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I.As related by the mathematician X, he was once visited by the brothers N, who, upon entering, took off their hats and hung them up on a rack in the hall. Later, when the guests were leaving and getting ready to put on their hats, it turned out to the host's chagrin that one hat was missing, although nobody had entered the hall during the time of the visit.

II.When the brothers N visited A'on another occasion, they again hung up their hats on the rack in the hall. Later, when the guests were leaving and getting ready to put on their hats, it turned out that there was an extra hat, although both the host and the guests were certain that there had been no hat on the rack when the guests arrived.

III. On the next visit, the guests put on their hats and left, and the host accompanied them to the street. Upon returning, he discovered the same number of hats on the rack as before the guests had left.

IV. Finally, on still another visit, the guests arrived without hats, and, upon leaving, put on the hats left over from the last visit. After accompanying the guests to the street, the host returned to discover once again the same number of hats on the rack as before the guests had left.

Explain all seemingly paradoxical events!

Found this riddle in Shilov's real and complex analysis. It is supposed to be an exercise on countable and uncountable sets. However I'm stuck!

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    $\begingroup$ Are you talking about the brothers $\mathbb N$? $\endgroup$ – Henry Feb 2 at 23:07
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    $\begingroup$ Are they staying at Hilbert's Hotel? $\endgroup$ – Henry Feb 2 at 23:08
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    $\begingroup$ I would say that these brothers have countably infinite siblings $\endgroup$ – Peter Foreman Feb 2 at 23:14
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    $\begingroup$ Same question was asked over at puzzling.SE (puzzling.stackexchange.com/questions/79207/…) where we've closed it as being more suitable for math.SE. Hope that's OK. (My guess is that the comments above are already enough for Dreamer123 to work out what's going on.) $\endgroup$ – Gareth McCaughan Feb 2 at 23:59
  • $\begingroup$ I hope there aren't uncountably many solutions. $\endgroup$ – timtfj Feb 3 at 0:23
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I. Give the first hat to the second person, 2nd hat to 3rd person and so on. First person is missing a hat.

II. Give 2nd hat to the 1st person, 3rd hat to the forth person, etc. 1st hat is left over.

III. The nth person takes the 2n-th hat.

IV. The nth person takes the (4n+1)-th hat.

Exercise. Show that no matter how many days the brothers N come without hats, that now they can all leave with hats.

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