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?

Answer 1

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.

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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.

Answer 2

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

Answer 3

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.

Answer 4

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).

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

Answer 5

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:

1. What is the smallest positive integer?
2. What is the largest positive integer?
3. 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.

Answer 6

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 $a<b.$