Tight strongly unbounded colorings

Use the ZFC, $\kappa ={\omega }_1$, and finite-string conventions of Luzin sets, stick, and almost-disjoint guessing at omega one. Given $c:\omega \times \kappa \to \omega$, its column at $\beta$ is $c_{\beta }(n)=c(n,\beta )$.

The coloring is strongly unbounded if for every uncountable set $B\subseteq \kappa$ there are $n<\omega$ and $t\in {\omega }^n$ such that

$\{c_{\beta }(n):\beta \in B,t\subseteq c_{\beta }\}$

is unbounded in $\omega$. Equivalently, for every $m<\omega$ some $\beta \in B$ extends $t$ and satisfies $c_{\beta }(n)>m$. Empty or countable $B$ are not tested. The case $n=0$ allows the empty prefix. Distinct column indices are not required to give distinct functions.

For any set $T\subseteq {\omega }^{<\omega }$, define

$[T{]}_c=\{\beta <\kappa :(\mathrm{\forall }n<\omega )c_{\beta }\upharpoonright n\in T\},{\mathcal{T}}_c=\{T\subseteq {\omega }^{<\omega }:[T{]}_c\text{ is uncountable}\}\mathrm{.}$

No closure-under-prefixes assumption is imposed on $T$. The full finite-string set lies in ${\mathcal{T}}_c$; the empty set does not, since even the length-zero prefix is required. The coloring is tight if there exists $\mathcal{U}\subseteq {\mathcal{T}}_c$, of size at most ${\mathrm{\aleph }}_1$, such that

$(\mathrm{\forall }T\in {\mathcal{T}}_c)(\mathrm{\exists }U\in \mathcal{U})U\subseteq T\mathrm{.}$

Thus the cofinality convention $\mathrm{cf}({\mathcal{T}}_c,\supseteq )\le {\mathrm{\aleph }}_1$ means downward cofinality under ordinary inclusion. The witnessing family cannot be empty, because ${\mathcal{T}}_c$ is nonempty. Tightness and strong unboundedness are properties, not existence assertions.

For distinct $x,y\in {\omega }^{\omega }$, let $\mathrm{\Delta }(x,y)=\min \{n:x(n)\mathrm{\ne }y(n)\}$. It is undefined for equal sequences. For any $x$ put $\widehat{x}(n)={\sum }_{m\le n}(x(m)+1)$. This is a natural-valued strictly increasing majorant: $\widehat{x}(0)=x(0)+1>x(0)$ and $\widehat{x}(n+1)=\widehat{x}(n)+x(n+1)+1>\widehat{x}(n)$, while the summand $x(n)+1$ gives $\widehat{x}(n)>x(n)$.