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Let $Y \supseteq^{*} X$ denote that all but finitely many members of X belong to Y. Recall that a tower is a linearly quasi-ordered subset of $(\mathcal{P}^{\omega}(\omega), \supseteq^{*})$ maximal with respect to upward extension (equivalently, with no infinite "almost intersections"; note that the notation $\supseteq^{*}$ corresponds here to reverse inclusion), and that the cardinal $\mathfrak{t}$ is the least cardinality of any tower. Van Douwen defines $\mathfrak{t}$ equivalently as the least cardinality of any well-ordered tower. It is not too hard to show that $\omega < \mathfrak{t} \leq \mathfrak{c}$ and that there exists a tower of cardinality $\mathfrak{c}$. However, it is less obvious to me that there exists a well-ordered tower of cardinality $\mathfrak{c}$.

In other words, does the poset of subsets of $\omega$ up to almost-equality, ordered with respect to almost-inclusion, admit a subset of order-type $\mathfrak{c}$?

My motivation for this question is that each of the cardinal characteristics of the continuum are usually defined as the minimum of sets of cardinals with trivial maximum, but it is not obvious to me whether the trivial maximum is attained in this otherwise natural alternative definition of $\mathfrak{t}$.

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    $\begingroup$ Are we supposed to be working without choice here? $\endgroup$ – Henning Makholm Aug 22 '18 at 22:10
  • $\begingroup$ Here I assume choice but not necessarily CH. $\endgroup$ – Scott Mutchnik Aug 22 '18 at 22:13
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    $\begingroup$ Perhaps I'm missing something fundamental, but if we well-order the continuum, the $\mathfrak c$ prefixes of that order are well-ordered by $\subseteq$. Their complements are well-ordered by $\supseteq$ and therefore also by $\supseteq^*$. This may not be a maximal subset, but one should think Zorn's lemma would allow us to extend it to one that is maximal. $\endgroup$ – Henning Makholm Aug 22 '18 at 22:14
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    $\begingroup$ But a tower is defined to consist of (infinite) subsets of the integers, not subsets of the continuum. For example we can get a (non-well-ordered) tower of cardinality that of the continuum by applying Zorn's lemma to Dedekind cuts of the rational numbers. $\endgroup$ – Scott Mutchnik Aug 22 '18 at 22:26
  • $\begingroup$ Ah. Silly mistake, sorry. Move along, nothing to see here. $\endgroup$ – Henning Makholm Aug 22 '18 at 22:29
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No, not necessarily. For instance, if you start with a model of CH and add any number of Cohen reals, there will never exist a well-ordered subset of $(\mathcal{P}^{\omega}(\omega), \supseteq^{*})$ of length $\omega_2$ (so in particular, if you add at least $\aleph_2$ Cohen reals, there will not exist a well-ordered tower of cardinality $\mathfrak{c}$). The idea of the proof is that given a set of more than $\aleph_1$ names for reals, we can find an automorphism of the forcing that swaps two of them (roughly because each name for a real depends on only countably many of our Cohen reals, but the forcing notion that adjoins countably many Cohen reals has only $\aleph_1$ different names for reals by CH). As a result, we cannot have a name for a set of more than $\aleph_1$ reals which is forced to be totally ordered by $\supseteq^*$ in a particular way (e.g., with order-type $\omega_2$), since that order would be violated by swapping any two of them.

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  • $\begingroup$ In finding the automorphism, we can identify each of the more than $\aleph_1$ names for reals with a name for a real given by the forcing extension adding countably many Cohen reals. Two of the names will be the same up to this identification, and we can easily find an automorphism exchanging them if they depend on disjoint sets of Cohen reals. Otherwise there is an obvious way of constructing automorphisms taking one name to the other, but not necessarily switching the two. What am I missing? $\endgroup$ – Scott Mutchnik Aug 24 '18 at 3:21
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    $\begingroup$ Right, this is nontrivial (and why I said "roughly"). The trick is to use the $\Delta$-system lemma: we have a set of $\aleph_2$ countable sets (the countable sets of Cohen reals for each of our names), and we can find a $\Delta$-system inside it (using CH). Then, after adjoining all the Cohen reals in the root of the $\Delta$-system, our countable sets become disjoint. $\endgroup$ – Eric Wofsey Aug 24 '18 at 3:32
  • $\begingroup$ Thanks, the delta-system lemma makes it a lot easier. $\endgroup$ – Scott Mutchnik Aug 24 '18 at 3:53

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