I have saved it for later! ]]>

let the axioms be those of NBG relativized to $u(x)$, where we replace “$y$ is a set” by “$y\in u(x)$” and “$y$ is a class” by “$y\in u(u(x))\wedge \forall z\in y. z\in u(x)$”, except for the axiom of infinity (which we won’t need, because $u(\emptyset)$ contains all the hereditarily finite sets anyway). ]]>

(a) the law of excluded middle is not quite provably false, but it is possible to prove (in ZF) that the only model of HoTT+LEM is trivial;

(b) the axiom of choice is a theorem, because the most natural wording of “nonempty set” is “pointed type”, for which AC obviously holds.

Then the argument fails to go through, because all functions are continuous, and any function from $S^Z$ to itself that sends equivalent elements (equal at infinitely many points) to the same element is heavily discontinuous under the product topology, as long as $S$ is disconnected.

There is actually a nice proof that all but the first prisoner (who cannot do better than by guessing) can survive:

Simply assign a group structure to the set of colors (there are arbitrarily large groups by LĂ¶wenheim-Skolem and the existence of $\Bbb Z$), select an element $E(C)$ of each equivalence class,

then prisoner $n$ computes $Q_n=\prod_{i=n+1}^\infty (C_iE(C)_i^{-1})$. This exists because all but finitely many factors are trivial.

Then set $A_0=Q_0$, $A_{n+1}=A_nQ_{n+1}^{-1}E_{n+1}$,

where prisoner $n$ can compute $A_n$ and announce it after hearing $A_{n-1}$, and $A_n=C_n$ if $n>0$ by simple computation.

The Axiom of Choice is Wrong | The Everything Seminar

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