Hilbert's hotel is utter nonsense

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Skepdick
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Hilbert's hotel is utter nonsense

Post by Skepdick »

Suppose that Hilbert's "infinite" hotel is full.
Suppose we want to make room for one more guest.
So we tell all guests to exit their rooms.
Fine...

There's a guest outside room 1; and a guest outside room 2; and a guest outside room 3...

There's a bijection (1:1 mapping) between guests and rooms.
Now we instruct every guest to move one room up (this alone is problematic - which guest moves first?)

Guest 1 is in room 2; guest 2 is in room 3; guest 3 is in room 4...

Room 1 is now empty.
There's now more room than guests.

This violates surjectivity and is no longer a bijection.
Magnus Anderson
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Re: Hilbert's hotel is utter nonsense

Post by Magnus Anderson »

He said the hotel is full. That means there are no empty rooms. And if there are no empty rooms, you can't accomodate a new guest. Moving guests around can't create new rooms.
Magnus Anderson
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Re: Hilbert's hotel is utter nonsense

Post by Magnus Anderson »

Skepdick wrote: Fri Nov 01, 2024 11:57 am Now we instruct every guest to move one room up (this alone is problematic - which guest moves first?)
Guest 1 moves in Room 2. That leaves Room 1 empty and makes Room 2 occuppied by Guest 1. Guest 2 is still in front of Room 2. The same process is repeated for Guest 2 and every subsequent guest.

The problem is that there are not enough rooms to accommodate all of the guests. But they mistakenly think there are due to the confusion created by the observation that if you add an element to an infinite set the set is still infinite.

Infinity is not a specific number, it's a category of numbers.
Skepdick
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Re: Hilbert's hotel is utter nonsense

Post by Skepdick »

Magnus Anderson wrote: Fri Nov 01, 2024 12:31 pm Guest 1 moves in Room 2. That leaves Room 1 empty and makes Room 2 occuppied by Guest 1. Guest 2 is still in front of Room 2. The same process is repeated for Guest 2 and every subsequent guest.
That's not possible. If there was a 1:1 bijection at the begining, and now you have an unoccupied room then some is necessarily left without a room.
Magnus Anderson wrote: Fri Nov 01, 2024 12:31 pm Infinity is not a specific number, it's a category of numbers.
It doesn't matter what it is. If there's a 1:1 bijection between rooms and guests, ANY operation which leaves empty rooms contradicts the premise.
Magnus Anderson
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Re: Hilbert's hotel is utter nonsense

Post by Magnus Anderson »

Skepdick wrote: Fri Nov 01, 2024 12:35 pm That's not possible. If there was a 1:1 bijection at the begining, and now you have an unoccupied room then some is necessarily left without a room.
Yes, that's what I said.
Skepdick wrote: Fri Nov 01, 2024 12:35 pm
Magnus Anderson wrote: Fri Nov 01, 2024 12:31 pm Infinity is not a specific number, it's a category of numbers.
It doesn't matter what it is. If there's a 1:1 bijection between rooms and guests, ANY operation which leaves empty rooms contradicts the premise.
If there's a bijection between ( 1, 2, 3, ... } and { 0, 1, 2, 3, . . .) then they can move all of the guests to the right and make the first room empty.

But no such bijection exists.
Skepdick
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Re: Hilbert's hotel is utter nonsense

Post by Skepdick »

Magnus Anderson wrote: Fri Nov 01, 2024 12:48 pm If there's a bijection between ( 1, 2, 3, ... } and { 0, 1, 2, 3, . . .) then they can move all of the guests to the right and make the first room empty.

But no such bijection exists.
Well, a Mathematician would tell you that the bijection is f(x) = x-1

1 -> 0
2 -> 1
3 -> 2
...

But what they are failing to disclose is that there's a bijection between f(x) = x -1 (when applied to {1, 2, 3, ... }) and g(x) = x (when applied to { 0, 1, 2, 3, . . .}).
Impenitent
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Re: Hilbert's hotel is utter nonsense

Post by Impenitent »

fine, I'll just sleep in the stairway...

-Imp
Magnus Anderson
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Re: Hilbert's hotel is utter nonsense

Post by Magnus Anderson »

Skepdick wrote: Fri Nov 01, 2024 1:11 pm Well, a Mathematician would tell you that the bijection is f(x) = x-1

1 -> 0
2 -> 1
3 -> 2
...
Yes, they would. But no such bijection exists. Bijection can only exist between sets of equal size. In this case, one of the sets is smaller.
Gary Childress
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Re: Hilbert's hotel is utter nonsense

Post by Gary Childress »

Skepdick wrote: Fri Nov 01, 2024 11:57 am Suppose that Hilbert's "infinite" hotel is full.
Suppose we want to make room for one more guest.
So we tell all guests to exit their rooms.
Fine...

There's a guest outside room 1; and a guest outside room 2; and a guest outside room 3...

There's a bijection (1:1 mapping) between guests and rooms.
Now we instruct every guest to move one room up (this alone is problematic - which guest moves first?)

Guest 1 is in room 2; guest 2 is in room 3; guest 3 is in room 4...

Room 1 is now empty.
There's now more room than guests.

This violates surjectivity and is no longer a bijection.
And just think, according to Zeno, before each guest moves to the next room, they must move an infinite number of halfway points to get there.
Magnus Anderson
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Re: Hilbert's hotel is utter nonsense

Post by Magnus Anderson »

Let the set of all hotel rooms be R = { R1, R2, R3, ... }.

Let the set of all guests that are currently in the hotel be G = { G1, G2, G3, ... }.

As stated at the start of the Hilbert's paradox, the hotel is full and each room has exactly one guest. This means the two sets are of equal size. For each room there is one guest.

R1 -> G1
R2 -> G2
R3 -> G3
...

But if we take the set R and remove from it room R1 to get the subset Rs = { R2, R3, R4, ... }, can we put Rs into one-to-one correspondence with the set of all guests G = { G1, G2, G3, ... } ?

In other words, can this subset of all hotel rooms accommodate all of our guests?

And that's pretty much what Hilbert's paradox is about. For if this is possible, we can make room R1 empty and accommodate a new guest.

So the question is:

Is there a one-to-one correspondence between the set Rs = { R2, R3, R4, ... } and the set G = { G1, G2, G3, ... } ?

They say there is.

R2 -> G1
R3 -> G2
R4 -> G3
...

But is that truly the case?

One cannot simply say, "Of course there is! There is this function f( x ) = x - 1 that maps every room index to every guest index."

How does one know that f( x ) = x - 1 actually exists? How does one know it's not a contradiction in terms, an oxymoron? Perhaps a useful one but still an oxymoron?

Two sets can be put into one-to-one correspondence if and only if they have the same exact number of elements.

Because the set Rs is basically the set R without room R1, and because the sets R and G are of the same size, the set Rs has one element less than R, which means, it is smaller than R. This means that there is no bijection between Rs and R. One guest must be roomless.

The common retort is that, because we can map ANY element from Rs to G, it follows we can map ALL elements from Rs to G. But just because we can map ANY element from Rs to G -- in fact, we can map any finite subset of elements from Rs to G -- it does not follow that we can map ALL of them.

The problem lies in the mistaken notion that if we take an infinite number and add 1 to it, we get the same exact number. This is based on the correct notion that if we add 1 to an infinite quantity, the result is still an infinite quantity, i.e. a quantity larger than every integer. But just because the resulting number is of the same type, it does not follow it's the same exact number.

That's all there is to it. The rest is just noise.
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