Timeline for Covering the Rationals -- A Paradox? [closed]
Current License: CC BY-SA 3.0
21 events
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| Jan 4, 2019 at 16:41 | comment | added | user112109 | @ Guillaume Brunerie: "as soon as you have an infinite number (even countable), weird things can (and do) happen." If you distrust concluding from the finite to the infinite, shouldn't you distrust bijections between infinite sets above all (and first of all)? | |
| Feb 12, 2012 at 22:07 | vote | accept | Ashley McNeile | ||
| Feb 12, 2012 at 21:15 | comment | added | Ashley McNeile | Steven: It seems to me, on reflection, that the Hilbert Hotel argument does not actually rely on an induction and so the moves, as you say, could happen simultaneously. I think this is a real difference between this and the "Covering the Rationals" example. That helps! Thanks again. Ashley | |
| Feb 12, 2012 at 20:06 | comment | added | Ashley McNeile | Steven: Surely claiming that the extra guest can be accommodated is the same as claiming that there is a time (state) when everyone has moved. If no such time (state) exists, the claim that the extra guest can be accommodated cannot be said to be true. Probably all you can actually claim is that you cannot establish for sure that the hotel is full (a different and weaker claim!) Put it this way, I would try and find a different hotel :-) Thanks for your comments! | |
| Feb 12, 2012 at 19:52 | comment | added | Yemon Choi | @Steven, on a slightly tangential note, what do you mean by "I expect you're going to be very good at math?" This seems to presume knowledge of Ashley's current status/development which I can't find mentioned above or in the profile. | |
| Feb 12, 2012 at 19:41 | comment | added | Steven Landsburg | Ashley: In the Hilbert Hotel, everyone moves simultaneously. Or, if you prefer to move them one at a time, that's okay too --- the observation then is that each person eventually moves (which is not the same as saying that at some particular time, everyone has moved). | |
| Feb 12, 2012 at 19:30 | comment | added | Ashley McNeile | Steven: I am quite happy that you cannot use this kind of inductive reasoning to make a conclusion about a transfinite state. This is what I was trying to express in the third of my possible resolutions. However, it seems to me that the usual Hilbert's Hotel reasoning about a new guest (The hotel has infinitely many rooms, we can move the guest occupying room 1 to room 2, the guest occupying room 2 to room 3 and so on, and fit the newcomer into room 1) also depends on arguing about a transfinite end state, as otherwise there are always 2 guests in some room. So this must be invalid too. | |
| Feb 12, 2012 at 18:53 | comment | added | Steven Landsburg | Ashley: Write down a list of the natural numbers. If your list is finite after writing down n numbers, then it's finite after writing down n+1. Do you want to conclude that the set of natural numbers is finite? | |
| Feb 12, 2012 at 17:45 | comment | added | Ashley McNeile | Thanks Guillaume. The reasoning I used that the number of gaps is less than the number of intervals placed is essentially mathematical induction. (If it is true after placing n intervals, and placing an interval cannot increase the number of gaps by more than 1, then it is true after placing n+1 intervals.) Does this mean that mathematical induction sometimes fails if applied to an infinite sequence of cases, because "weird things happen"? The "Hilbert's Hotel" example uses a very similar kind of inductive reasoning but is generally held not to be fallacious. Why not? | |
| Feb 12, 2012 at 17:18 | comment | added | Guillaume Brunerie | "The number of gaps is less than the number of rationals" : No, it is true if you have a finite number of intervals, but as soon as you have an infinite number (even countable), weird things can (and do) happen. | |
| Feb 12, 2012 at 16:23 | comment | added | Ashley McNeile | Thank you all for your help. From your answers, I think the following are all true: - The number of rationals in [0,1] is countable - The number of gaps is less than the number of rationals (as the number of intervals placed is the same as the number of rationals, and placing an interval never creates more than one gap) - The number of gaps is uncountable. Have I got this right? Many thanks Ashley | |
| Feb 12, 2012 at 13:42 | history | closed |
Joel David Hamkins Bill Johnson Simon Thomas Dmitri Pavlov François G. Dorais |
off topic | |
| Feb 12, 2012 at 13:12 | comment | added | Joel David Hamkins | Even though I voted to close, I agree with Steven's previous comment. | |
| Feb 12, 2012 at 13:04 | answer | added | Gerald Edgar | timeline score: 1 | |
| Feb 12, 2012 at 12:22 | comment | added | Steven Landsburg | Sridhar Ramesh has given you exactly the right answer. I want to add that although this question will surely (and appropriately) be closed as too elementary for MO, and although you are in fact making what could fairly be called a "rookie mistake", your careful writeup suggests to me that you've got a lot of talent, a lot of insight, and a lot of genuine intellectual curiosity. I expect you're going to be very good at math. | |
| Feb 12, 2012 at 11:57 | comment | added | Sridhar Ramesh | So the problem is mainly your second bullet point, but it does not involve a new uncountable infinity (and what would it mean to be an uncountable infinity smaller than $\aleph_0$?). You are simply wrong in supposing that the number of gaps will be less than the number of rationals; you have no means of constructing a partial surjection from the latter to the former. | |
| Feb 12, 2012 at 11:53 | comment | added | Sridhar Ramesh | Indeed, instead of using mini-intervals, we might imagine removing single points: removing one rational at a time from [0, 1], we end up with n + 1 many connected components left after the first n many rationals have been removed. But after removing every rational, we are not left with a countable collection of connected components; instead, we are left with an uncountable collection of single points (the irrationals). | |
| Feb 12, 2012 at 11:49 | comment | added | Sridhar Ramesh | It is not true that the number of gaps must be countable. Your argument is simply "The number of gaps when the first n mini-intervals have been placed is <= n + 1; therefore, the number of gaps when all the mini-intervals have been placed is countable". But this argument is fallacious. | |
| Feb 12, 2012 at 11:41 | comment | added | Asaf Karagila♦ | I have no idea what to write about that, but usually when coming up with an uncountable $\aleph_{-1}$ which is "smaller than $\aleph_0$" should hint you that you lack the needed understanding of infinities. You should probably start with reading about basic cardinal arithmetic and $\aleph$-numbers. | |
| Feb 12, 2012 at 11:28 | history | edited | Ashley McNeile | CC BY-SA 3.0 |
added 9 characters in body
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| Feb 12, 2012 at 11:22 | history | asked | Ashley McNeile | CC BY-SA 3.0 |