- #1

mathmari

Gold Member

MHB

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I look at the problem with hilbert's hotel and the busses. First one bus with infinitely many guests, then four busses with infinitely many guests and then infinitely many busses with infinitely many guests.

At each case we move all the guests that are already in the hotel to the odd room numbers.

So the even room numbers are free for the new guests.

In the first with one bus we use the formula $n = 2(i-1)+2$ where $i\in \mathbb{N}\setminus\{0\}$ is the number of place in the bus and $n$ is the even room number that this guest will get. In the second case with the four busses we use the following:

\begin{equation*}\begin{matrix} & 1. \text{ Bus } & 2. \text{ Bus } & 3. \text{ Bus } & 4. \text{ Bus } \\ 1. \text{ Person } & \text{ Room } 2 & \text{ Room } 4 & \text{ Room } 6 & \text{ Room } 8 \\ 2. \text{ Person } & \text{ Room } 10 & \text{ Room } 12 & \text{ Room } 14 & \text{ Room } 16 \\ 3. \text{ Person } & \text{ Room } 18 & \text{ Room } 20 & \text{ Room } 22 & \text{ Room } 24 \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ i. \text{ Person } & \text{ Room } 8(i-1)+2 & \text{ Room } 8(i-1)+4 & \text{ Room } 8(i-1)+6 & \text{ Room } 8(i-1)+8\end{matrix}\end{equation*} In the last case with the infinitely many busses we use the formula $2^{i}\left (2(j-1)+1\right )$ where $i$ is the place in the bus and $j$ is the number of the bus.

Using this we get \begin{equation*}\begin{matrix} \text{Place}/\text{Bus} & 1. \text{ Bus} & 2. \text{ Bus} & 3. \text{ Bus} & 4. \text{ Bus} & 5. \text{ Bus} & \ldots \\ 1 & 2 & 6 & 10 & 14 & 18 & \ldots \\ 2 & 4 & 12 & 20 & 28 & 36 & \ldots \\ 3 & 8 & 24 & 40 & 56 & 72 & \ldots \\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \end{matrix}\end{equation*}

I want to describe the injectivity and the surjectivity of the maps, and if the maps for the old and the new guests are injective and surjective.

The map for the old guests is injective, since it means that the new room thet they get either had no previous guest or one.

The same holds also for the map for the new guests. Their new rooms either had no previous guests or one.

As for the surjectivity. As for the rooms of the old guests, only the odd room umbers have a preimage and so the map is not surjective. As for the rooms of the new guests, only the even room umbers have a preimage and so the map is not surjective. Is that correct? (Wondering)