Basics on towers on the natural numbers

For A, B \subseteq \omega we write A \subseteq^* B if A\setminus B is finite, and we write A\simeq^* B if A\subseteq^* B and B\subseteq^* A.

A tower is a collection {\cal T} of co-infinite subsets of \omega such that for all A\neq B\in {\cal T} we have A\not\simeq^*B and either A\subseteq^* B or B \subseteq^* A. (A\subseteq \omega is co-infinite if \omega\setminus A is infinite.)

If {\cal S}, {\cal T} are towers, we say that {\cal S}\leq_t {\cal T} if A\in {\cal S} and B\in{\cal T} with B\subseteq^* A jointly imply that B\in{\cal S}. (In other words, this means that {\cal S} is a down-set or initial segment of {\cal T} with respect to \subseteq^*). It is easy to prove that \leq_t is a partial order on the collection of all towers on \omega.

The remainder of this post is about maximal towers with respect to \leq_t. The proof of the following lemma is routine.

Lemma 1. If {\frak T} is a collection of towers, such that for all {\cal S}, {\cal T}\in {\frak T} we have either {\cal S}\leq_t {\cal T} or {\cal T}\leq_t {\cal S}, then

1. \bigcup {\frak T} is a tower, and
2. {\cal T}\leq_t \bigcup{\frak T} for all {\cal T}\in{\frak T}.

Corollary 1. Zorn’s Lemma and Lemma 1 imply that there is a tower in \omega that is maximal with respect to \leq_t.

Lemma 2. If {\cal A} is a countable tower then {\cal A} is not maximal.

Proof. Let (A_n)_{n\in\omega} be a sequence of co-infinite subsets of \omega such that for all n\in\omega we have A_n \subseteq^* A_{n+1} . We want to show there is A\subseteq \omega co-infinite with A_n\subseteq^* A for all n\in\omega.

Step 1. If k\in \omega, then \bigcup_{i=0}^k A_i is co-infinite.

Step 2. There is f:\omega\to\omega strictly increasing such that f(n) \in \omega\setminus\big(\bigcup_{i=0}^n A_i\big) for all n\in\omega.

Step 3. Set A = \bigcup_{k\in\omega}\big(A_k\setminus [0,\ldots,f(k)]\big).

Then it follows that

1: A_k\subseteq^* A for all k\in \omega since (A_k\setminus A) \subseteq [0,\ldots,f(k)] which is a finite set.

2: A is co-infinite, since for all k\in \omega we have f(k)\notin A, so A\cap \{f(k):k\in\omega\}=\emptyset, and since f is strictly increasing we have \{f(k):k\in\omega\} is infinite, so A is co-infinite.

Letting {\cal A} =\{A_n:n\in\omega\} we get that {\cal A}' = {\cal A} \cup\{A\} is a tower with {\cal A} \leq_t {\cal A}' but {\cal A}' \not\leq_t {\cal A}, so the countable tower {\cal A} is not maximal. \Box


About dominiczypen

I'm interested in general topology, order theory, and graph theory. This link takes you to my preprints on arXiv.
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