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Off topic: Will McDuck go bankrupt?
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wolfgang.m...@hs-augsburg.de
Apr 24, 2018, 12:39:33 PM4/24/18
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David Hilbert once said: "Physics is to difficult for physicists." Meanwhile it appears that mathematics is too difficult for mathematicians. Therefore I would like to ask this mathematical question here.
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 as a partner or index.
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?
Regards, WM
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 as a partner or index.
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?
Regards, WM
wolfgang.m...@hs-augsburg.de
Apr 28, 2018, 8:20:01 AM4/28/18
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The solution of McDuck's paradox
Every received dollar is returned. This fact can be considered in actual and in potential infinity.
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 n in |N: after step n there are 9n dollars not returned. This proves for every step that the set of not returned dollars after step n is not empty. Since dollars can only be returned at finite steps n (there is no action possible "between all n and omega"), there is a contradiction with an empty set of not returned dollars.
In potential infinity there is no "all", neither of dollars nor of steps.
For every n in |N: The dollars 1 to n are returned. No contradiction, since every n belongs to a finite initial segment which is followed by an infinity of others.
In most concentrated form:
- McDuck's wealth Wn = |Sn| can only change with n.
- For every n, Wn 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 indicate that all received dollars are returned somewhere. This explanation fails already because unions and intersections to calculate the set limit range over k to oo. Further if interpreted in actual infinity, the failure of completing the return at any finite step disproves the complete return. If interpreted in potential infinity, it is correct but forbids to prove equinumerosity of received and returned dollars.
Another attempt to argue by the limits of analysis is besides the point: Let qn = 0.0...01...10... be the sequence of rational numbers having digit 1 in kth position iff Scrooge possesses dollar note number k at day n. So qn 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 imit 0 of the sequence (qn) is nothing that the terms qn 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 qn in mathematics is simply the improper limit oo, i.e., for every number there is a larger one but never omega or aleph_0 is reached.
For the sequence of sets things are quite different. There one set Sn is transformed into its successor 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, is impossible per se. Simple to see: For every step n there are elements. In the "limit" there are none. Contradiction. Mathematically reasonable is only the limit lim(n-->oo) 1/Wn = 0. It does not require an actually infinite or complete sequence (Wn).
Regards, WM
Every received dollar is returned. This fact can be considered in actual and in potential infinity.
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 n in |N: after step n there are 9n dollars not returned. This proves for every step that the set of not returned dollars after step n is not empty. Since dollars can only be returned at finite steps n (there is no action possible "between all n and omega"), there is a contradiction with an empty set of not returned dollars.
In potential infinity there is no "all", neither of dollars nor of steps.
For every n in |N: The dollars 1 to n are returned. No contradiction, since every n belongs to a finite initial segment which is followed by an infinity of others.
In most concentrated form:
- McDuck's wealth Wn = |Sn| can only change with n.
- For every n, Wn 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 indicate that all received dollars are returned somewhere. This explanation fails already because unions and intersections to calculate the set limit range over k to oo. Further if interpreted in actual infinity, the failure of completing the return at any finite step disproves the complete return. If interpreted in potential infinity, it is correct but forbids to prove equinumerosity of received and returned dollars.
Another attempt to argue by the limits of analysis is besides the point: Let qn = 0.0...01...10... be the sequence of rational numbers having digit 1 in kth position iff Scrooge possesses dollar note number k at day n. So qn 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 imit 0 of the sequence (qn) is nothing that the terms qn 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 qn in mathematics is simply the improper limit oo, i.e., for every number there is a larger one but never omega or aleph_0 is reached.
For the sequence of sets things are quite different. There one set Sn is transformed into its successor 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, is impossible per se. Simple to see: For every step n there are elements. In the "limit" there are none. Contradiction. Mathematically reasonable is only the limit lim(n-->oo) 1/Wn = 0. It does not require an actually infinite or complete sequence (Wn).
Regards, WM
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