Hilbert’s Hotel 2.0: Is Infinity Odd or Even?

Even more thought experiments on infinity

In Achilles, the Tortoise and the Hotel: Exploring the Infinite, we discussed different kinds of infinity. We saw how an hotel with a countably infinite number of rooms (Hilbert’s Hotel) can always make room for a finite number of guests, even if every room is already occupied. What’s more, we saw how this hotel can fit every guest from a bus with countably infinite seats, and even every guest from a ferry with a countably infinite amount of such buses, and every guest from a countably infinite amount of such ferries, and so on. The only thing Hilbert’s Hotel can’t do is fit everyone from a bus with an uncountably infinite number of guests.

Today, I want to expand on this thought experiment. Let’s say Hilbert’s Hotel has a special lamp: one that is controlled by a countably infinite amount of switches, one in every room of the hotel. For each switch it’s true that

• if the lamp is on, flipping the switch turns it off
• if the lamp is off, flipping the switch turns it on.

So far, so good. Now let the guests do the following experiment, starting with the lamp being off:

1. Guest 1 (the guest staying in Room 1) flips the switch, turning the lamp on.
2. 1 hour later, Guest 2 flips the switch, turning the lamp off.
3. Another 1/2 hour later, Guest 3 flips the switch, turning the lamp on again.
4. Another 1/4 hour later, Guest 4 flips the switch, turning the lamp off again…
5. And so on, and so on.

We can prove that, if the guests go on like this, after 2 hours, every guest has flipped the switch. (Yes, 1 + 1/2 + 1/4 + 1/8 + … = 2, exactly. We will in fact prove this later.)

We could say off. After all, Guest 1 turned the lamp on, and each switching on of the lamp is followed by switching the lamp off.

However, this argument works both ways. After Guest 1 turns the lamp on, Guest 2 turns it off, and each switching off of the lamp is followed by switching the lamp on again. Infinity is weird (and really awesome) this way.

So what’s the answer? Well, this problem is known as Thomson’s lamp, and there is no obvious answer. The on and off switching of the lamp is analogous to the sequence 1–1 + 1–1 + 1 — … which is divergent: it doesn’t converge towards (as in, get closer and closer to) a particular value.

To my mind, though, the question is Thomson’s lamp on or off? is equivalent to another question:

And it’s not like we can’t in principle answer such questions about infinity. For example consider the following problem.

As usual, Hilbert’s Hotel is fully occupied. But starting from there, every week, 2 buses show up. At Week 1, Bus A takes Guests 1 to 10 from the hotel back home, whereas Bus B brings exactly 1 guest to stay in Room 1. At Week 2, Bus A takes Guests 11 to 20 from the hotel back home, and Bus B brings a guest for Room 2. In general, at Week n, Bus A takes Guests n to n + 9, and Bus B brings Guest n. After a countably infinite amount of such weeks, is Hilbert’s Hotel fully occupied?

There are 9 more people going away than coming in each week, so you might resolve this problem —derived from the Ross-Littlewood paradox — as “no”. But, well, infinity is being weird again. We can prove that after a countably infinite number of weeks, every room is occupied.

How?

Well, ask yourself this: is Room 1 occupied? Yes, at Week 1, the room was temporarily freed by Bus A, but Bus B brought a new guest for Room 1.

Is Room 2 occupied? Yes, it got reoccupied at Week 2.

Is Room 3… Well, you get the picture. We can show that for every natural number n, Room n is occupied. So every room is occupied.

Can’t we use a similar argument for Thomson’s lamp?

Well, let’s try. Change the bus problem to one where in Week n, Bus A takes Guest n and Bus B brings in a new Guest n. Taking a guest is analogous to switching the lamp on, whereas bringing in a guest is analogous to switching it off. Since we can prove — as before — that every hotel room will be occupied after a countably infinite number of weeks, that would suggest the lamp is off.

But the problem here is this: what if taking a guest is analogous to switching the lamp off, and bringing in a guest is analogous to switching the lamp on? Then, after the lamp is switched on the first time, the above argument proves the exact opposite conclusion: the lamp is on!

So Thomson’s lamp has no obvious answer, but we can prove that 1 + 1/2 + 1/4 + 1/8 + … = 2. Let’s.

First, we want to show that

Let’s first subtract 1 from both sides:

Note that

And we can go on like this forever. Note that the last 2 terms on the left are always equal. So in general, we get

which is equal to

Note that, for n approaching infinity, the left half of the equation is equal to the left hand side of what we want to prove:

And — again for n approaching infinity — the right hand side of the same equation

approaches 1, which is the right hand side of the equation we want to prove. In other words,

which — for n approaching infinity — implies

and therefore

Thanks for reading! And a special thank you to Ahmed S. Attaalla for his explanation of the proof here.