20
$\begingroup$

I do not remember the name/source of this paradox,but I remember I have discussed this with mathematicians and non mathematicians at least 5 times.
It goes like this:

"Every point of a line has length $0$ and every line segment consists of points. So the length of the line segment is the sum $0+0+\ldots=0$"

which seems like a paradox.
Of course the number of points on a line segment is uncountable so the "summation" actually does not contain every point,so there is an obvious mistake there and we are done.
My question is:
Was this paradox discussed (or answered) before Cantor?

$\endgroup$

migrated from mathoverflow.net Apr 2 '15 at 16:43

This question came from our site for professional mathematicians.

  • 5
    $\begingroup$ I'm not sure why you say "before Cantor". Cantor's ideas, while not irrelevant to this question, certainly do not resolve it on their own. $\endgroup$ – Eric Wofsey Apr 2 '15 at 14:40
  • $\begingroup$ @EricWofsey why not? $\endgroup$ – Konstantinos Gaitanas Apr 2 '15 at 14:41
  • 6
    $\begingroup$ "A line segment consists of points" is a very modern conception. In Euclid there are lines, there are points, and there is incidence (whether a given point is on a given line). But certainly there is no assertion that the line segment consists of points. $\endgroup$ – Gerald Edgar Apr 2 '15 at 14:45
  • $\begingroup$ It is unfortunate that you should have mentioned Cantor in the title. While I don't have an answer to the question it seems to me that this "paradox" is more related to measure theory than to set theory. The fact is that the measure of a countable_ disjoint union of measurable sets is the sum of their measures, this is not true for an uncountable union (even if that sum is defined). Also note that the fact that there are uncountably many points on the line segment means you cannot write the sum as $0+0+0+\cdots$: there is no way to "line up" the contributions into such a (countable!) sum. $\endgroup$ – Marc van Leeuwen Apr 3 '15 at 13:14
  • $\begingroup$ Although most people today do think of a line as a point set, that isn't the only possibility. In smooth infinitesimal analysis (SIA), for example, a line isn't a point set. $\endgroup$ – Ben Crowell Apr 5 '15 at 13:44
15
$\begingroup$

Zeno (around 500 BC) raised this paradox to argue against the notion of "plurality", arguing that a belief in the existence of many things rather than only one leads to absurd conclusions: If there are many things, they must be both small and large; so small as not to have size, but so large as to be unlimited.

See section 2.2 of Zeno's paradoxes in the Stanford Encyclopedia of Philosophy. The following quote is from a discussion of Zeno's book on paradoxes by Simplicius (who writing one thousand year after Zeno apparently still had his book, which now is lost):

If a thing has no magnitude or bulk or mass, it would not exist. For if it should be added to something else that exists, it would not make it any bigger. For if it were of no size and was added, it cannot increase in size. And so it follows immediately that what is added is nothing. But if when it is subtracted, the other thing is no smaller, nor is it increased when it is added, clearly the thing being added or subtracted is nothing.

This seems to be the earliest source for the paradox, but not its resolution, which had to wait several millennia for the development of the notion of a distance function, which allows you to compute the length of an uncountable infinity of points.

$\endgroup$
  • $\begingroup$ So,did anyone try to give an answer to this paradox before Set theory was introduced? $\endgroup$ – Konstantinos Gaitanas Apr 2 '15 at 14:40
11
$\begingroup$

Aristotle gave the first systematic rebuttal of Zeno, in particular he wrote in Physics: "…a line cannot be composed of points, the line being continuous and the point indivisible". According to Aristotle, a line can be composed only of smaller, indefinitely divisible lines, and not of points without magnitude. This was the mainstream view until the "dissociation" of continuum by Cantor and Dedekind at the end of 19-th century. Under the new paradigm the paradox was resolved by the Lebesgue measure theory, which postulated that length is only countably additive, so the continuum wide summation of point lengths is invalid. This vindicated Aristotle's view: in terms of magnitude continuum can not be assembled from points.

The idea that continuum is a "set of points" leads to a number of conceptual problems, this being one of them. Others include its well-orderability, which strains credulity and implies existence of Lebesgue non-measurable sets, and the undecidability of the continuum hypothesis. A lot of effort was expended, by Zermelo, Lebesgue and others, to reconcile the intuition of continuum with "arithmetical" set theory. However, despite the objections of intuitionists like Hermann Weyl, in the end the benefits to analysis far outweighed the costs for most practitioners. Weyl's views echoed Aristotle's:"The notion that a set is a “gathering” brought together by infinitely many individual arbitrary acts of selection, assembled and then surveyed as a whole by consciousness, is nonsensical; “inexhaustibility” is essential to the infinite... Exact time- or space-points are not the ultimate, underlying atomic elements of the duration or extension given to us in experience."

$\endgroup$
  • $\begingroup$ There is a wonderful discussion of this topic in ch. 3 of Weyl's The Open World. I disagree with your claim that the benefits outweigh the costs; this was the common prejudice which Weyl eventually came round to, but Bishop's Constructive Analysis was hailed precisely because it showed how modern analysis could be carried through from this point of view. $\endgroup$ – Marius Kempe Apr 3 '15 at 0:26
  • $\begingroup$ I only meant that large majority of analysts came to prefer Cantor-Dedekind's continuum over the alternatives. The most common reason cited was that alternatives, including constructivism, imposed too many "inconvenient" restrictions on what can be done in analysis. $\endgroup$ – Conifold Apr 3 '15 at 1:48
  • $\begingroup$ Also, I have read that a well-ordering of the reals cannot be exhibited (in an essay of Thurston's) - there are some interesting related Mathoverflow threads (e.g. this one). So the reader may decide for themselves the extent to which they believe in today's "foundations of mathematics". $\endgroup$ – Marius Kempe Apr 3 '15 at 17:26
  • $\begingroup$ The same is true of Lebesgue non-measurable sets, Hamel bases, non-linear additive functions, everywhere defined unbounded linear operators, etc., none of them can be exhibited. This is a price to pay for things like every field has algebraic closure, every ideal is contained in a maximal one, every bounded linear functional extends to the entire space, etc. These are axiom of choice trade-offs. But there is a way to keep arithmetical continuum and Lebesgue measure theory (with all sets being Lebesgue measurable!) while discarding axiom of choice en.wikipedia.org/wiki/Solovay_model $\endgroup$ – Conifold Apr 3 '15 at 18:25
4
$\begingroup$

You might look at Galileo's discussion of infinity in "Discourses and Mathematical Demonstrations Relating to Two New Sciences": e.g. look up Galileo's paradox".

$\endgroup$
  • $\begingroup$ This isn't far from a link-only answer. $\endgroup$ – David Richerby Apr 3 '15 at 9:13
  • $\begingroup$ I don't think this addresses the issue of the question. As I noted in a comment to the question, the question is measure theoretic, not set theoretic. $\endgroup$ – Marc van Leeuwen Apr 3 '15 at 13:17
2
$\begingroup$

See the book discussed in this post .
Amir Alexander, Infinitesimal: How a Dangerous Mathematical Theory Shaped the Modern World

In the 17th century there was a dispute about whether a plane region consists of an infinite family of parallel line segments.

Argument against:

Do these line segments have width zero? Then the whole region would have area zero. Do they have positive width? Then there can be only finitely many of them.

The whole book is interesting.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.