I have a question about the Chocolatier's game, which I had introduced in my recent answer to a question of Richard Stanley.

To recap the game quickly, the Chocolatier offers up at each stage a finite assortment of chocolates, and the Glutton chooses one to eat. At each stage of play, the Chocolatier can extend finitely the chocolate assortment on offer, and the Glutton chooses from those currently available. After infinite play, the Glutton wins if every single chocolate that was offered is eventually consumed.

In this post, I am interested in the version of the game where the Chocolatier is not allowed to repeat chocolate types — each new chocolate on offer is a uniquely exquisite new creation.

As I explain at the other the post, the Glutton clearly has a winning strategy, simply by keeping track of when new chocolates are added and making sure to organize the consumption so that every chocolate is eventually eaten. (And this idea works even when the Chocolatier is allowed to extend the offers countably infinitely, and not necessarily just finitely.)

Furthermore, in the case where there are only countably many possible chocolate types, then the Glutton has a winning strategy that depends only on the chocolates currently on offer, not requiring any knowledge of the game history. The strategy is simply to fix an enumeration of all the possible chocolate types and then eat the chocolate available that appears earliest in that enumeration. None could be left at the limit, since it would have been chosen once the earlier ones had been consumed.

Meanwhile, in the case where there are uncountably many chocolate types available, I had proved that the Glutton has no such strategy that depends only on the chocolates currently on offer.

My question is whether we can extend this argument also to allow the strategy to depend on the set of chocolates already eaten.

**Question.** Does the Glutton have a winning strategy in the
Chocolatier's game which depends only on the set of chocolates
currently on offer and the set of chocolates already eaten?

I conjecture a negative answer on sufficiently large uncountable sets and perhaps on all uncountable sets.

Here is an alternative equivalent formulation of the game, the catch-up covering game on a set $X$. The first player plays an increasing chain of finite subsets $$A_0\subset A_1\subset A_2\subset\cdots\subset X$$ and the second player chooses elements $a_n\in A_n$. After infinite play, the second player wins if $\bigcup_n A_n=\{a_0,a_1,a_2,\ldots\}$. Of course, the second player can win, by looking at the history of how elements were added to the sets, but the question is whether there is a winning strategy that at move $n$ depends only on the current set $A_n$ and the set of already-chosen elements $\{a_k\mid k<n\}$. The argument on the previous post shows that on an uncountable set $X$ there can be no such winning strategy that depends only on the difference set $A_n\setminus\{a_k\mid k<n\}$, which is the set of elements currently available for choosing anew.

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