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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 $i \in A$, $f_i \neq f \upharpoonright i$.

Thanks!

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!

Suppose $\kappa$ is regular uncountable (say even weakly Mahlo). 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 $i \in A$, $f_i \neq f \upharpoonright i$.

Thanks!

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 $i \in A$, $f_i \neq f \upharpoonright i$.

Thanks!