I was reading a book called Calculus Basic Concepts for High Schools, and, under the topic limit, it was discussed that one cannot have two limits for a given sequence, provided the sequence has a limit. And the immediate implication of this is that there cannot be a neighboring number because one can always find a number between any selected numbers; however, close they may be. So in an open interval $(a,b)$, one cannot find the largest number.

I want someone to elaborate on this piece of text from "Calculus Basic Concepts for High Schools: L. V. Tarasov":

However, if there were a point neighboring 1, after the removal of the latter this "neighbor" would have become the largest number. I would like to note here that many "delicate" points and many "secrets" in the calculus theorems are ultimately associated with the impossibility of identifying two neighboring points on the real line, or of specifying the greatest or least number on an open interval of the real line.

  • What would happen if we can find a neighboring number?
  • How is calculus associated with the impossibility of identifying two neighboring points on the real line?
  • $\begingroup$ in an open interval (a,b), one cannot find the largest number If there existed a largest number in $(a,b)$, and if that largest number were $b'$ then the interval would be identical to $(a, b']$. Moreover, the interval $(b',b)$ would be empty, so $(b'+b)/2$ could not exist, for example. Sorry if that sounds obvious or naive, but it's not clear what level of responses you expect. $\endgroup$ – dxiv Dec 31 '17 at 5:51
  • $\begingroup$ $(x+y)/2$ is between $x$ and $y$ when $x\ne y.$ There are no gaps. $\endgroup$ – DanielWainfleet Dec 31 '17 at 7:27
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    $\begingroup$ I am sure that your book is wrong or misleading. Rationals are also dense and open intervals of rationals do not have greatest of least elements, as Paramanand stated in his answer as well. So ultimately this has little to do with the so-called secrets of calculus theorems. $\endgroup$ – user21820 Dec 31 '17 at 11:27
  • $\begingroup$ A more important property is that that there are no "infinitesimals". That is, there is no positive real that's less then every positive $rational$. This is implied by the $definition$ of $\Bbb R$. Without a definition you can't prove anything; it's like trying to prove that widgets are blue. $\endgroup$ – DanielWainfleet Dec 31 '17 at 16:22
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    $\begingroup$ @user21820 : I saw your comment a bit late and its spot on. Those so called secrets (mainly the deep theorems which are stated without proofs) are related to completeness, but the author has somehow mixed it with denseness. IMHO textbooks shouldn't dilute the concepts to the point being incorrect. $\endgroup$ – Paramanand Singh Jan 1 '18 at 11:30

Well, the point being talked about is called denseness/density and it applies to rational numbers also. The case of rational numbers is simpler and ideally should be in the mind of a seventh grader. It is rather unfortunate that most textbooks don't emphasize this concept of density at the right time and later on students have to struggle while studying calculus.

We start with the following:

Theorem 1: Between any two (distinct) rational numbers lies another rational number.

This is an immediate consequence of the following:

Theorem 2: Given any positive rational number there is another smaller positive rational number.

This is easy to understand and prove as well. If $m/n$ is a positive rational number then $m/(n+1)$ is a positive and smaller rational number. So the whole thing is ultimately dependent on the existence of a larger positive integer $n+1$ given a positive integer $n$. Moreover this also shows that there is no least positive rational number.

Next we can use theorem 2 to prove theorem 1. If $a, b$ are two rationals with $a<b$ then the number $d=b-a>0$ and we just need to find a smaller positive rational number $d'$ (which exists via theorem 2) and we can take $c=a+d'$ as our rational number lying between $a, b$.

Both the theorems above can also be proved using midpoint technique. Thus if $d$ is a positive rational number $d/2$ is a smaller one. And clearly $(a+b) /2$ is a number which lies between $a, b$. It is important to note both the approaches towards these theorems. Note also that the result in theorem 1 can be repeated to get as many rational numbers as we please between any two given rationals.

Now the idea of a neighbor (successor or predecessor for the case of integers) breaks down for rational numbers. To put it more precisely given any rational number $r$, there is no least rational number which exceeds $r$ and there is no greatest rational number which is exceeded by $r$.

This is expressed informally by saying that given any rational number $r$ we can find a rational number as close (near) to $r$ and less (or greater) than $r$ as we please.

The definitions of key concepts (limits) in calculus use the ideas mentioned above and are crucially dependent on the fact that there is no least positive number and there is no largest positive integer. However the fact that there is no next neighbor for a given rational number is not a drawback. It's a feature which is used everywhere in the definitions in calculus. A consequence of these facts is that we have an infinite supply of smaller and smaller positive numbers and greater and greater positive integers. And calculus in general deals with and therefore needs such infinite things.

Another key and slightly difficult ingredient in calculus is what we call completeness and that deals with the fact that although rationals are dense, they do lack something and this inadequacy is fulfilled by creating real numbers. The idea of completeness is not given sufficient focus in most textbooks of calculus (like in this answer too, but that's only to keep the length of the answer in control and can be discussed in another answer if you wish) but one can summarize the picture as follows:

The idea of denseness is essential to formulate the concepts of calculus and the idea of completeness is necessary so that these concepts don't operate in a vacuum (ie they do have some non-trivial consequences).

  1. If we can find a neighboring number, then the basic rules of arithmetic have collapsed and been disproven. If this happened, the universe would be collapsing anyway. If $a,b$ are consecutive real numbers, with $a<b$, then set $c=\frac{a+b}{2}$. We have $$a=\frac{a}{2}+\frac{a}{2}<\frac{a}{2}+\frac{b}{2} = c < \frac{b}{2}+\frac{b}{2}=b$$ and we have disproven that $a,b$ are consecutive (since $c$ is between them, and is also a real number.

    1. Calculus is all about not single numbers, but sequences of numbers getting closer and closer to another number. In the interval $(3,5)$, we take a sequence $4,4.5, 4.9, 4.99, \ldots$ that are getting closer to the top of the interval but never reaching it.

Essentially, all this boils down to the special property of the real numbers that every possible gap is filled, so to say. This is what makes calculus possible, for it is this property of real numbers that assures us that we can be confident that the set of real numbers is compact.

The rational numbers do not have this property, despite being dense. That is, consider that between any two distinct integers, there are always a finite number of integers; however, between any two distinct rational numbers you can always find a rational number, so that between two different rational numbers there are infinitely many rationals.

But interesting as this is, which might lead one to expect that the rationals exhaust every possible gap, we find that some gaps still exist, and that is what the irrational numbers fill. Thus, the very definition of reals as being limits of sequences of rational numbers makes the set complete. This is why you can always get arbitrarily close to an irrational number without ever landing on it, so to say.

So, if there are no numbers between two integers, they are not necessarily distinct. But this is already satisfied by the rationals. The irrationals complete this by adding those points where no rational could possibly be.

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    $\begingroup$ I think the rationals do have that property. The problem with rationals is something else which is not related to denseness at all $\endgroup$ – Paramanand Singh Dec 31 '17 at 7:19
  • $\begingroup$ Your last paragraph is also wrong. If the difference between two numbers is zero then they are same no matter whether they are integers, rationals, reals or for that matter even complex. $\endgroup$ – Paramanand Singh Dec 31 '17 at 7:25
  • $\begingroup$ @ParamanandSingh What are you talking about. Where did I say the 'problem' of the rationals is that they aren't dense? $\endgroup$ – Allawonder Dec 31 '17 at 7:26
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    $\begingroup$ Explanations can avoid difficult aspects but not at the cost of correctness. I will wait for your fix and then reverse my downvote. Moreover the key point which OP wants to discuss is density so discussing completeness is optional. I have given it just a cursory glance in my answer at the end and if OP wishes to discusses completeness it can be done in a different question. $\endgroup$ – Paramanand Singh Dec 31 '17 at 7:37
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    $\begingroup$ Well i will remove the downvote, but it still needs some cleaning. Mainly the idea that you can get arbitrarily close to a real number without landing on it is related to density and not completeness. $\endgroup$ – Paramanand Singh Dec 31 '17 at 7:43

Let's assume that for some real number $a$ there existed a next-larger real number $a'$. Now even the most basic definitions of calculus assume that you can take the sum and difference of any two real numbers. Therefore there would have to exist the number $\epsilon := a'-a$. Now $\epsilon$ would be the smallest positive real number, since if there were a number $0<\epsilon'<\epsilon$, then we'd have $a < a+\epsilon' < a+\epsilon=a'$ in contradiction to the original assumption that $a'$ is the next number after $a$.

However, if $\epsilon$ is the smallest possible number, what about $\epsilon/2$? Clearly that should also be positive, but smaller than $\epsilon$! So the only way to make that claim consistent is to declare that $\epsilon$ cannot be divided. Now this is already a quite serious restriction; but then we accept the same for integers (in the integers, $1$ cannot be divided by $2$), so let's for the moment just accept that. Then everything is good, right?

Well, no. First, we notice that $\epsilon$ cannot be a rational number, as any rational number is divisible by $2$. But $\epsilon$ is the smallest positive number, so it is smaller than all rational numbers. Still doesn't sound like a big deal, right?

But then, consider the sequence $a_n = \frac{1}{n}$. In the rational numbers this converges to $0$. But all $a_n$ are rational and positive, therefore for all $n$, we find that $\epsilon < |a_n-0| = a_n$. So maybe it converges to $\epsilon$ instead? No, because we demanded that you can add and subtract terms with $\epsilon$ (and without that, we could not even define convergence!), and therefore there also has to exist the number $2\epsilon = \epsilon+\epsilon$. And it's easy to see that $2\epsilon$ cannot be rational either, otherwise half of it would be, too, but half of $2\epsilon$ is $\epsilon$, which we already identified as non-rational. And then we find that $a_n > 2\epsilon$ and therefore $\epsilon < |a_n-\epsilon|$. We can continue this game, and find that we can find no number to which $a_n$ converges!

But the reason why we introduce the real numbers is exactly to get limits for sequences that do not converge in the rationals but "look like" convergent sequences. But now with the assumption of a next number, for any real number, gives us quite the opposite result: Now even sequences that converge quite fine in the rational numbers would no longer converge in the real numbers!

Fortunately in the actual definition of the real numbers there does not exist a next number to any real number (just as there doesn't exist a next number to any rational number), and therefore all those problems do not arise.

Also note that the ultimate source of the observed problems was that from the assumption we could derive a positive number smaller than every positive rational number. Therefore any construction that introduces such numbers (so-called infinitesimal numbers) runs into the same problems.

Now there does exist something called non-standard analysis which does indeed introduce such infinitesimal numbers (but not a next number to a number). But it can do so only at the cost of completely getting rid of limits and re-defining virtually every concept of analysis.

  • $\begingroup$ Well your description about dealing with $\epsilon/2$ seems a bit contrived. If we agree to deal with a positive number $\epsilon$ then we have to accept the existence of $\epsilon/2$ which is smaller and positive as well. We don't need any of that sequence / limit stuff to understand this. It is nothing but a result of how the arithmetic operations and order relations work with numbers. $\endgroup$ – Paramanand Singh Dec 31 '17 at 11:08
  • $\begingroup$ @ParamanandSingh: With the integers, we do accept that there are some numbers we cannot divide by $2$, so it's not a big stretch to imagine that when extending the rational numbers with new ones, we might be willing to accept similar restrictions to those new numbers, if it helps with other issues (just as we accept for complex numbers that we no longer can order them). Well, it turns out it doesn't help; quite the opposite. $\endgroup$ – celtschk Dec 31 '17 at 11:43
  • $\begingroup$ But we do know that we are constructing the new system of reals as an ordered field. And the failure of ordering in the field of complex numbers is quite well known and advertised in most introductory textbooks. I think the concept of denseness has to be treated trivially and emphasized early on in math curriculum. It is the completeness which is subtle and needs to be discussed before introducing calculus. $\endgroup$ – Paramanand Singh Dec 31 '17 at 13:05

First question: What would happen if we can find a neighboring number?

If we can find a neighboring number then the set of real numbers will be countable which is not true.
Second question: How is calculus associated with the impossibility of identifying two neighboring points on the real line?

Calculus is mostly about differentiation and integration both of which depend on the concept of limit. The real line with the existence of neighboring number becomes a discrete set in which the concept of limit as we know it breaks down.


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