Question

Hilbert’s hotel and different sized infinities

Answer 1

Surely no. Hilbert's hotel is necessarily countable, so has nothing to say about uncountable infinity. While it's true that each member could label themselves with a real number as well, bear in mind that the hotel is already infinitely large at the point that we construct a new member R.

It's precisely the point of the diagonal argument that even if you construct an infinite list of real numbers, you can always expand upon it arbitrarily often, hence showing that uncountable > countable

Answer 2

> We can easily obtain a new real number by using the diagonalization proof technique.

Yes we can. And after we obtain that new real, we can make a new list the same way we move the guests in Hilbert's Hotel.

So now we have a new countable list of real numbers. And Cantor tells us it's incomplete, we can always find another.

You're saying that somehow we can get all the real numbers into rooms this way, but you're assuming what you're trying to prove. Cantor's proof shows us that you can never have all the real numbers assigned to a room, there's always another one you left off the list.

Answer 3

Not really. Let's see.

Assume, without loss of generality, that all guests are real numbers within [0, 1], just to apply the diagonalization properly.

If the (countably) infinite hotel is full, one can use diagonalization to create a new guest and put it in the hotel.

But... How many guests can be created, given a specific guest list? The answer is lim (n -> oo) 9^n = 9^N, since, for each digit of the diagonal number, there are 10 - 1 = 9 choices of a substitute digit. A set of cardinality 9^N is already known to be uncountable, because 9^N > 2^N, the cardinality of base-2 numbers from 0 to 1.

In short: you can use diagonalization to add ever more guests to the Hilbert Hotel, but only a countable set; the vast majority of the waiting list won't ever enter.