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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?

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    $\begingroup$ Sorry, I was being silly - your original formulation was fine, I parsed "$\bigcap$" as "$\cap$." $\endgroup$ – Noah Schweber Mar 13 at 16:07
  • $\begingroup$ No problem. I changed it back, because if you want to take complements, you have to restrict that there is only one successor... $\endgroup$ – Johannes Schürz Mar 13 at 16:16
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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 $\Sigma^1_1$.

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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).

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  • $\begingroup$ But there is absoluteness when the cardinal structure up to $\kappa^+$ is preserved, right? $\endgroup$ – Asaf Karagila Mar 14 at 7:52
  • $\begingroup$ That's what I am primarily interested in. $\kappa$ should remain inaccessible. $\endgroup$ – Johannes Schürz Mar 14 at 9:53
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    $\begingroup$ 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. $\endgroup$ – Goldstern Mar 15 at 10:14

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