A good clopen family for summable slaloms

Enumerate all clopen subsets of $2^{\omega }$ as $(C_l{)}_{l\in \omega }$, with every clopen occurring infinitely often. Such sets are finite unions of basic cylinders: compactness of $2^{\omega }$ gives a finite subcover by cylinders, and compactness follows directly from the finite-branching binary tree. Fix a repeating enumeration $(U_n)$ of the nonempty basic cylinders. All enumerations can be obtained by listing finite binary words and finite lists, so their codes are fixed without a choice.

Fix $n$. For each $k$, let $A_k$ consist of indices $l>k$ such that for every $I\subseteq \{0,\dots ,k\}$, $U_n\cap {\bigcap }_{i\in I}{C}_i\mathrm{\ne }\mathrm{\varnothing }⟹U_n\cap C_l\cap {\bigcap }_{i\in I}{C}_i\mathrm{\ne }\mathrm{\varnothing }\mathrm{.}$ The empty $I$ is included. These are finite tests on clopen codes, so each $A_k$ is a fixed, decidable set of indices.

If $D$ is dense open, then $A_k$ contains arbitrarily large indices $l$ with $C_l\subseteq D$. Indeed, there are only finitely many nonempty clopen sets $W_I=U_n\cap {\bigcap }_{i\in I}{C}_i$ in step 2.1. For each such $I$ choose the least coded basic cylinder $B_I\subseteq D\cap W_I$. Their finite union $C$ is clopen, lies in $D$, and meets every nonempty $W_I$. The repeating clopen enumeration lists $C$ beyond every prescribed $k$. Thus the required $l$ exists, and all choices were finite least-index choices.

Put $q=2^n$. List, with repetitions if necessary, every clopen set of the form $C_{m_0}\cup \cdots \cup C_{m_q}$ where $U_n\cap C_{m_0}\mathrm{\ne }\mathrm{\varnothing }$ and $m_{i+1}\in A_{m_i}$ for $i<q$; call the resulting enumeration $(S_m^n{)}_m$. There are infinitely many such tuples by step 3.1 with $D=2^{\omega }$. Every listed union meets $U_n$ through its first term. For a dense open $D$, choose $m_0$ with $C_{m_0}\subseteq D\cap U_n$, then recursively choose $m_{i+1}\in A_{m_i}$ with $C_{m_{i+1}}\subseteq D$ by step 3.1. The resulting $S_m^n$ lies in $D$.

Take $1\le r\le q$ listed unions, writing the $a$-th one as $V_a={\bigcup }_{i=0}^q{C}_{m_i^a}$ with $m_{i+1}^a\in A_{m_i^a}$. Select distinct rows $a_0,\dots ,a_{r-1}$ as follows: at stage $j$, among rows not yet selected, choose one with the least $j$-th index $m_j^a$. We claim by induction that $U_n\cap {\bigcap }_{i\le j}{C}_{m_i^{a_i}}\mathrm{\ne }\mathrm{\varnothing }(j<r)\mathrm{.}$ At $j=0$ this is the condition on $m_0^{a_0}$. For $j>0$, row $a_j$ was available at every earlier stage $i<j$, so $m_i^{a_i}\le m_i^{a_j}\le m_{j-1}^{a_j}$. Hence all previously selected indices belong to $\{0,\dots ,m_{j-1}^{a_j}\}$. As $m_j^{a_j}\in A_{m_{j-1}^{a_j}}$, the defining implication of step 2.1 preserves the nonempty intersection when $C_{m_j^{a_j}}$ is added.

Each selected diagonal clopen $C_{m_j^{a_j}}$ lies in its row union $V_{a_j}$. Thus step 5.1 gives $U_n\cap {\bigcap }_{a<r}{V}_a\mathrm{\ne }\mathrm{\varnothing }$ for $1\le r\le q$; the empty intersection is $U_n$ and is nonempty. Removing repeated members from a family of at most $q$ sets only reduces $r$, so this proves the third property. The first two properties were proved in step 4.1. ∎