_2$ is an example of a number $\epsilon > 0$, and the premise of the theorem applies with $\epsilon_2$ in place of $\epsilon$:

$$ a \leq b + \epsilon_2 . $$

Now figure out that $\epsilon_2 < \epsilon_1$ (strictly!) and therefore $a \leq b + \epsilon_2$ implies that $a < b + \epsilon_1$ (strict inequality).

So where is the $\epsilon > 0$ such that $a = b + \epsilon$? It doesn't exist, because no matter which $\epsilon$ you try, the premise (which said something about every possible choice of a positive number $\epsilon$) tells us there's always some other choice of $\epsilon$ that says this choice doesn't give us equality.

And so we see that someone was clever: they knew that they could prove $a \leq b$ from the premise that $a < b+\epsilon$ for all $\epsilon >0$, but they also saw that they could give even less information in the premise -- use the symbol $\leq$ instead of the more informative $\lt$ -- and they could still prove the same statement in the end. Getting the same result from less information is good mathematics, so they did it.

Why it is impossible to let $a=b+\epsilon/2 ?$ I mean since $a\le b+\epsilon$ for all $\epsilon$, but $\epsilon/2<\epsilon$. — iefjkfdhfure, Dec 14 '19 at 09:17
Remember that in the "for all $\epsilon > 0$" part, the $\epsilon$ stands for all positive numbers. There is no such thing as a positive $\epsilon/2$ that is smaller than all other positive numbers. As you said, $a$ and $b$ can be treated as fixed numbers here, which means if it were true that $a=b+\text{(something positive)}$, it could only be true for one "something positive" and there is always a smaller positive number such that $a\leq b+\text{(the smaller positive number)}$. — David K, Dec 14 '19 at 15:44
Part of the reason I wrote $\epsilon_1$ and $\epsilon_2$ instead of just $\epsilon$ is to emphasize that if $a = b+\epsilon$ the "$\epsilon$" in that equation has to be a particular number and not the "all $\epsilon>0$" that is in the premise. So do you think such a number exists? Fine, let's call that number $\epsilon_1$, but then if you don't like the result you don't get to say, "Wait, that's not actually the number I meant to call $\epsilon_1$, I really meant a different number." — David K, Dec 14 '19 at 15:50
Perhaps Charles Hudgins used better notation. By definition, $a = b + (a - b)$, so if there is any value of $\epsilon>0$ for which $a = b + \epsilon,$ that particular value of $\epsilon$ is $(a - b).$ But if $(a-b)>0$ then also $(a-b)/2>0$, from which we find that when we try the particular value of $\epsilon$ that is $(a-b)/2$ we get $a > b + \epsilon,$ so it is not true that $a\leq b+\epsilon$ for all $\epsilon>0.$ — David K, Dec 14 '19 at 16:19

score 1 · Answer 2 · answered Dec 13 '19 at 03:10

Yes, the inequality is strict, but using the symbol $\le$ doesn't mean both the $\lt$ and $=$ are simultaneously true. Indeed, what it means is that exactly one of the relations is true. In this case it is the inequality that is true. So we may still truly write $$a\le b,$$ since we have that $a<b.$

score 0 · Answer 3 · answered Dec 12 '19 at 19:03

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Without equality, we have $$\Bigl((\forall \epsilon>0) \;\; a<b+\epsilon \Bigr)\;\; \; \implies \;\; a\le b$$ or $$(\forall n\in \Bbb N )\;\; 1<1+\frac{1}{1+n}$$

answered Dec 12 '19 at 19:03

hamam_Abdallah

63,719

score 0 · Answer 4 · answered Dec 12 '19 at 19:22

0

Suppose that it is not the case that $a \leq b$. That is, $a > b$. Set $\epsilon = \frac{a - b}{2} > 0$. Then $$ b + \epsilon = \frac{a + b}{2} < \frac{a + a}{2} = a $$ That is, there exists $\epsilon > 0$ such that $b + \epsilon < a$. In other words, it is not the case that $a \leq b + \epsilon $ for all $\epsilon > 0$.

We have shown $$ \lnot (a \leq b) \implies \lnot (\forall \epsilon > 0 \ \ a \leq b + \epsilon) $$ This is equivalent to the contrapositive $$ (\forall \epsilon > 0 \ \ a \leq b + \epsilon ) \implies a \leq b $$ which is what needed to be shown.

answered Dec 12 '19 at 19:22

Charles Hudgins

5,907

I have stayed away from proof by contradiction here because I find that people just starting to write math proofs tend to rely to heavily on contradiction. Contradiction is a fine proof technique, but it needlessly complicates things when a proof by contraposition would suffice. A proof intended for a more experienced audience would simply remark "we shall prove the contrapositive" and end with the second line of text. – Charles Hudgins Dec 12 '19 at 23:51

score 0 · Answer 5 · edited Jun 12 '20 at 10:38

Let's compare the proofs of the following two statements:

(1) "$a < b + \epsilon$ for all $\epsilon > 0$ implies $a \leq b$"

(2) "$a \leq b + \epsilon$ for all $\epsilon > 0$ implies $a \leq b$"

Suppose that your assumption was instead "Assume that for all $\epsilon > 0$, $a < b + \epsilon$". You would prove that $a \leq b$ as follows:${}^{1}$

Proof 1: Suppose not. Then $a > b$ so let $\epsilon = a - b$. By assumption, $a < b + \epsilon$ so $a < b + \epsilon = b + (a - b) = a$. Thus $a < a$, giving us a contradiction.

However, your actual assumption is the following: "Assume that for all $\epsilon > 0$, $a \leq b + \epsilon$". If we were to use the above proof with the only alteration being changing $a < b + \epsilon$" into $a \leq b + \epsilon$", then we would conclude that $a \leq a$, which does not give us a contradiction. So how can be change the proof so that we do get a contradiction? Just change "$\epsilon = a - b$" to "$\epsilon = \frac{1}{2} (a - b)$":

Proof 2: Suppose not. Then $a > b$ so let $\epsilon = \frac{1}{2} (a - b)$. By assumption, $a \leq b + \epsilon$ so using the fact that $b < a$ we have $a \leq b + \epsilon = b + \frac{1}{2} (a - b) = \frac{1}{2} a + \frac{1}{2} b < \frac{1}{2} a + \frac{1}{2} a = a$. Thus $a < a$, giving us a contradiction.

Importantly, note that this second proof can be used to prove both statements (1) and (2) (since $a < b + \epsilon$ implies $a \leq b + \epsilon$).

So if you're still confused then since you mentioned that you're comfortable with the reason why statement (1) is true, I'd recommend that you first try to understand why this second proof proves statement (1) and then directly transfer this knowledge to reason about why it also proves (2).

In my opinion this is the most natural way to prove the conclusion $a \leq b$.

score 0 · Answer 6 · answered Dec 13 '19 at 02:08

The confusion here is mostly a by-product of using too much symbolism. Translating the problem into your natural language makes it trivial.

Convince yourself that you can answer the following questions:

What is the smallest positive integer?
What is the largest positive integer?
What is the smallest positive rational number?

The first question is trivial and put there for contrast with the next questions. The next two questions are related. Once you observe that there are no numbers which can answer the last two questions in affirmative then you have got the crux of the problem at hand.

If you have understood the above questions and their answers then it should be obvious to you that if a rational number $x$ does not exceed any positive rational number then $x$ must be zero or negative. In your question $x$ is $a-b$.

The previous argument holds if the word "rational" is replaced by "real" because order relations on real numbers follow the same laws as those on rationals.

Intuition of $a\le b$ if $a\le b+\epsilon$

6 Answers6