Murder at Hilbert's Hotel!

Solution 1:

Well, let's look at the structure of the problem:

There is a set $S$ of suspects (three in the original problem, a countably infinite number of them in Hilbert's hotel).

There's a subset $G\subset S$ of guilty suspects.

And there's a mapping $f:S\to P(S)$ where $P(S)$ is the power set (set of subsets) of $S$, where $M\in f(s)$ means "If $s$ says the truth, it is possible that $G=M$". $f(s)$ is specified by a logical form $L_s$, that is, $f(s) = \{M\in P(S): L_s(M)\}$.

Now we can formulate the separate questions as follows:

a) Is $\bigcap_{s\in S} f(s) \ne \emptyset$?

b) For which pairs $(s,t)\in S\times S$ do we have $f(s)\subseteq f(t)$?

c) Assuming $G=\emptyset$, what is $\{s\in S: G\notin f(s)\}$?

d) What is $\bigcap_{s\in S} f(s)$? (Actually, the question as formulated already presumes that this set has exactly one element; especially it assumes that the answer to (a) is "yes").

e) What is $\bigcap_{s\in G} (P(S)\setminus f(s)) \cap \bigcup_{s\in S\setminus G} f(s)$?

So to generalize the problem to Hilbert's hotel, you have to find a function $f(n)$ specified by a logical formula dependent on $n$ such that $\bigcap_{s\in n} f(n)$ has exactly one element, and (to be a generalization of the original problem) reduces to the original problem when restricted to the set $\{0,1,2\}$

Let's look closer at the original testimonies:

• Brown gives an explicit list of who's guilty or innocent.
• Jones gives an implication connecting the other two.
• Smith makes a testimony about himself, and the claim that someone is guilty.

Associating $0$ with Brown, $1$ with Jones and $2$ with Smith, we could write those as follows in the formalism derived above, with $S=\{1,2,3\}$ $$\begin{align} f(0) &= \{M\in P(S): 1 \in M \land 2\notin M\}\\ f(1) &= \{M\in P(S): 0 \in M \implies 2\in M\}\\ f(2) &= \{M\in P(S): 2 \notin S \land M\ne\emptyset\} \end{align}$$

There are of course many ways to generalize that, but let's try the following: $$f(n) = \begin{cases} \{M\in P(\mathbb N): \forall m > n, m\in M\iff m \equiv 1\ (\mod 2)\} & n \equiv 0\ (\mod 3)\\ \{M\in P(\mathbb N): \forall i < n, \forall k > n, m\in M\implies k\in M\} & n \equiv 1\ (\mod 3)\\ \{M\in P(\mathbb N): n\notin M\land M\ne\emptyset\} & n\equiv 2\ (\mod 3) \end{cases}$$ However this gives an inconsistent set of conditions (i.e. $\bigcup_{n\in\mathbb N} f(n)=\emptyset$), since from $f(0)$, one concludes $5\in G$, but from $f(5)$ one concludes $5\notin G$. This is a deviation from the original problem where the statements are indeed consistent.

I'm not going to spend the time to actually find a proper generalization now (I already spent far more time on this answer than originally planned), but I think the mathematical concepts involved should now be clear.

Solution 2:

It sounds to me like you are asking about Infinitary logic. I've pondered this idea myself a fair bit. For instance, we can make sense of the 'limit object' of this sequence $$ a_1 \wedge a_2, (a_1 \wedge a_2) \wedge a_3, (((a_1 \wedge a_2) \wedge a_3) \wedge a_4$$ where $\wedge $ denotes logical and. In this case the limit object has a value of true if and only if every $ a_n $ is true, otherwise it is false. But what if we replace those and's with logical implication, $\Rightarrow $?