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I'm taking calculus right now.

If the difference between each real number and the next is an infintesimal, then wouldn't the following sequence $\{0\,dx, 1\,dx, -1\,dx, 2\,dx, -2\,dx, \ldots\}$ be a set of all real numbers?

But I was schooled that you can't make a sequence of all real numbers. Is somebody a liar here?

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Veta
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    Why do you think that ${ 1dx, 2dx, 3dx,\ldots }$ would include all real numbers ? – Mauro ALLEGRANZA Mar 28 '15 at 12:59
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    For that matter, why do you think ${1\mathrm{d}x, 2\mathrm{d}x, …}$ includes any real number? Why would $\mathrm{d}x$ be a real number? – user3493525 Mar 28 '15 at 13:04
  • If it's an infinitesimally small number and the difference between each real number and the next is an infinitesimal, the sequence of real numbers should be the above no?

    Is the issue that there's no such thing as an infinitesimal?

     – Veta Mar 28 '15 at 13:12
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    Your notation is not well chosen. $dx$ is not an infinitesimal. – Frieder Mar 28 '15 at 13:38

3 Answers3

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It appears to be suggested here that the sequence $n\,dx$ for $n=1,2,3,4,\ldots$ should contain all positive real numbers. To present-day mathematicians this will seem alien and at first sight absurd, for several reasons:

  • Despite being an immensely useful heuristic, well worth learning and using, to think of $dx$ as a infinitesimal, that is usually not taken literally in doing mathematical reasoning today.

  • Infinitesimals are not real numbers, and $dx, 2\,dx, 3\,dx, \ldots$ are infinitesimals.

  • Not all infinitesimals are included in the sequence; e.g. one could ask about $1.3\,dx$ or $dx/20$, etc.

  • Among the numbers $n\,dx$ apparently proposed to be included in the sequence, every integer $n$ is finite, hence every member of the sequence is infinitesimal, so no non-infinitesimal number, such as $1$, will ever be reached.

  • If one were to allow some sort of infinite integer in the role of $n$, so that for some infinitely large value of $n$ the number $1$ were equal to $n\,dx$, then the "sequence" would not be a sequence in the sense in which that term is intended when it is said that no sequence contains all real numbers. Every term in a sequence, in that sense of the word sequence, has only finitely many terms before it.

There are in fact some systems in which one can do calculus with infinitesimals that are consistent with present-day standard modes of thinking. One of those is Robinson's non-standard analysis. In Robinson's system, one could consider the set $\{n\varepsilon: n=1,2,3,\ldots \}$, where $n$ goes through the list of non-standard positive integers, which includes some infinite integers. Then every real number would at least be infinitely close to some terms in this "sequence" -- more than one of them. Suppose one chooses an infinitely large integer $n$ such that $n\varepsilon$ differs from $\pi$ by an infinitesimal. Then $(n\pm1)\varepsilon$, $(n\pm2)\varepsilon$, etc., would also differ from $\pi$ by an infinitely small amount, whenever the number after "$\pm$" is finite. That doesn't mean $\pi$ would appear in the sequence; rather $\pi$ might be somewhere between $n\varepsilon$ and $(n+1)\varepsilon$. And for sufficiently large values of $n$, the number $n\varepsilon$ itself would be infinitely large -- thus bigger than all real numbers. Robinson and his followers do call something like this a "sequence" and are able to apply some of the same modes of reasoning to it that we normally apply to what we normally call sequences, but within Robinson's system no sequence of this kind exhausts the set of nonstandard real numbers, since for example many nonstandard real numbers are between $n\varepsilon$ and $(n+1)\varepsilon$.

Appendix, quoting from an earlier answer I posted:

What is $dx$ in integration?

begin quote

Leibniz, who introduced this notation in the 17th century, thought of $dx$ as an infinitely small increment of $x$, and at least as a heuristic, that is an immensely useful idea.

However, note some other points:

  • $\displaystyle\int f(x,y)\,dx$ differs from $\displaystyle\int f(x,y)\,dy$. In one case, one integrates a function of $x$, and $y$ is constant; in the other these roles are reversed and one might be integrating a very different function.
  • If $f(x)$ is in meters per second and $dx$ is in seconds, then $f(x)\,dx$ is in meters, and so is the integral. These things should be dimensionally correct, and are not so without the "$dx$".
  • Sometimes one has a dot-product or a cross-product or a matrix product or some other sort of product between $f(x)$ and $dx$. How would one specify that without the "$dx$" written there?
  • When doing substitutions, it becomes important to distinguish between $dx$ and $du$, etc.

end quote

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Michael Hardy
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The correct statement is not:

  1. The real numbers cannot be put into a list. (Ambiguous)

it is really more like:

  1. The real numbers cannot be put into a list that is indexed by $\mathbb{N}$.

What you've found is a counterexample to certain formalizations of (1); you have not found a counterexample to statement (2).

In fact, you have done more than this. Here's what you've done: Let $h$ denote an infinitesimal number in whatever extension-by-infinitesimals $\mathbb{R}^*$ of the real line $\mathbb{R}$ that you prefer. Then $\mathbb{R}^*$ will also tend to have a subset $\mathbb{Z}^*$ of integer-like elements. Then it will tend to be the case that for all $x \in \mathbb{R}$, there is some $n \in \mathbb{Z}^*$ we have that $nh$ is infinitesimally close to $x$. This implies that the elements of $\mathbb{Z}^*$ cannot be put into a list that is indexed by $\mathbb{N}$. In other words, you've shown that $\mathbb{Z}^*$ has quite a lot of elements!

Now for a bit of history. Two of the most surprising discoveries of early set theory were:

  • Some sets have the property that their elements cannot be put into an $\mathbb{N}$-indexed list. (See also: Cantor's theorem.)

  • There's a generalization of the natural numbers called the ordinal numbers, and if we allow our lists to be indexed by ordinal rather than natural numbers, then amazingly, every set $X$ can be put into list form! (See also: well-ordering theorem).

Both Cantor's theorem and the well-ordering theorem were very controversial in their days. I suggest Googling the history of these ideas, its all really quite interesting.

goblin GONE
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    I totally agree with 1. being ambiguous. People seem to perceive something that I don't with the term 'list' and associated properties. – Git Gud Mar 28 '15 at 13:33
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    Thanks this was super interesting, I had read a little bit about Cantor which was part of the impetus for this post. I'm keen to read more about it. – Veta Mar 28 '15 at 14:22
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    @veta, glad to be of service. One observation that might help you get started in learning more: an $I$-indexed list of elements of $X$ is really the same thing as a function $I \rightarrow X$. This is also sometimes times called an $I$-indexed sequence in $X$. Note, however, that since such a thing really needn't look like a list or sequence at all, we usually call this an "$I$-indexed family." This avoids unnecessary connotations regarding how $I$ is (or is not) ordered. Hope that helps :) – goblin GONE Mar 28 '15 at 14:28
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    @veta, also, note that to say that an $I$-indexed "sequence" is injective is just to say that no element occurs in two different places in the "sequence." And to say that an $I$-indexed sequence in $X$ is surjective is just to say that every element of $X$ is represented at one or more places in the "sequence." Hope that helps. – goblin GONE Mar 28 '15 at 14:30
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    For an infinitesimal $h$, the $\mathbb Z^*$-indexed sequence $n\mapsto nh$ will not contain $\mathbb R$. It will, however, contain elements within infinitesimal distance from any given real number. – Andreas Blass Mar 28 '15 at 14:33
  • @AndreasBlass, true, and that is the problem with answering questions about topics about which one knows next to nothing. Is there a way to recover this answer? In particular, can we modify the argument so that we still deduce that $\mathbb{Z}^*$ will tend to be uncountable? – goblin GONE Mar 28 '15 at 14:36
  • @AndreasBlass, I've made it community wiki, so feel free to "take control" of the answer so to speak, if you can see how to fix it. – goblin GONE Mar 28 '15 at 14:37
  • it is helpful :) – Veta Mar 28 '15 at 14:38
  • Yes, the uncountability of $\mathbb Z^*$ still follows. Every (standard) real number $r$ has a (nonstandard) $nh$ infinitely near it, and different $r$'s will need different $n$'s because different $r$'s are not infinitely near each other. – Andreas Blass Mar 28 '15 at 14:40
  • @veta, check out this video to learn a few basic facts. But note that some of the stuff that is said after $8:30$ is kind of incorrect. A little bit. This is also good stuff. – goblin GONE Mar 28 '15 at 14:47
  • wow goblin, that was fascinating and so appropriate, thank you – Veta Mar 30 '15 at 17:27
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The very idea of a "next" real number (under the usual ordering) is self-contradictory: If $x$ and $y$ are arbitrary real numbers with $x < y$, then $x < \frac{1}{2}(x + y) < y$. That is, $y$ is not "adjacent" to $x$.

Andrew D. Hwang
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