Suppose $\kappa$ is regular uncountable (assume $2^{< \kappa} = \kappa$ and $\kappa$ is weakly Mahlo if needed). Is the following true?


For every sequence $\langle f\_i: i \to 2 \mid i \in A \rangle$ where $A \subseteq \kappa$ is non stationary, there exists a diagonalizing function $f: \kappa \to 2$ which means: For all $i \in A$, $f\_i \neq f \upharpoonright i$.


Edit: Will has shown that this is hopelessly wrong. I was trying to understand step (c) in Fremlin's proof that diamond holds at continnum if there is a sigma saturated normal ideal over continuum. See theorem 5N on page 47 here: <https://www.essex.ac.uk/maths/people/fremlin/rvmc.pdf>


Can you explain why $A'\_{\xi}$'s as defined in part (c) form a diamond sequence?


Thanks!



Diagonalizing against a non stationary set of functions
Sorry this is late and you may already have it all sorted out more attractively. I think an argument might go as follows. It rests on K. Devlin, *Variations on* $\diamondsuit$, JSL 44 (1979), modified for the case of various diamonds on $\mathbb{c}$ instead of on $\omega\_1$.


Let $\diamondsuit^{Fr}(\mathbb{c})$ be Fremlin's principle of $5N$: there exists $\langle A\_\xi : \xi < \mathbb{c} \rangle$ such that $(\forall A \subseteq \mathbb{c})(\exists 0 < \xi \in lim(\mathbb{c}))(A \cap \xi = A\_\xi)$.


Observe $\diamondsuit^{Fr}(\mathbb{c})$ implies $\diamondsuit\_4(\mathbb{c})$: there exists $\langle S\_\xi \subseteq P(\xi) : \xi < \mathbb{c} \rangle$ such that $(\vert S\_\xi \vert \leq \aleph\_0)$ and $(\forall A \subseteq \mathbb{c})(\exists 0 < \xi \in lim(\mathbb{c}))(A \cap \xi \in S\_\xi)$. This needs no comment, save that the notation $\diamondsuit\_4$ follows Devlin's paper.


Now by the proof of Theorem 2.3 in that paper, modified to the case $\mathbb{c}$ instead of $\omega\_1$, $\diamondsuit\_4(\mathbb{c})$ implies $\diamondsuit\_1(\mathbb{c})$ (Kunen's diamond on $\mathbb{c}$) which in turn implies $\diamondsuit(\mathbb{c})$.


