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Let $\mathcal I$ be the set of increasing injections from $\mathbb N$ to $\mathbb N$, and $\alpha$ a well ordered uncontable set that ordinal type is regular. Suppose that $f : \beta \mapsto f_{\beta}$ is a fonction from $\alpha$ to $\mathcal I$ such that for any $(i,j)\in \alpha\times \alpha$, such that $i<j$, there exists $k\in \mathbb N$ such that for all $n>k$,

$$f_i(n)<f_j(n)\tag{1}$$

Now suppose that there exists $g\in \mathcal I$ such that $f_i(n)<g(n)$ for all $i\in \alpha$ and $n\in \mathbb N$.


Question.

Does there exists an infinite $A\subset \mathbb N$ and $\gamma\subset \alpha$ such that $|\gamma|=|\alpha|$ and such that for any $c\in \gamma$, $|A\setminus f_c(\mathbb N)|<\infty$.

(I'm not assuming $CH$)

Note about the title : The condition $(1)$ gives a quasi order on $\mathcal I$, according to whom I use the words "decreasing" and "low-bounded" in the title.

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  • $\begingroup$ That is a very odd way of saying $f_i(n)<f_j(n)$ for all $n\geq k$. $\endgroup$ – Asaf Karagila Feb 16 '18 at 9:13
  • $\begingroup$ Also, I don't really see how your "Now suppose that there exists $g\in\cal I$ ..." can be true. Certainly it can be true if $g$ is only required to eventually dominate each of the $f_i$'s, but all of them and everywhere? $\endgroup$ – Asaf Karagila Feb 16 '18 at 9:14
  • $\begingroup$ I'am not sure to understand your last comment... Do you mean that such a $g$ would'n exists except if you ask that for each $i\in \alpha$ there exists $k_i\in \mathbb N$ such that for all $n>k$ we have $f_i(n)<g(n)$. But if such a $g$ exists, then (because $\alpha$ is uncountable, there should exist $k\in \mathbb N$ such that the set of $i\in \alpha$ such that $k_i=k$ has the cardinality $|\alpha|$... but this is only true if $\alpha$ is regular, so I should precise it in the post! Did I understand correctly your comment by the way? $\endgroup$ – jcdornano Feb 16 '18 at 14:08
  • $\begingroup$ to be clear about what I just commented : if the "new" $g$ exists, than the "old" $g$ exists up to a translation of $k_i$.. but this is only true if $|\alpha|$ is regular and I edited in the post (as I edited the "odd" formulation thanks to your first comment) $\endgroup$ – jcdornano Feb 16 '18 at 14:21
  • $\begingroup$ Yes, you understood correctly my concern. You're right about your claim. It requires only that $\alpha$ is uncountable to obtain $|\alpha|$ many $f_i$'s with the same $k$, and if you want also the order type to be preserved, then you need that the cofinality of $\alpha$ is uncountable, I think. Although an assumption of regularity will do little damage here. Not that you might also have to change the values of all the $f_i$'s below that fixed $k$ so they are all below $g$ everywhere. $\endgroup$ – Asaf Karagila Feb 16 '18 at 14:37

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