# On the absoluteness of higher Borel sets?

Consider the higher Cantor space $$2^\kappa$$ with the $${<}\kappa$$-box topology ($$\kappa$$ at least inaccessible). This canonically defines the notion of higher Borel sets.

A higher Borel code $$\mathbf{B}$$ is a wellfounded tree of size $$\kappa$$ consisting of finite sequences, such that terminal nodes are label with basic clopen sets and non terminal nodes are label with either $$\bigcup$$ or $$\bigcap$$ .

Let $$V \subsetneq V'$$ be models of ZFC* (large enough fragment of ZFC), let $$V \vDash$$ " $$\mathbf{B}_1 , \mathbf{B}_2$$ are higher Borel codes " , and $$V \vDash$$ " $$\mathbf{B}_1 \cap \mathbf{B}_2 = \emptyset$$ " . Does this also hold in $$V'$$ if $$\kappa$$ remains a large cardinal?

• Sorry, I was being silly - your original formulation was fine, I parsed "$\bigcap$" as "$\cap$." – Noah Schweber Mar 13 '19 at 16:07
• No problem. I changed it back, because if you want to take complements, you have to restrict that there is only one successor... – Johannes Schürz Mar 13 '19 at 16:16

In the case of forcing extensions by $${<}\kappa$$-complete posets (which you probably meant), we have $$\Sigma^1_1$$-absoluteness. (This is well known / folklore, see e.g. Friedman Khomskii Kulikov, Lem 2.7). Strategic closure is sufficient. And Borelcode evaluates to something nonempty'' is $$\Sigma^1_1$$.

The answer is no. The motivation is this: Borel codes in $$2^\kappa$$ are essentially formulas in the propositional infinitary logic $$L_{\kappa^+,0}$$ with propositional symbols $$\langle P_\alpha : \alpha < \kappa\rangle$$, but for uncountable $$\kappa$$, the satisfiability of an $$L_{\kappa^+,0}$$ formula is not absolute because $$\kappa$$'s cardinality can be changed. I'll just translate this into your context.

To avoid coding, we work with $$2^{\omega\times \kappa}$$ instead. Let $$A_{n,\alpha}$$ denote the clopen set of $$\chi\in 2^{\omega\times \kappa}$$ with $$\chi(n,\alpha) = 1$$. There is a higher Borel code $$B$$ that in any "extension of the universe" is interpreted as the collection of $$\chi\in 2^{\omega\times \kappa}$$ such that the set $$f= \{(n,\alpha) : \chi(n,\alpha) = 1\}$$ is a partial function from $$\omega$$ onto $$\kappa$$. To say that $$f$$ is a partial function: $$\bigwedge_n\bigwedge_\alpha\bigwedge_{\beta\neq \alpha} (A_{n,\alpha}\Rightarrow \neg A_{n,\beta})$$. To say $$f$$ is surjective: $$\bigwedge_{\alpha < \kappa}\bigvee_{n < \omega} A_{n,\alpha}$$. Then $$B$$ is the conjunction of these clauses. (I used partial functions only to avoid having to include the clause that says $$f$$ is total.)

In $$V$$, $$B$$ is interpreted as the empty set since $$\kappa$$ is uncountable, but in any extension of the universe where $$\kappa$$ is countable, $$B$$ is interpreted to be nonempty.

Edit: if you don't want to change cardinal structure (and you assume $$\kappa >\mathfrak c$$) then you can Borel code the set of sequences $$\chi\in 2^\kappa$$ such that $$\chi\restriction \omega\notin V$$. This is even easier: $$\bigwedge_{x\in 2^\omega} \bigvee_{n < \omega} \chi(n) \neq x(n)$$.

The question becomes interesting if you insist that in the extension $$V\subseteq V'$$, no bounded subsets of $$\kappa$$ are added and $$\kappa$$ remains inaccessible (if you are allowed to singularize $$\kappa$$, then Prikry forcing works).

• But there is absoluteness when the cardinal structure up to $\kappa^+$ is preserved, right? – Asaf Karagila Mar 14 '19 at 7:52
• That's what I am primarily interested in. $\kappa$ should remain inaccessible. – Johannes Schürz Mar 14 '19 at 9:53
• Moreover: the typical situation is that you have a forcing extensions where the forcing is somewhat complete (say: strategically $\kappa$-complete), and in particular adds no bounded sets. – Goldstern Mar 15 '19 at 10:14