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I just finished watching this video and I feel like I'm missing something very obvious. I've read quite a number of the questions here but haven't been able to figure out what.

I assume Hilberts hotel is relatively well known. So I won't go into the details. My question is regarding two cases:

case $1$: a bus with infinitely many people shows up. Everyone is accommodated by moving to the room with double their current room number, leaving infinite empty rooms.

case $2:$ a buss with infinitely many people shows up, this time everyone is labelled with an infinitely long string consisting out of $A$ and $B:$

$- AAAA...$

$- BAAA...$

$- ABAA...$

  • etc.

The video explains that these people cannot be accommodated because Cantor's diagonal proof

In my mind, I would flip everyone's names, and switch out A for 0 and B for 1, then translate this from binary to a regular number:

$- AAAA... → ...000 → 0$

$- BAAA... → ...001 → 1$

$- ABAA... → ...010 → 2$

  • etc.

And now you could apply the original method again of moving everyone up to a room of double their current room number. I fail to see why this wouldn't work.

The argument in the video is that there cannot be an exhaustive list of the passengers of the second bus because of earlier mentioned Cantor's diagonal. However, it seems to me you could just as well use Cantor's diagonal in the first bus, as long as you allow for leading zeros.

Could someone explain what the difference is between the first and second bus that I'm missing?

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    $\begingroup$ Some people in the second bus may have an infinite number of Bs (or $1$s) so you do not get a finite integer from your flipping $\endgroup$
    – Henry
    Commented Aug 10, 2022 at 16:51
  • $\begingroup$ @Henry that's it, devil is in the details... If you post it as answer I'll accept it. $\endgroup$
    – Nathan
    Commented Aug 10, 2022 at 16:54
  • $\begingroup$ Quite... you would be able to assign a legal "number" only to those people whose names contained finitely many B's. That is a very small proportion of the people on your bus. Any one else from that bus would not have an actual "number" given to them but rather some object more complicated than a number. $\endgroup$
    – JMoravitz
    Commented Aug 10, 2022 at 16:55

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From comment:

Some people in the second bus may have an infinite number of Bs (i.e. $1$s when translated)

so for them you would not get a finite integer from your flipping

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