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A tree $A\subseteq \omega^{<\omega}$ - possibly with dead ends - is pointed iff every path $p\in[A]$ has $p\ge_TA$. This lifts to two distinct notions of pointedness for closed sets in Baire space:

  • A closed $S\subseteq \omega^\omega$ is pointable if there is some pointed $A$ with $S=[A]$.

  • A closed $S\subseteq \omega^\omega$ is pre-pointable if for each $p\in S$ there is some tree $A_p\subseteq\omega^{<\omega}$ with $p\ge_TA_p$ and $[A_p]=S$.

These definitions also make sense for closed sets in Cantor space.

Now it's not hard to whip up a closed set in Baire space which is pre-pointable but not pointable ; however, the only argument I have at the moment breaks down for Cantor space. (Specifically, the set in Baire space we get has two elements, but in Cantor space compactness implies that pointed = pre-pointed for finite sets.) So my question is:

Is there a closed set in Cantor space which is pre-pointable but not pointable?

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  • $\begingroup$ What does $\leq_T$ mean? $\endgroup$ Commented Jul 13, 2019 at 16:05
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    $\begingroup$ @JamesHanson Turing reducibility. $\endgroup$ Commented Jul 13, 2019 at 16:51
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    $\begingroup$ Ah, I thought the subscript $T$ was the tree $T$. $\endgroup$ Commented Jul 13, 2019 at 18:02
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    $\begingroup$ @JamesHanson That is a bit annoying - I've changed the notation now. $\endgroup$ Commented Jul 13, 2019 at 22:24
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    $\begingroup$ @BjørnKjos-Hanssen No real motivation (as per usual, honestly) - I was just idly thinking about mass problems. A set of points determines two mass problems: the problem of finding an element of the set and the problem of finding a code (in whatever appropriate sense) for the set as a whole. Pre-pointability is then just saying that one problem reduces to the other, which is a neat structural property, and showing that it was nontrivial was a fun exercise (it's a sort of "toy tagged tree" forcing; nothing actually complicated, but a forcing condition is a finite tree with some ordinalish tags). $\endgroup$ Commented Jul 21, 2019 at 4:07

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Let $\mu$ be a measure on $2^\omega$ which doesn't have a least Turing degree.

This exists by Theorem 4.2 of

Day, Adam R.; Miller, Joseph S., Randomness for non-computable measures, Trans. Am. Math. Soc. 365, No. 7, 3575-3591 (2013). ZBL1307.03026.. Trans. Amer. Math. Soc., 365:3575–3591, 2013.

Let $S$ be the class of infinite sequences of reals $X_i$ that are Martin-Lof random with respect to the infinite product measure of $\mu$, $$\mu\times\mu\times\dots$$ with a fixed randomness deficiency constant $c$.

This $S$ is pre-pointable because by the Law of Large Numbers each element of $S$ can compute a representation of $\mu$.

If $S$ is pointable, it implies something close to saying that $\mu$ has a least Turing degree after all: there is a representation of $\mu$ that all the randoms compute.

Then we may use a de Leeuw, Moore, Shannon, Shapiro / Sacks Theorem:

Computability by Probabilistic Machines K. de Leeuw, E. F. Moore, C. E. Shannon & N. Shapiro Journal of Symbolic Logic 35 (3):481-482 (1970)

for $\mu$ to the effect that if a real $A$ [in particular $A$ could be a representation of $\mu$ or $S$] is computed by all sequences of mutually $\mu$-randoms, then $A$ is computed by all representations of $\mu$. This would be a generalization of a result for Bernoulli measures (which is are already infinite product measures) from

Kjos-Hanssen, Bjørn, Permutations of the integers induce only the trivial automorphism of the Turing degrees, ZBL06916706.

We also want a lemma to the effect that computing a representation of $\mu$ is equivalent to computing a representation of $S$. Computing a representation of $S$ from a representation of $\mu$ is I guess the familiar direction, similar to the case of Bernoulli measures. Computing a representation of $\mu$ from a representation of $S$ runs into the problem that some strings are dead ends and may mislead us. However we may again use a certain effectivity in the LLN to (nonuniformly) bound the rate of convergence and hopefully overcome this.

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