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A $\kappa$-tower in $\mathbb{N}^\mathbb{N}$ is a sequence $\langle a_\alpha : \alpha<\kappa\rangle$ in $\mathbb{N}^\mathbb{N}$ that is $\le^*$-increasing with $\alpha$ and has no $\le^*$-upper bound.

Piotr Szewczak and I need a reference for the consistency of the existence of a $\kappa$-tower in $\mathbb{N}^\mathbb{N}$ for some (any) cardinal number $\kappa>\mathfrak{b}$.

It seems that this consistency (and much more) is established in Peter Dordal's paper on towers, but we guess there is a more elementary and earlier reference for this specific result?

The motivation comes from selection principles (products of Menger spaces).

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A theorem of Hechler says that, given any poset in which every countable subset has a strict upper bound, you can arrange for that poset to be cofinal in the $\leq^*$ ordering of $\mathbb N^{\mathbb N}$. Apply that to the poset $\omega_1\times\omega_2$ (ordered componentwise). Then $\mathfrak b=\aleph_1$, but $\mathbb N^{\mathbb N}$ also contains a tower of length $\omega_2$.

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  • $\begingroup$ The reference for Hechler's result is "On the existence of certain cofinal subsets of $\omega^\omega$" (Proc. Symp. Pure Math 13(2) Amer. Math. Soc. (1974) pp. 155-173). Although published in 1974, this is from the proceedings of a 1967 conference, and I believe it's written in the old style of ramified forcing over $L$. I also believe that this is the work for which Hechler invented what is now called Hechler forcing. $\endgroup$
    – Andreas Blass
    May 20, 2016 at 17:00
  • $\begingroup$ I seem to miss something. Do you mean, perhaps lexicographic order? If not, then what is the cofinal $\omega_2$-chain in $\omega_1\times\omega_2$? $\endgroup$
    – Boaz Tsaban
    May 21, 2016 at 20:43
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    $\begingroup$ @BoazTsaban It's not a cofinal chain (that would require $\mathfrak b=\mathfrak d$). It's an increasing $\omega_2$-chain with no upper bound (with respect to $\leq^*$), i.e., what you called a tower. It corresponds to the subset $\{0\}\times\omega_2$ of $\omega_1\times\omega_2$. $\endgroup$
    – Andreas Blass
    May 22, 2016 at 22:07
  • $\begingroup$ Awesome! To summarize: The set $\{0\}\times\omega_2$ is unbounded in the set $\omega_1\times\omega_2$, which is mapped on a cofinal set, and unbounded in a cofinal implies unbounded. $\endgroup$
    – Boaz Tsaban
    May 23, 2016 at 9:16

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