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

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!

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    $\begingroup$ There are two trivial ways this can fail. If $0 \in A$ then $f_0$ and $f \upharpoonright 0$ are both the empty function (hence equal). A little less trivially, if $0 \notin A$ but $n \in A$ for every $0 < n < \omega$, then suppose $f_{n+1}$ is the function mapping $m$ to $0$ for $m < n$ and $n$ to $1$. If your condition is satisfied, this forces $f$ to map every $n < \omega$ to $0$. But if $f_\omega$ is constantly $0$, your condition fails at $\omega$. I'm writing this as a comment instead of an answer because I'm sure it's not really what you're wanting. Maybe put some restrictions on $A$? $\endgroup$
    – Will Brian
    Commented Jul 7, 2015 at 17:59
  • $\begingroup$ Thanks Will, that pretty much nails it. I was trying to finish a proof of Fremlin's. I will edit to explain where i am stuck. $\endgroup$
    – Ashutosh
    Commented Jul 7, 2015 at 18:45
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    $\begingroup$ I have to admit that I won't really understand the link you give without putting a lot more effort in. That said, I'll share one more observation. If $A$ is such that there is an injective regressive function $\varphi: A \rightarrow \kappa$, then the answer to your original question is yes. (On the image of $\varphi$, define $f$ so that $f(\varphi(i)) \neq f_i(\varphi(i))$ for every $i \in A$. Off of the image of $\varphi$ define $f$ arbitrarily.) Since Fremlin mentions injective regressive functions in his argument, maybe this is relevant? $\endgroup$
    – Will Brian
    Commented Jul 7, 2015 at 19:30
  • $\begingroup$ In nutshell, step c in Fremlin's argument seems to be saying that if you can guess every set at a limit stage then you can modify your guessing sequence to guess everything at stationary many stages. I don't think this is always true and i dont see how to do this modification in this specific case. $\endgroup$
    – Ashutosh
    Commented Jul 7, 2015 at 19:41
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    $\begingroup$ Assaf pointed me to his blog post where he mentioned a result of Matet that explains this. Here's the link: blog.assafrinot.com/?p=1867 $\endgroup$
    – Ashutosh
    Commented Jul 7, 2015 at 20:08

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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 $\diamondsuit(\mathbb{c})$.

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