If $dx$ is an infinitesimal, can't you list all real numbers as the sequence each whole number times dx?

Question

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?

Answer 1

The correct statement is not:

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

it is really more like:

2. 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.

Answer 2

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 $x$.

Answer 3

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