Hilbert’s Hotel
Consider a hotel with inifinitely many rooms, all of which are occupied. If we move the guest in room k to room k+m (for k=1,2,…), then we will get m new empty rooms. If we move the guest in room k to room 2k, we will get infinitely many new empty rooms. It seems empty rooms can be created from nowhere.
The original status of the hotel can be modeld by the series:
\[(1-1)+(1-1)+\cdots=0.\]And the status after the “m-step move” becomes:
\[1+1 \cdots +1 + (1-1) + (1-1) +\cdots = m.\]Although the latter can be obtained from reordering the former, these two series yield different sums.
The main issue is: When a series is not absolutely convergent, one cannot manipulate the terms and expect the sum to remain unchanged.
Thomson’s Lamp
Consider this one-hour task: flip the switch of a lamp after 30 minutes, then flip it again after another 15 minutes, then after 7.5 minutes, and so on. These time intervals form a geometric series, and thus, after one hour, the switch will have been flipped infinitely many times. The question is: what is the status of the lamp at the end?
The procedure can be described by the series $+1-1+1-1\cdots$, which diverges. Hence, the final status, as a limit, is not well-defined.
The main issue is: The “geometric time compression” trick rescales infinity to finite time. Our intuition (built on finite-times tasks) tends to suggest the lamp should have a deterministic, well-defined final state.
Ross’s Vase
Consider infinitely many balls numbered $1, 2, 3, \ldots$ and this one-hour task: After 30 minutes: add balls 1-10 to the vase, remove ball 1; After 15 more minutes: add balls 11-20 to the vase, remove ball 2; And so on. The question is: How many balls remain in the vase after one hour?
Two contradictory answers emerge: (1) ininfity, as at each step 9 more balls are added; (2) zero, as every ball is removed after finitely many steps.
This problem can be modeled by the following infinite matrix:
| Step | Ball 1 | Ball 2 | Ball 3 | ⋯ |
|---|---|---|---|---|
| 1 | 0 | 1 | 1 | ⋯ |
| 2 | 0 | -1 | 0 | ⋯ |
| 3 | 0 | 0 | -1 | ⋯ |
| ⋮ | ⋮ | ⋮ | ⋮ | ⋱ |
where entries are $+1$ for adding a ball, $-1$ for removing a ball, and $0$ for no action.
The first answer corresponds to summing rows first, then columns (net change per step). The second answer corresponds to summing columns first, then rows (fate of individual balls).
The main issue is: The infinite double sum is not absolutely convergent. Hence, different orders yield different results. Meanwhile, under the geometric time compression trick, our intuition tends to suggest this process should have a definite outcome.
Final thought: our intuition often fails us in the infinite regime, as we’ve never actually encountered infinity in the physical world. So, be rigorous when handling infinity.