# Can Hilbert's grand hotel accommodate *infinite* layers of infinity?

Suppose you have a hotel with infinite rooms, housing an infinite number of guests. The hotel can never be full no matter how many $$n$$ new guests arrive, because they can always kick out each current guest staying in room $$i$$ and make them move to room $$i+n$$. Even if an infinite number of new guests show up, the hotel can never get full, because you could just move the guest at room $$i$$ to room $$2i$$ and have the $$j$$th new guest move to room $$2i-1$$.

Now, this hotel can never get full even if there are an infinite number of coaches, each containing an infinite number of passengers (2 layers of infinity). There are several ways to do this, whether by:

• ...taking the powers of prime numbers...
• ...interleaving the digits of the coach numbers and seat numbers, or...
• ...treating the hotel like a triangle and putting the passengers on each coach (and the current guests) in different diagonals (with each guest already in the hotel at room $$i$$ moving to the room numbered as the $$i$$th triangular number).

These algorithms can be extended to further layers of nesting, e.g. an infinite number of ships, each containing an infinite number of coaches, each containing an infinite number of passengers (3 layers of infinity).

However, my question is, does this apply to infinite layers of nesting? Like, every passenger, coach, ship, etc. is "on" something, no matter how many layers you "go up"?

The reason I ask is because the Wikipedia article on Hilbert's paradox of the grand hotel used to say this, unsourced:

#### Infinite layers of nesting

Although a room can be found for any finite number of nested infinities of people, the same is not always true for an infinite number of layers, even if there are a finite number of people that exist at each layer.

However, I was just reading that article again, and noticed that this bit got removed only a few days ago (2 Sep. 2024), for being inaccurate. Here is the edit summary of the removal:

[A]ctually, remove this as unsourced and likely wrong. Each person in this situation would still have a finite "address", the set of which is countable. (1.234, 1.23456, etc.) The Cartesian product of countably many countable sets is still countable as well.

(The same editor added "because Cantor's diagonal argument becomes applicable" to the original bit, but then did away with it entirely.)

I don't find the editor's explanation too convincing, considering how, for instance, $$1.23456$$ is really just $$1.234560000000000...$$, and so on. There are countably infinite natural and whole numbers (in other words, the cardinality of these sets is $$\aleph_0$$). And there are countably infinite rational numbers (as demonstrated by the square lattice, or the Calkin-Wilf tree or the Stern-Brocot tree if you don't like duplicates), and the situation of that is reminiscent of the 2-layered infinity situation (infinite coaches, each with infinite guests): passenger $$p$$ of coach $$q$$ can be seen as analogous to the ratio $$\frac{p}{q}$$. Yet there are uncountably infinite real numbers (cardinality = $$c$$, and this can be shown through Cantor's aforementioned diagonal argument.

So who's right?