In [this question](https://mathoverflow.net/questions/151286/probabilities-in-a-riddle-involving-axiom-of-choice) the following observation was made:


Consider a sequence of boxes numbered 0, 1, ... each containing one real number. The real number cannot be seen unless the box is opened. 


Define a *play* to be a series of *steps* followed by a *guess*. A *step* opens a set of boxes. A *guess* guesses the contents of an unopened box. A *strategy* is a rule that determines the steps and guess in a play, where each step or guess depends only on the values of the previously opened boxes of that play. 


Then for every positive integer $k$, there is a set $S$ of $k$ strategies such that, for any sequence of (closed) boxes, there is at at most one strategy in $S$ that guesses incorrectly.


My question is this: Can $k$ be countably infinite (instead of a positive integer)? If not, is there a proof?


[Edit: the original question also asked whether $k$ can be uncountable; this was answered by Dan Turetsky in the negative in comments].


The best I have been able to show is that, if the function $f:\mathbb{N}\to\mathbb{R}$ defined by the contents of the initial sequence of boxes is recursive (viewing elements of $\mathbb{R}$ as binary sequences), then $k$ can be countably infinite. To see this, call a subset $X$ of $\mathbb{N}$ *signature* if two recursive functions on $\mathbb{N}$ that eventually agree on $X$ also eventually agree on $\mathbb{N}$. (Two functions "eventually agree" if they differ in finitely many places). Call two Turing Machines *equivalent* if their associated functions on $\mathbb{N}$ are equivalent (that is, eventually agree). A diagonalization argument on the class representatives of the Turing Machines yields an infinite partition $U$ of $\mathbb{N}$ into signature subsets. The $i$'th strategy in $S$ first opens all the boxes whose indices are not in the $i$'th element $U\_i$ of $U$, determines the class representative Turing Machine T that generates the resulting values on the opened boxes for boxes whose indices are greater than $N$ (for some positive $N$), and guesses that a box with index greater than $N$ and in $U\_i$ has a value specified by $T$. 


However, I have not been able to modify this for the non-computable case.



Can an infinite number of mathematicians guess the number in a box with only one error?
It is possible to have every mathematician guess the number in one of the boxes with at most one error.




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Partition the natural numbers into countably many sets, $\{S\_i\}\_{i=0}^\infty$, where each $S\_i=\{n\_{i\_1},n\_{i\_2},\dots,\}$ is countably infinite. (There are many ways to do this) Since we have countably many mathematicians, we may list them, and assign $S\_i$ to the $i^{th}$ mathematician.


If $u\_k$ denotes the real number in the $k^{th}$ box, then the $i^{th}$ mathematician will be assigned the sequence of real numbers $u\_{n\_{i\_j}}$, for $j=1,2,3\dots$. Using the axiom of choice, we may chose a representative for each equivalence class of sequences of real numbers under the equivalence relation $\{u\_n\}\_{n=1}^\infty\equiv\{v\_n\}\_{n=1}^\infty$ if there exists $M>0$ such that $v\_n=u\_n$ for all $n>M$. Thus, for the $i^{th}$ mathematician there will exist an integer $M\_i$ such that for all $j>M\_i$, the sequence $u\_{n\_{i\_j}}$ is equal to the representative of its equivalence class. The goal is to have mathematician $i$ guess an integer $H\_i>M\_i$ by looking at every box except those in the set $S\_i$. If this happens, then mathematician $i$ may look at all of the elements of $u\_{n\_{i\_j}}$ with $j\geq H\_i+1$, determine the equivalence class, and guess the box with $j=H\_i$. Since $H\_i>M\_i$, his guess will be correct. It follows that we need all but possibly one mathematician to guess an integer $H\_i>M\_i$. If the sequence $M\_i$ is bounded, then the problem is easy. The difficulty is handling an unbounded sequence $M\_i$. 


Under the same system of representatives, the sequence $\{M\_i\}\_{i=0}^\infty$ lies in some equivalence class of real numbers. Since mathematician $i$ knows the value of $M\_l$ for all $l\neq i$, each mathematician can determine the equivalence class of the sequence $\{M\_i\}\_{i=0}^\infty$. Let $\{v\_i\}\_{i=0}^\infty$ denote the representative of this equivalence class. Then there exists an integer $N$ such that for every $i>N$, $M\_i=v\_i$. Mathematician $i$ with $i\leq N$ can determine $N$, however each mathematician with $i>N$ only knows that $N\leq i$. The strategy for guessing is as follows: Assign to mathematician $i$ with $i>N$ the integer $$H\_i=1+\max\{v\_i,M\_{i-1},M\_{i-2},\dots,M\_1,M\_0\},$$ and to each mathematician with $i\leq N$, the integer $$H\_i=1+\max\{M\_{N},M\_{N-1},\dots,M\_{i+1},M\_{i-1},\dots,M\_1,M\_0\}.$$ Then we must have $H\_i>M\_i$ for every $i$ except possibly one. Thus, we have set up a strategy which allows every mathematician except possibly one to guess some box correctly.


