1
$\begingroup$

In Set Theory Jech defines a cone to be a subset of the Baire Space $\mathcal{N}$ of the form $$\operatorname{cone}(x_0)= \{x : x_0 \in L[x]\}$$ where $x_0 \in \mathcal{N}$. Jech then defines the equivalence relation $\equiv$ with $$x \equiv y \iff (x \in L[y] \land y \in L[x]).$$ In a similar fashion, in Set Theory, Schindler defines a Turing cone to be a subset of the Baire Space of the form $$\operatorname{cone}_T(x_0) =\{x : x_0 \leq_T x \}$$ where $\leq_T$ denotes Turing reductibility. Schindler then defines the equivalence relation $\equiv_T$ with $$x \equiv_T y \iff (x \leq_T y \land y \leq_T x).$$ They both want to prove essentially the same thing, which is:

Assume $AD$. Then if $A \subset \mathcal{N}$ is a $\equiv$(resp. $\equiv_T$)-closed subset, then either it or its complement contains a (resp.Turing)cone.

However, what they both do is pick for instance $\sigma$ a winning strategy for $I$ in $G_A$, and then show that $\operatorname{cone}(\sigma) \subset A$. Now of course, this isn't well defined, as $\sigma \notin \mathcal{N}$ (obviously). Now you can of course widen the notion of cones, but then the risk is losing a few characteristics of the filter you're trying to build (as for instance its non principality). How do you do this well, to solve this issue and not lose the properties that you require ?

Improve this question
$\endgroup$
15
  • 4
    $\begingroup$ Strategies $\sigma$ are easily coded by elements of the Baire space and are usually identified with the corresponding element. (In a similar fashion, elements of the Baire space are usually called reals.) $\endgroup$
    – Simon Thomas
    Commented Dec 20, 2016 at 13:53
  • 2
    $\begingroup$ It's recursive. $\endgroup$
    – Simon Thomas
    Commented Dec 20, 2016 at 13:59
  • 1
    $\begingroup$ Also note that $L[x]$ is not the least inner model containing $x$ - the least inner model containing $x$ is $L(x)$. $L[x]$, in contrast, is the least inner model closed under $y \mapsto y \cap x$. However, for $x \in \mathcal{N}$, these notions agree, i.e. $L[x]= L(x)$. $\endgroup$
    – Stefan Mesken
    Commented Dec 20, 2016 at 15:01
  • 3
    $\begingroup$ @Stefan: For sets of ordinals the two models coincide. $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 20, 2016 at 15:33
  • 1
    $\begingroup$ @Asaf Yes, I'm aware. However, since $x \in \mathcal{N}$ is not a set of ordinals in Ralf's book, this seemed more relevant to OP. The underlying reason though, that $L[x] = L(x)$, is - as you hinted - that it can be recursively coded as a subset of $\omega$. $\endgroup$
    – Stefan Mesken
    Commented Dec 20, 2016 at 15:44

1 Answer 1

Reset to default
4
$\begingroup$

Simon Thomas already mentioned that a winning strategy can be recursively coded into a real. I thought it might be a good idea to write down one such coding explicitly:

Let $A \subseteq \mathcal{N}$ and let $\Sigma$ be a winning strategy for $G_{A}$ (say for player $II$, the other case is virtually the same). Then $$ \Sigma \colon \bigcup_{n < \omega} \mathbb{N}^{2n+1} \to \mathbb N. $$ Let $\mathbb P = \{ p_i \mid i < \omega \}$ be the increasing enumeration of all primes and let $$ \Sigma^c\colon \mathbb N \to \mathbb N, \prod_{p \in \mathbb P} p^{n_p} \mapsto \Sigma(n_3, \ldots, n_{p_{(2 \cdot n_2 + 1)}}). $$ Then $\Sigma^c \in \mathcal{N}$ is a recursive coding of $\Sigma$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .