Will McDuck go bankrupt?

Question

The present question is of interest because it is answered in very different ways by different groups (mathematicians, physicists, students, professionals of non-mathematical occupations). I ask here (1) because I have no experience yet with answers of philosophers, and (2) because this question is fundamental for Cantor's set theory, who also published a lot of his work in philosophical journals. With respect to mathematical papers he said: "the fact that my presently written work is issued in mathematical journals does not modify the metaphysical contents and character of this work." [G. Cantor, letter to T. Esser (15 Feb 1896)]

The basis of set theory is the proof of equinumerosity or equicardinality of infinite sets by one-to-one mappings. This tool proves for instance that the natural numbers and the fractions are equinumerous sets: Every natural number has its own fraction as a partner and every fraction has its own natural number.

This surprising result was explained by A.A. Fraenkel who told the story of Tristram Shandy. [Laurence Sterne: "The life and opinions of Tristram Shandy, gentleman" (1759-1767)]

"Well known is the story of Tristram Shandy who undertakes to write his biography, in fact so pedantically, that the description of each day takes him a full year. Of course he will never get ready if continuing that way. But if he would live infinitely long then his biography would get 'ready', because every day in his life, how late ever, finally would get its description. No part of his biography would remain unwritten, for to each day of his life a year devoted to that day's description would correspond." [A. Fraenkel: "Einleitung in die Mengenlehre", 3rd ed., Springer, Berlin (1928) p. 24. A.A. Fraenkel, A. Levy: "Abstract set theory", North Holland, Amsterdam (1976) p. 30]

A shorter and simpler variant is the story of Scrooge McDuck: Every day Scrooge McDuck earns 10 enumerated dollars and returns 1 enumerated dollar. If, as a cartoon character, he lives forever and if he happens to return always the dollar with the least number, he will go bankrupt because for every dollar we know when it is issued.

The question is: Is the latter argument sufficient to conclude that Tristram Shandy will get ready and that McDuck will go bankrupt?

Answer 1

The quotation about Tristram Shandy appeals to the fact that every page will eventually be finished. This condition is not sufficient to ensure that the task of writing the biography will ever be complete. In an infinite series like this, there is no inconsistency between saying that Shandy will never finish his biography, and saying that any given part of the biography will eventually be written. Intuitively this occurs because the written component increases without bound, but the unwritten component also increases without bound. This is explained in greater detail on mathematics.SE.

Answer 2

Both stories are examples of Supertasks.

The Scrooge McDuck story in particular is better known as the Ross-Littlewood Paradox or tennis-ball problem or ping-pong conundrum: if you start with no tennis balls, and every day you add 10 but remove 1, you can argue that you end up with none, since if you number the balls, and if you remove 1 after adding 1 through 10, remove 2 after adding 11 through 20, etc., then 'in the end' every tennis ball will get removed. Then again, if you remove 2 after adding 1 through 10, remove 3 after adding 11 through 20, etc., then 'in the end' you will end up with 1 tennis ball. And if you always remove the lowest even numbered tennis ball, you end up with an infinite tennis balls (all the odd numbered ones) after all. Which is what makes this a paradox.

One big problem (and possible resolution to the paradox) of course is that there seems to be no such thing as 'in the end', since we're talking about an infinite number of sequential actions. Some people will therefore say that such supertasks are logically impossible: Tristan's biography will never get completed, and Scrooge will never go bankrupt.

Others, however, maintain that a supertask can be completed by taking the first action in 1 second, and each successive action in half the time it takes to do the previous action - after 2 seconds, you will then have done an infinite number of actions.

Of course, this latter argument seems to be problematic as well: first of all it assumes that between now and 2 seconds from now there are an infinite number of points of time ... and as such it assumes that the flow of time itself is performing a supertask, and thus it effectively begs the question as to whether supertasks are possible. Indeed, arguments like Zeno's paradox argue against the infinite divisibility of space and time.

And finally, Benacerraf argued that these kinds of descriptions of supertasks only describe what happens while the task is being performed ... but the descriptions are silent on what should be the case when the task is done. Thus, even if supertasks are possible, it is still not clear what facts would hold when it is 'completed'.

Needless to say, the possibility of supertasks, and what we can conclude when supertasks would in fact be performed, is still hotly debated.

Answer 3

A philosopher would note that mathematics has a long history of being able to transcribe a word problem into a mathematical form where each of the elements of the word problem is translated to a corresponding mathematical form, but the resulting form is semantically different. A very common example is translating "If P then Q" into P=>Q. Many have great trouble with the fact that P=>Q is true if P is false, regardless of Q's value. The English phrasing "If P then Q" rarely has this connotation, but the math does. Thus care must be taken when transcribing a word problem into mathematics.

The philosopher would argue that the answers to these questions are obvious, and it is only the mathematical formalisms that cause issues. Thus, a philosopher would challenge the formalism.

Consider that one must translate phrases like "live forever" into a valid mathematical phrasing. In your self-answer, you used the notation S(n) to be the set of dollars in Scrooge's account after n steps, and we can use that to explore this difficult phrasing. We cannot say S(∞) in traditional ways because ∞ is not a natural number and steps are enumerated by ℕ. This is what I believe you were getting at with "potential infinity." We could use the "limit as n approaches infinity of S(n)," and by the definition of limits, we would see that the intuitive answer holds.

The issue you raise is that there should be a one to one mapping between the infinite sets of numbers. However, nothing precludes this. You can say that you added a set of enumerated bills of size ℵ0, and subtracted from it a set of enumerated bills of size ℵ0, and be left with a set of remaining bills of size ℵ0. Just because |A| = |B| does not mean that |A - B| = 0. For an obvious example, consider the difference between the natural numbers and the set of even numbers. Obviously the difference produces the set of odd numbers, yet all three sets have a size of ℵ0.

If I were to think of the problem in terms of a set of bills added, A, and a set of bills removed, R, at the end of "forever," I could say that |A| = |ℕ|, and that |R| = |ℕ|, but that does not mean |A-R| = 0 automatically (as seen above). I would need some proof that A=ℕ and R=ℕ or perhaps more generally A={f(n) for n in ℕ} and B{f(n) for n in ℕ}. Coming up with the mathematical phrasing which permits such equivalence (rather than just equal cardinalities) is where you will find the paradoxical results can arise. The notation regarding limits has been recognized for hundreds of years as an acceptable way to resolve these issues, and a philosopher will be very suspect of any other argument. After all, that's the job of a philosopher: to question things everyone else takes for granted.

Answer 4

The basis of set theory is the proof of the equinumerosity of infinite sets

This isn't correct. The basis of set theory is in the name, ie sets. What is true is that investigation of cardinality is an important part of modern set theory and this proof was the seed of that.

The story about Tristram Shandy narrated by Fraenkel is a rhetorical sleight of hand used to introduce the paradoxes of infinite sets via a story; in fact, the story makes no sense when thought through because the actions it relates are recursive, and this is not the kind of world we live in. It's an entertaining try by Fraenkel though.

The story about Scrooge is not a variant of this; it's already addressed in the theory of infinite sums: the partial sums are divergent and thus he won't get bankrupt.

Answer 5

This is a step-by-step procedure, a construction, and every received dollar is returned.

Considered in potential infinity there is no "all", neither of dollars nor of steps. For every natural number n: The dollars 1 to n are returned. Every n belongs to a finite initial segment which is followed by a potential infinity of others. No contradiction appears. But potential infinity disallows to prove equinumerosity of received and returned dollars.

Considered in actual infinity, there are all dollars but also all steps.

• All dollars are received and returned.

• All steps are not sufficient for that task, because for all natural numbers n: after step n there are 9n dollars not returned. Therefore the set of not returned dollars is not empty. Transactions are only possible at finite steps n (there is no action possible "after all finit steps" or "between all n and the limit"). Therefore, even if a discontinuity in the cardinality function is tolerated, there is a contradiction with an empty set of not returned dollars.

In most concentrated form:

• McDuck's wealth W(n) = |S(n)| can only change with n.

• For every n, W(n) is positive and increasing.

• In the limit after all n, W = |{ }|. – Contradiction.

An attempt to save transfinite set theory is to claim that the empty limit set does not mean a state after all natural numbers but only indicates that all received dollars are returned somewhere. This explanation fails already because unions and intersections to calculate the set limit range over an infinite domain. Further, if interpreted in actual infinity, the failure of completing the return at any finite step disproves the complete return.

Another attempt, to argue by the limits of analysis, is besides the point: Let q(n) = 0.0...01...10... be the sequence of rational numbers having digit 1 in kth position iff Scrooge McDuck possesses dollar note number k at day n. So q(n) has 9n digits 1, but in the limit of that sequence all these digits 1 are gone. Is this is a contradiction in the notion of limit?

No. The analytical limit 0 of the sequence (q(n)) is nothing that the terms q(n) would "evolve to". It is simply a real number that is approximated better and better by the terms of the potentially infinite sequence. The limit of the number of digits 1 of the q(n) in mathematics is simply the improper infinite, i.e., for every number there is a larger one (but never omega is reached).

For the sequence of sets things are quite different. There one set S(n) is transformed into its successor S(n+1) by adding and removing dollars. If their infinity is actual such that it is possible to complete the set, but if no dollar remains forever, then the complete loss of all dollars leaves the empty set. That is not mathematically possible.

A "limit" with quantized numbers, integers or cardinalities, different from all terms of the sequence, is impossible per se. Simple to see in the present case: For every step n there are elements. In the "limit" there are none. Contradiction. Mathematically reasonable is only the limit 0 of 1/W(n). It does not require an actually infinite or complete sequence (W(n)).

Summary: McDuck will never get bankrupt, where "never" means never and not "at omega".

Answer 6

Cantor's broken logic comes from a broken definition of "number." Cantor's numbers are allowed to do what other numbers cannot. For example, infinity + 1 equals infinity. A more sensible take on things is to recognize that some limitless things do not correspond to numbers (the boundless quantity of natural numbers is not itself a number), while infinite numbers (defined as numbers which are larger than any natural number) follow the "transfer principle" and obey the same rules as finite numbers, such as x+1 != x.

Given this viewpoint, after an infinite amount of time (let's call it t, with t being an infinite number) all the dollars labeled by natural numbers will have been returned yet an even greater amount of dollars labeled by infinite numbers (such as t-1) will remain in McDuck's possession.

Answer 7

Let me add another aspect which might deserve attention because it is not common knowledge but very important for set theory: In set theory we can count to ω. See the evidence collected in the footnote below.

Therefore it is not only "useful" to organize the meaning of "will go bankrupt" into a set to be interpreted as how things look 'at infinity', it is in fact necessary. Only an infinite bijection (between returned dollars and days) can be complete.

Therefore McDuck has gone bankrupt only when the set of his dollars is empty as soon as the set of days is ω. Since provably the set of his dollars is never empty, the bijection is never complete. The question "Are there any dollars not returned?" is irrelevant since the number of received and not returned dollars increases to 9ω = ω without bound.

Footnote: Evidence for counting to ω

Hausdorff describes Cantor's original method of well-ordering: "From the set A to be well-ordered take by arbitrary choice an element and denote it as a0, then from the set A \ {a0} an element a1, then an element from the set A \ {a0, a1} and so on. If the set {a0, a1, a2, ...} is not yet the complete set A, we can choose from A \ {a0, a1, a2, ...} an element aω, then an element a(ω+1), and so on. This procedure must come to an end, because beyond the set W of ordinal numbers which are mapped on elements of A, there are greater numbers; these obviously cannot be mapped on elements of A." [F. Hausdorff: "Grundzüge der Mengenlehre", Veit, Leipzig (1914); reprinted: Chelsea Publishing Company, New York (1965) p. 133f]

We see that Hausdorff counts from 0 to 0, 1, 2, ... and beyond to ω and further. This method is not endorsed by Fraenkel and Zermelo but by leading present set theorists:

"It is a perfectly valid and commonly used construction." Emil Jeřábek

"I endorse this method." J.D. Hamkins

Cast in the same mould: "We repeat this process countably many times." [Paul J. Cohen: "The discovery of forcing", Rocky Mountain Journal of Mathematics 32,4 (2002) p. 1076f] "If this procedure is iterated aleph_2 times ..." [Karel Hrbacek, Thomas Jech: "Introduction to set theory", 2nd ed., Marcel Dekker, New York (1984) p. 232ff]