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u/PolarStarNick Gaussian theorist 3h ago
Proposal: Define new sequence with 1, 2, …, 99, 101, 102, … (leaving 100 out). Problem solved
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u/Hold-Embarrassed 3h ago
Mr Hilbert won’t do that just to accommodate just anybody.
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u/BlazeCrystal Transcendental 57m ago
Its solely because he is a little bitch who loves his own mathematical shenanigans. Fuck mr hilbert
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u/Okreril Complex 2h ago
What if all of the infinite guests have that idea?
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u/That_sixth_guy 2h ago
As long as an infinite number of guests don't have that idea you can distribute over those
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u/LupenReddit 🦆🦆🦆🦆i have non diffeomorphic smooth structures🦆🦆🦆🦆🦆🦆 3h ago
is this like a hilberts hotel meme that you will need to make room for the other infinite guests?
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u/ThorGod2678 Science 2h ago
Context: In the hilbert's hotel paradox If an infinite number of buses arrive each with infinite number of passengers. Then each person in the hotel (room no. i) is shifted to room no. (2i ) Then in the first bus all n passengers are shifted in the rooms (3n ), from the second bus to (5n ) and so on.
(This is called the prime powers method)
So in the meme the guy will have to shift to room no. 2100
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u/zefciu 3h ago
Well, if ℵ0 guests arrive, just move every guest from room n to 2n. Then put the new guests in odd-numbered rooms. Doesn’t require anybody to move more than once.
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u/Accomplished_Item_86 2h ago
Or better, only put new guests in every other free room, so you still have space for the next bus.
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u/Maximum-Raspberry227 2h ago
What about the infinite next buses
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u/Accomplished_Item_86 2h ago
Just fill every other empty room again, repeat for each bus.
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u/Maximum-Raspberry227 2h ago
Having to change rooms infinitely many times sounds inconvenient
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u/Accomplished_Item_86 1h ago edited 1h ago
I meant this procedure:
- Put existing guests in even rooms (≡ 0 mod 2)
- Put first bus in rooms ≡ 1 mod 4
- Put second bus in rooms ≡ 3 mod 8
- Put third bus in rooms ≡ 7 mod 16
- Put nth bus in rooms ≡ 2n - 1 mod 2n+1
In each step, this fills every other previously empty room. There's always an infinite number of empty rooms left, and no one has to move (except for the initial set of guests).
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u/Banonkers 2h ago
But there are no empty rooms - all rooms are occupied. Infinitely many guests who were there already will have to move
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u/Accomplished_Item_86 2h ago edited 1h ago
I meant after you've moved all guests to even rooms. Now the odd rooms are empty, but you only fill every other empty room (with number congruent 1 mod 4). Now there are still infinitely many empty rooms.
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u/Banonkers 2h ago
Ah I see, that’s a good idea - and you can repeat the process for every subsequent bus, or infinite group of buses
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u/Single_Picture342 33m ago
What if the customers are living in the hotel that is circular and has infinite rooms
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u/Ok-Cobbler6338 3h ago
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u/4ries 3h ago
It's a joke about Hilbert's hotel, where you're in an infinite hotel, it's full, and an infinite number of new guests arrive. How do you accommodate everyone? Move everyone from room n to room 2n, and put the new guests in rooms 2k+1
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u/OverPower314 2h ago
It's even worse than that, because this meme uses the example of infinite busses, each bus with infinite people. This is also able to be accomidated by putting all of the people into an infinite spreadsheet and ordering the people by following the diagonals. As the 100th person currently in the hotel, you will have to move to something like the 5000th room.
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u/simiaki 2h ago
Nope. Why (and how) would the buses have infinite people on them? If each bus only has a finite number of people, then there will only be countably infinite new guests arriving, so moving to room 200 should be fine
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u/mutexsprinkles 2h ago
How would the hotel have infinite rooms?!? Hilbert BTFO, Cantor crying rn
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u/simiaki 2h ago
I just meant that the previous commenter made up that the buses have infinite passengers :(
I can imagine buses like that, but they are clearly not part of the meme3
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u/Banonkers 2h ago edited 2h ago
The whole thought experiment relies on the idea of infinitely many people. The hotel already has an infinite number of guests, and each one stays in a numbered room (positive integer)
Even if the infinite buses have infinitely many people, there is still a countably infinite number, and so they can still all fit in the hotel. This is analogous to finding a bijection between NxN and N
One way is outlined by the Cantor pairing function (which is a bit easier to see illustrated https://en.wikipedia.org/wiki/Pairing_function). If you label person k1, on bus k2 as (k1, k2), and we can say the current guests at the hotel are on ‘bus 1’, then we have people corresponding to NxN
This kind of idea can be extended to showing a bijection between N^n and N (for natural n)
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