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A family of functions $\langle f_\alpha:\alpha<\kappa\rangle$ from $\omega_1$ to $\omega_1$ is called a strong chain if $\alpha<\beta<\kappa\Longrightarrow \{\xi<\omega_1: f_\beta(\xi)\leq f_\alpha(\xi)\}$ is finite. It is a result of Koszmider [1] that the existence of a strong chain of length $\omega_2$ is consistent. (Another presentation of this result is in [2], using a technique due to Neeman.)

The existence of a strong chain of length $\kappa$ implies that $\kappa\leq\frak{c}$ (as the restrictions of the functions to the first $\omega$ places are distinct), so in particular, if there is a strong chain of length $\omega_2$ then the Continuum Hypothesis fails.

Does the existence of a strong chain of length $\omega_2$ have any influence on the configuration of standard cardinal characteristics of the continuum?

[1] Koszmider, Piotr, On strong chains of uncountable functions, Isr. J. Math. 118, 289-315 (2000). ZBL0961.03039.

[2] Veličković, Boban; Venturi, Giorgio, Proper forcing remastered, Cummings, James (ed.) et al., Appalachian set theory 2006–2012. Based on the Appalachian set theory workshop series during the period 2006–2012. Cambridge: Cambridge University Press (ISBN 978-1-107-60850-4/pbk). London Mathematical Society Lecture Note Series 406, 331-362 (2013). ZBL1367.03094.

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  • $\begingroup$ so Chang's Conjecture keeps this $\kappa$ small ($\aleph_1$) and CC is c.c.c indestructible so basically this $\kappa$ cannot bound any cardinal invariant. For the other direction, is it known if it is possible to have $\kappa\geq \omega_3$? $\endgroup$
    – Jing Zhang
    Commented Mar 3, 2020 at 16:09
  • $\begingroup$ I don't the answer to the $\omega_3$ question. What I'm really curious about is if the existence of the strong $\omega_2$-chain implies that some cardinal characteristics must be greater than $\omega_1$ (we know that $\mathfrak{c}>\omega_1$, for example). $\endgroup$
    – Todd Eisworth
    Commented Mar 3, 2020 at 21:00
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    $\begingroup$ It seems that the forcing to add such an $\omega_2$ strong chain is strongly proper, so at least we can exclude invariants like $\mathfrak{b}$ (and many others) since no strongly proper forcing can add a dominating real (any real lives in a Cohen sub-extension but Cohen forcing can't add a dominating real). $\endgroup$
    – Jing Zhang
    Commented Mar 4, 2020 at 5:37
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    $\begingroup$ Yes. And once you have a long strong chain, it will still exist in further ccc extensions, so you can achieve further configurations, say by adding a small dominating family. $\endgroup$
    – Todd Eisworth
    Commented Mar 4, 2020 at 14:21

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