# More on infinitary decision strategies

I’ve mostly lost interest in further studying the infinitary version of the problem. The simple conclusion seems to be that the structure of the uncountable case is significantly more complicated than the structure of the set of measures on a sigma-algebra, so there’s probably not much more classification one can do. Here are some brief notes on examples:

Lemma: Every set admits a subset consistent strategy
Proof: Every set can be well-ordered. Let $$s(U)(V) = 1$$ if $$\mathrm{min}(U) \in V$$, else $$0$$.

Theorem: Every measure $$m$$ can be extended to a subset consistent strategy. That is, there is some subset consistent strategy with $$s(S) = m$$.
Proof:

Let $$s$$ be a subset consistent strategy on $$S$$. Define $$s'(U)(V) = \frac{m(U \cap V)}{m(U)}$$ if $$m(U) > 0$$, else $$s(U)(V) = s'(U)(V)$$.

That this works is a simple matter of case checking.
Let $$U \subseteq V \subseteq W$$. We want to show $$s(W)(U) = s(W)(V) s(V)(U)$$.

If $$m(W) = 0$$ then $$s = s’$$ on all the sets in question, so this follows from $$s’$$ being subset consistent. Else if $$m(V) = 0$$ then $$S(W)(V) = S(W)(U) = 0$$ so both sides are 0. Else $$m(W)(U) = \frac{m(U)}{m(W)} = \frac{m(U)}{m(V)} \frac{m(V)}{m(W)} = s(U)(V) s(V)(W)$$ as desired.

QED

In fact every subset consistent strategy may be decomposed nearly this way. The only caveat is that the subset consistent strategy $$s’$$ ends up being on the set $$S’ = \{x : m(\{x\}) \neq 0\}$$.

As proof, just take $$m = s(S)$$ and $$s’ = s|_{S’}$$.

However somewhat self-evidently this operation isn’t idempotent: Because in the construction we could have taken $$s’$$ to be just about anything we wanted, we can do this again and again.

Here’s an interesting example: Let $$A$$ be a well ordered set. Let $$S = \bigcup I_a$$ where $$I_a$$ is a copy of the unit interval $$[0,1]$$. Additionally, fix some well-ordering of $$S$$. Give this the $$\sigma$$-algebra generated by the measurable sets of each $$I_a$$.

Lemma: Every measurable set in this $$\sigma$$ algebra intersects either countably or coucountably $$I_a$$.
Proof: The set of sets which do that is closed under countable unions and complements.

Define $$s(U)(V) = \frac{m(U \cap I_a \cap V)}{m(U \cap I_a)} where \(a$$ is minimal such that $$I_a \cap U \neq \emptyset$$. If there is no such $$a$$ then instead use the well-ordering’s strategy.

Claim: If $$A = \omega_2$$ then given this $$s$$ there is no well-ordered sequence of decompositions as above that gives $$s$$.

Proof: Well, I haven’t really proved it to be honest. It has plausibility though. Here’s my handwavey reasoning:

The decomposition of a set in this involves just knocking off the first initial interval. So you can’t decompose it in fewer than $$\omega_2$$ steps. There are then $$\omega_1$$ values of $$I_a$$ which the $$\omega_1$$th step does not intersect, which is not co-countable.

QED

I’ve done some additional work on the partial orders on sets that such strategies define, but it doesn’t seem to go anywhere very fruitful. I’m going to consider this problem closed for now.

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