Let me recall that, for a positive integer n, Cut(n) is the statement that, for each sequence of n-element sets, the union of all sets of this sequence is at most countable. Cut(fin) is the statement that countable unions of finite sets are at most countable. I am unable to deduce whether it is true in ZF that if Cut(n) holds for each positive integer n, then Cut(fin) also holds. Perhaps, there exists a mathematician who knows a model for ZF in which Cut(fin) fails and, simultaneously, Cut(n) holds for each positive integer n. I would be grateful for any helpful hint to give a satisfactory answer to my question. Regards, Eliza Wajch
Dear Eliza,
Perhaps Levy's Model I, Model N6 in the Howard\Rubin book on consequences of the axiom of choice will do the job.
Thank you for your answers, especially for the hint concerning Levy's Model I. I shall have a look at this model in Howard/Rubin book, alhough I am still uncertain while working with ZF models. Dear Peter, when one deals with an infinite countable collection $\{ X_n: n\in\omega\}$ of sets X_n such that each X_n has exactly two elements, there may not exist a sequence (
When you start to write marks of numbers 1, 2,... today and you will contiune it tomorrow, the day after tomorrow, the day afer after tomorrow and so on, at every moment of your life from today, you will have only a finite number of written marks of numbers. Please, imagine to yourself two identical balls, one of them kept in room A, the other one kept in room B. Suppose that, when you are out of the house inside which there are the rooms, someone has changed the places of the balls: the one from room A one has moved to room B, and the one from room B one has moved to room A. When you come back home and see the balls, you will be unable to notice the changes. You are unable to distinguish the balls even when you know that they are two balls. The situation is much more complicated when there are infinitely many places for, in some sense, undistinguishable objects. I think that the best for me is just to buy myself the famous book ``Consequences of the Axiom of Choice" by Howard and Rubin and, then, read it in deep. Regards, Eliza
Please see the book - Consequence of the Axiom of Choice, Howard and Rubin, American Mathematical Society, Mathematical Survey and Monograph, V-59, 1998
I know that CUT(Fin) is equivalent to AC_omega(Fin), that is, all countable families of finite sets have a choice function. I guess that AC_omega(fatorial of n) implies CUT(n) - there are at most fatorial of n enumerations of a finite set with at most n elements, so choosing one enumeration for each one of the finite sets it is possible to proceed with some well ordering argument. So I guess that a equivalent questions is: is it possible to have simultaneously AC_omega(n) for every n and the failure of AC_omega(Fin) ? However, I am not aware of a model with such properties. I hope my remark helps somehow.
Dear Samuel,
Thank you for your interest in my question. When I buy myself the book by Howard and Rubin, I hope to become cleverer I am looking foward to meeting you at Toposym. I have already submitted my slides for my talk to toposym. Several poblems relevant to this question will be announced by me. I am happy that I can go to Prague where I shall meet so many active and advanced topologists.I hope to learn a lot there. Eliza
Yes, we will meet at Prague, it will be a very interesting conference, I am sure of it !
Doesn't the usual compactness argument work here? Add a constant symbol c to the language, and extend ZF with a statement saying that c is a natural number such that Cut(c) fails, and every statement of the form Cut(n), where n is a natural number. Apply compactness to get a non-standard model of ZF such that Cut(n) holds for every n but Cut(fin) fails.
Edited to add: I realize I may have misread your question.
Dear Taneli Huuskonen,
I thank you for your answer very much. I shall think of it in deep when I learn more about non-standard models of ZF. I think that you have understood my question very well. So, I have a lot of work to do - to study models of ZF - standard and non-standard.
Regards, Eliza
Thank you for your interest in my question; however, I think that your answer is not professional. It is not a satisfactory answer to my question. Eliza Wajch
Hi, I have found something which could be related to this question. In a paper of Eleftherios Tachtsis (JSL 81, 1, 2016, 384-394) it is commented in page 385 that Hall and Shelah have proved that for each integer m not smaller than 2 it is not provable in ZF that "every countable family of m-element sets has a partial choice function" implies "every countable family of m-element sets has a choice function" (and it refers to Fundamenta Mathematica 220 (2013), 207-216).
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